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Prentice Hall Algebra 1 • Teaching Resources

Copyright © by Pearson Education, Inc., or its affiliates. All Rights Reserved.1

Name Class Date

8-1ELL Support

Adding and Subtracting Polynomials

Concept List

binomial constant cubicdegree fourth degree linear

monomial quadratic trinomial

Choose the concept or concepts from the list above that best represent the

item in each box.

1. 2 x 31 5 2. 5 x 1 4 x

2 3. 8

4. 9 5. 3 x 21 6 x 1 4 6. 3 x

21 6 x

7. 4 x 41 6 x

31 2 x

2 8. 7 x 21 x 9. 5 x

4

2

constant/monomial

constant/monomial

fourth degree/trinomial fourth degree/monomial

quadratic/trinomial

degree

quadratic/binomial

cubic/binomial binomial/quadratic

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8-1Think About a Plan


Geometry Te perimeter o a trapezoid is 39a 2 7. Tree sides have the ollowing

lengths: 9a, 5a 1 1, and 17a 2 6. What is the length o the ourth side?

Understanding the Problem

1. What is the perimeter o the trapezoid?

2. What are the lengths o the sides you are given? , ,

3. How many sides does a trapezoid have?

4. How do you fnd the perimeter o a trapezoid?

5. What is the problem asking you to determine?

Planning the Solution

6. Draw a diagram o the trapezoid and label theinormation you know.

7. Write an equation that can be used to determine the length o the ourth side.

Getting an Answer

8. Solve your equation to fnd the length o the ourth side o the trapezoid.

39a 2 7

9a

4 sides

add the side lengths

the length of the fourth side

s 5 (39a 2 7) 2 (9a 1 5a 1 1 1 17a 2 6)

8a 2 2

s

5a 1 1

17a 2 6

9a

5a 1 1 17a 2 6

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Prentice Hall Gold Algebra 1 • Teaching Resources


8-1Practice Form G


Find the degree of each monomial.

1. 2b2c 2 2. 5 x 3. 7 y 5 4. 19ab

5. 12 6.12 z 2 7. t 8. 4d 4e

Simplify.

9. 2a3b 1 4a3b 10. 5 x 3 2 4 x 3 11. 3m6n32 5m6n3

12. 26ab 1 3ab 13. 4c 2d 6 2 7c 2d 6 14. 315 x 2 2 30 x 2

Write each polynomial in standard form. Ten name each polynomial based on

its degree and number of terms.

15. 15 x 2 x 3 1 3 16. 5 x 1 2 x 2 2 x 1 3 x 4 17. 9 x 3

18. 7b2 1 4b 19. 23 x 2 1 11 1 10 x 20. 12t 2 1 1 2 3 x 1 8 2 2 x

Simplify.

21.8z 2 12

1 6z 1 90 22.9 x 3 1 3

1 4 x 3 1 7 23.

6 j 2 2 2 j 1 5

1 3 j 2 1 4 j 2 6

24. (3k 2 1 5) 1 (16 x 2 1 7) 25. ( g 4 2 4 g 2 1 11) 1 (2 g 3 1 8 g )

26. A local deli kept track of the sandwiches it sold for three months. Te

polynomials below model the number of sandwiches sold, where s

represents days.

Ham and Cheese: 4s3 2 28s2 1 33s 1 250

Pastrami: 27.4s2 1 32s 1 180

Write a polynomial that models the total number of these sandwiches that

were sold.

4

0

6a3b

2 x 3 1 15 x 1 3; cubictrinomial

7b21 4b; quadratic

binomial

14 z 2 3

3k 2 1 16 x 2 1 12

4 s32 35.4 s2

1 65 s 1 430

g42 g3

2 4g21 8g 1 11

13 x 3 1 10 9 j 2 1 2 j 2 1

23 x 2 1 10 x 1 11;quadratic trinomial

12t 2 2 5 x 1 9;quadratic trinomial

3 x 4 1 2 x 2 1 4 x ; fourthdegree trinomial

9 x 3; cubic monomial

x 3 22m6n3

23ab 23c 2d 6 285 x 2

2 1 5

51 2

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Simplify.

27.11

n2 4

2 (5n 1 2) 28.7 x

41 9

2 (8 x 4 1 2) 29. 3d 21

8d 2

22 (2d 2 2 7d 1 6)

30. (28e 3 1 3e 2) 1 (19e 3 1 e 2) 31. (212h4 1 h) 2 (26h4 1 3h2 2 4h)

32. A small town wants to compare the number o students enrolled in public and

private schools. Te polynomials below show the enrollment or each:

Public School: 219c 2 1 980c 1 48,989

Private School: 40c 1 4046

Write a polynomial or how many more students are enrolled in public school

than private school.

Simplify. Write each answer in standard form.

33. (3a2 1 a 1 5) 2 (2a 2 5) 34. (6d 2 10d 3 1 3d 2) 2 (5d 3 1 3d 2 4)

35. (24s3 1 2s 2 3) 1 (22s2 1 s 1 7) 36. (8p3 2 6p 1 2p2) 1 (9p2 2 5p 2 11)

37. Te ence around a quadrilateral-shaped pasture is3a2 1 15a 1 9 long. Tree sides o the ence have the

ollowing lengths: 5a, 10a 2 2, a2 2 7. What is the length

o the ourth side o the ence?

38. Error Analysis Describe and correct the error in

simpliying the sum shown at the right.

39. Open-Ended Write three diferent examples o the sum o aquadratic trinomial and a cubic monomial.

8-1Practice (continued) Form G


a2ź7

5a

?

10a ź2

6x3 + 4x – 10

+ (–3x2 + 2x + 8)

3x3 + 6x – 2

6n 2 6

47e31 4e2

26h42 3h2

1 5h

219c 2 1 940c 1 44,943

3a22 a 1 10

24 s32 2 s2

1 3 s 1 4

2a21 18

two unlike terms, 6 x 3 and 23 x 2, were added;

6 x 3 2 3 x 2 1 6 x 2 2

Answers may vary. Sample: ( x 2 1 2 x 1 1) 1 x 3;

(2 x 2 1 5 x 1 6) 1 3 x 3; (r 2 1 r 1 1) 1 8r 3

8 p31 11 p2

2 11 p 2 11

215d 3 1 3d 2 1 3d 1 4

2 x 4 1 7 d 2 1 15d 2 8

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Prentice Hall Foundations Algebra 1 • Teaching Resources


8-1Practice Form K



1. 3s3t 3 2. 3n 3. 5 xy

4. 7 5.14k

5 6. d

Simplify.

7. 3mn41 6mn4 8. 12 g 2 2 7 g 2

9. 211c 4d 1 12c 4d 10. 42z 3 2 15z 3

Write each polynomial in standard form. Ten name each polynomial based on

its degree and number of terms.

11. 7a 1 4 2 a2 12. 5b21 2n

13. 211d 4 14. 2 x 3 2 9 1 2 x 1 8 2 4 x

15. A pizza shop owner is monitoring the amount of cheese he uses each week.

Te polynomials below model the pounds of cheese ordered in the past,

wherep represents pounds.

Mozzarella: 3p32 6p2

1 14p 1 125

Cheddar: 12.5p21 18p 1 75

Write a polynomial that models the total number of pounds of cheese that

were ordered.

6 1 2

0 5 1

9mn4 5g2

c 4d 27 z 3

2a21 7a 1 4; quadratic trinomial 5b2

1 2n; quadratic binomial

211d 4; 4th degree monomial 2 x 3 2 2 x 2 1; cubic trinomial

3 p31 6.5 p2

1 32 p 1 200

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8-1Practice (continued) Form K


Simplify.

16.

3r 1 5

1 7r 1 3 17. (t

42

4t

21

9)1

(2

t

31

3t )

18.7b2 1 6

1 4b2 1 5 19.

4z 1 7

2 (6z 1 1)

20. (26k 3 1 3k ) 2 (25k 3 1 3k 2 2 8k ) 21.3p4 1 1

2 (9p4 1 5)

22. A city wants to compare the number o people who own their own home and

who rent their home. Te polynomials below show expressions or each. In

each polynomial, p 5 0 corresponds to the frst year.

Own: 4p2 1 360p 1 22,178

Rent: 6p21 125p 1 5286

Write a polynomial or how many more own their home than rent their home.

23. Te wallpaper border that runs all the way around a room is 5 f 21 19 f 1 11

long. Tree sides o the room have the ollowing lengths o border: 6 f , 5 f 2 7,

2 f 21 2. What is the length o the ourth side o the room?

24. Open-Ended Write two dierent polynomials with a dierence o

23 x 2 1 5 x 2 7.

10r 1 8 t 4 2 t 3 2 4t 2 1 3t 1 9

11b21 11 22 z 1 6

2k 3 2 3k 2 1 11k 26 p42 4

22 p21 235 p 1 16,892

3f 2 1 8f 1 16

Answers may vary. Sample: (21 x 2 1 6 x 2 4) 2 (2 x 2 1 x 1 3) and

(24 x 2 1 7 x 2 5) 2 (2 x 2 1 2 x 1 2)

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8-1Standardized Test Prep


Multiple Choice

For Exercises 1–6, choose the correct letter.

1. What is the degree o the monomial 3 x 2 y 3?

A. 2 B. 3 C. 5 D. 6

2. What is the simplifed orm o 8b3c 2 1 4b3c 2?

F. 12bc G. 12b3c 2 H. 12b6c 4 I. 12b9c 4

3. How is 6d 2 8 1 4d 2 written in standard orm?

A. 4d 2 1 6d 2 8 B. 4d 2 1 6d 1 8 C. 4d 2 2 6d 2 8 D. 4d 2 2 6d 1 8

4. What is the simplifed orm o (4 j 2 1 6) 1 (2 j 2 2 3)?

F. 6 j 2 2 3 G. 6 j 2 1 3 H. 6 j 2 1 9 I. 4 j 4 1 3

5. What is the dierence o the ollowing polynomials?

6 x 3 2 2 x 2 1 4

2 (2 x 3 1 4 x 2 2 5)

A. 4 x 3 2 2 x 2 2 1 B. 8 x 3 1 6 x 2 2 1 C. 4 x 3 2 2 x 2 1 1 D. 4 x 3 2 6 x 2 1 9

6. What is the simplifed orm o (3 x 2 2 4 x 1 6 x ) 1 (5 x 3 1 2 x 2 2 3 x ) in

standard orm?

F. 5 x 3 1 10 x 2 2 x G. 8 x 3 2 2 x 2 1 3 x H. 5 x 3 1 10 x 2 2 5 x I. 5 x 3 1 5 x 2 2 x

Short Response

7. Suppose you have been given this polynomial.

5b 1 4b22 3b4

1 3

a. How can you write this polynomial in standard orm?

b. What is the degree o this polynomial? Explain.

C

G

G

D

I

A

23b41 4b2

1 5b 1 3

4; b4 is the term with the greatest degree

[2] Both parts answered correctly with full explanations

[1] One part answered correctly or both parts answered correctly with incomplete

explanations

[0] Neither part answered correctly

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8-1Enrichment


Packing boxes and packing sheets in diferent sizes are given by the expressions

below. o nd the number o packing boxes and sheets that will t in a larger

shipping box, add or subtract the polynomials. ell the total number o boxes andsheets. Ten tell how many medium and large boxes and sheets you could t into

the shipping box. Te rst one has been started or you.

Boxes

a35 small box b

35 medium box c

35 large box

6 small boxes 5 1 medium box 4 medium boxes 5 1 large box

Sheets

a25 small sheet b

25 medium sheet c

25 large sheet

4 small sheets 5 1 medium sheet 8 medium sheets 5 1 large sheet

1. 7a3 1 5b3 1 5a3 2 3b3 5 12a3 2 2b3

5 12 small boxes and 2 medium boxes

5 4 medium boxes

5 large box(es)

2. 6a21 3b2

2 8c 2 1 12b22 2a2

1 10c 2 5

5 small sheets, medium sheets, and large sheets

5 medium sheets and large sheets

5 large sheets

3. (8a32 3b3

1 6c 3) 2 (2a32 14b3

1 2c 3) 5

5 small boxes, medium boxes, and large boxes

5 medium boxes and large boxes

5 large boxes

4. (15c 2 1 12a3 2 9b2) 1 (214c 2 1 6a3 1 5b3 1 25b2) 5

5

5 medium boxes medium sheets, and large sheets

5 large boxes and large sheets

1

4

16

4

4

6

7

11

12

16 18

2 3

4

5 medium boxes, 16 medium sheets, and 1 large sheet 18a31 5b3

1 16b21 c 2

2

15 2

6a31 11b3

1 4c 3

4a21 15b2

1 2c 2

18 small boxes,

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You can add and subtract polynomials by lining up like terms and then adding or

subtracting each part separately.

Problem

What is the simplifed orm o (3 x 2 2 4 x 1 5) 1 (5 x 2 1 2 x 2 8)?

Write the problem vertically, lining up the like terms.

Ten add each pair o like terms.

Solve Add the x 2 terms. Add the x terms. Add the constant terms.

3 x 2 1 5 x 2 5 8 x 2 24 x 1 2 x 5 22 x 5 1 (28) 5 23

3 x 2 2 4 x 1 5

1 5 x 2 1 2 x 2 8

8 x 2 2 2 x 2 3

Add the sums.

Check Check your solution using subtraction.

8 x 2 2 5 x 2 5 3 x 2 22 x 2 2 x 5 24 x 23 2 (28) 5 5

Solution: (3 x 2 2 4 x 1 5) 1 (5 x 2 1 2 x 2 8) 5 8 x 2 2 2 x 2 3

ExercisesSimplify.

1.5b2

1 3b

1 2b22 5b

2.3c 2 1 3c

1 4c 2 1 2c 3.

4d 2 2 3d 1 6

1 2d 2 1 5d 2 3

4.23e 2 2 5e 1 2

1 e 2 1 2e 2 7 5.

4 f 31 2 f 2 1 5 f

1 2 f 32 4 f 22 3 f 6.

5 g 3 2 2 g 2 1 3 g

1 2 g 3 1 5 g 2 2 2 g

7. (3h21 5) 1 (25h2

2 3) 8. (2 j 2 1 4 j 2 6) 1 (4 j 2 2 3 j 2 3)

8-1Reteaching


3 x 2 2 4 x 1 5

1 5 x 2 1 2 x 2 8

7b22 2b

22e22 3e 2 5

22h21 2 6 j 2 1 j 2 9

6f 32 2f 21 2f 7g31 3g2

1 g

7c 2 1 5c 6d 2 1 2d 1 3

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o subtract polynomials, ollow the same steps as in addition.

Problem

What is the simplifed orm o (6 x 3 1 4 x 2 2 3 x ) 2 (2 x 3 1 3 x 2 2 5 x )?

Write the problem vertically, lining up the like terms.

Ten subtract each pair o like terms.

Solve

Subtract the x 3 terms. Subtract the x 2 terms. Subtract the x terms.

6 x 3 2 2 x 3 5 4 x 3 4 x 2 2 3 x 2 5 x 2 23 x 2 (25 x ) 5 2 x

6 x 3 1 4 x 2 2 3 x

2 (2 x 3 1 3 x 2 2 5 x )

4 x 3 1 x 2 1 2 x

Add the differences.

Check Check your solution using subtraction.

4 x 3 1 2 x 3 5 6 x 3 x 2 1 3 x 2 5 4 x 2 2 x 1 (25 x ) 5 23 x

Solution: (6 x 3 1 4 x 2 2 3 x ) 2 (2 x 3 1 3 x 2 2 5 x ) 5 4 x 3 1 x 2 1 2 x

Exercises

Simplify.

9.4k 2 1 5k

2 (3k 2 1 2k ) 10.

5m22 4m

2 (2m21 3m)

11.7n2

1 4n 1 9

2 (4n21 3n 1 5)

12.5p2

1 6p 1 4

2 (7p21 4p 1 8)

13.3q3

1 2q21 7q

2 (6q32 4q2

2 5q) 14.

2r 3 2 2r 2 1 5r

2 (4r 3 1 5r 2 1 3r )

15. (6s22 5s) 2 (22s2

1 3s) 16. (3w 2 1 6w 2 5) 2 (5w 2 2 4w 1 2)

8-1Reteaching (continued)


6 x 3 1 4 x 2 2 3 x

2 (2 x 3 1 3 x 2 2 5 x )

k 2 1 3k

22 p21 2 p 2 4

8 s22 8 s 22w 2 1 10w 2 7

3m22 7m

2

3q31

6q21

12q 2

2r 3 2

7r 2 1

2r

3n21 n 1 4

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8-2ELL Support

Multiplying and Factoring

Tere are two sets of note cards below that show how Brittany factors the

polynomial 5 x 51 15 x

31 4 x

2. Te set on the left explains the thinking. Te

set on the right shows the steps. Write the thinking and the steps in the correctorder.

Think Cards Write Cards

Think Write

Factor each term of thepolynomial.

x2(5x31 15x 1 4)

The GCF is x ? x, or x2.

5x5 1 15x3 1 4x2 5

x2(5x3) 1 x2(15x) 1 x2(4)

5x55 5 ? x ? x ? x ? x ? x

15x35 3 ? 5 ? x ? x ? x

4x25 2 ? 2 ? x ? x

Simplify.

Find the GCF of the threeterms.

Factor out the GCF from eachterm.

Step 1

Step 2

Step 3

Step 4

First, she should factor each term ofthe polynomial.

5 x 5 5 5 ? x ? x ? x ? x ? x

15 x 3 5 3 ? 5 ? x ? x ? x

4 x 2 5 2 ? 2 ? x ? x

The GCF is x ? x , or x 2.

5 x 5 1 15 x 3 1 4 x 2 5

x

2(5 x

3) 1 x

2(15) 1 x

2(4)

x 2(5 x 3 1 15 x 1 4)

Second, she should find the GCF ofthe three terms.

Next, she should factor out the GCF

from each term. Then factor it out ofthe polynomial.

Finally, she should simplify.

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a. Factor n21 n.

b. Writing Supposen is an integer. Is n21 n always, sometimes, or never an

even integer? Justify your answer.

1. Factor out n from the expression.

nau 1ub

2. What are the two factors? ,

3. What is an integer?

4. Are n and n + 1 consecutive integers? Explain.

5. What do you know about the product of odd and even integers?

EVEN 3 EVEN 5

ODD3

ODD5

EVEN 3 ODD 5

ODD 3 EVEN 5

6. Out of two consecutive integers, how many are odd?

7. Is the product of consecutive integers odd or even? Explain.

8. n21 n is an even integer because

.



n

n

positive and negative whole numbers and zero

yes; the next number after n is n 1 1

even

even

even

always

it is the product of two consecutive integers

1

even; Two consecutive integers

odd

n 1 1

1

must be an odd integer and an even integer. If 1 factor is even, the product will be even.

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Simplify each product.

1. 2 x ( x 1 8) 2. (n 1 7)5n 3. 6h2(7 1 h)

4. 2b2(b 2 10) 5. 23c (8 1 2c 2 c 3) 6. y (2 y 2 2 3 y 1 6)

7. 4t (t 2 2 6t 1 2) 8. 2m(4m32 8m2

1 m) 9. 7 j (22 j 2 2 8 j 2 3)

10. 2t 2(2t 4 1 4t 2 8) 11. 2k (23k 3 1 k 2 2 10) 12. 8a2(2a7 1 7a 2 7)

13. 4v 3(2v 2 2 3v 1 5) 14. 5d (2d 3 1 2d 2 2 3d ) 15. 11w (w 2 1 2w 1 6)

Find the GCF of the terms of each polynomial.

16. 15 x 1 27 17. 6w 3 2 14w 18. 63s 1 45

19. 72 y 5 1 18 y 2 20. 218q3 2 6q2 21. 108 f 3 2 54

22. b3 1 5b2 2 20b 23. 9m31 30m 2 24 24. 4p3 1 12p2 2 18p

25. 2e 2 1 12e 2 22 26. 14b3 1 21b2 2 42b 27. 212 x 3 1 24 x 2 2 16 x

28. 8a4 1 24a3 2 40a2 29. 36 j 3 2 3 j 2 2 15 j 30. 12 j 8 1 30 j 4 2 6 j 3

Factor each polynomial.

31. 12 x 2 9 32. 18s2 1 54 33. 108t 2 2 60t

34. 220w 2 1 16w 35. 32 y 3 1 8 y 2 36. 300d 2 2 175d

37. 12n3 2 36n2 1 18 38. 40t 3 1 25t 2 1 80t 39. 42 x 4 2 56 x 3 1 28 x 2

40. 15c 4 1 24c 3 2 6c 2 1 12c 41. 8m31 14m2

1 6m 42. 10 x 2 1 50 x 2 25

43. 36p4 1 14p3 1 35p2 44. 9a5 1 27a4 1 63a2 45. 4b4 1 20b3 1 12b

46. x 6 2 x 4 1 x 2 47. 34 g 3 1 51 g 2 1 17 g 48. 18h4 2 27h2 1 18h

8-2Practice Form G


2 x 2 1 16 x

2b31 10b2

4t 3 2 24t 2 1 8t

22t 6 2 4t 3 1 8t 2

8v 5 2 12v 4 1 20v 3

3

18 y 2

b

2

3(4 x 2 3)

24w (5w 2 4)

6(2n32 6n2

1 3)

3c (5c 3 1 8c 2 2 2c 1 4)

p2(36 p21 14 p 1 35)

x 2( x 4 2 x 2 1 1)

2m(4m21 7m 1 3)

9a2(a31 3a2

1 7)

17g(2g21 3g 1 1)

8a2

2w

26q2

3

7b

18( s21 3)

8 y 2(4 y 1 1)

5t (8t 2 1 5t 1 16)

25d (12d 2 7)

14 x 2(3 x 2 2 4 x 1 2)

5(2 x 2 1 10 x 2 5)

4b(b31 5b2

1 3)

9h(2h32 3h 1 2)

3 j

9

54

2 p

4 x

12t (9t 2 5)

6 j 3

26k 4 1 2k 3 2 20k

25d 4 1 10d 3 2 15d 2

28a91 56a3

2 56a2

11w 3 1 22w 2 1 66w

24m41 8m3

2 m2214 j 3 2 56 j 2 2 21 j

3c 4 2 6c 2 2 24c

5n21 35n 6h3

1 42h2

2 y 3 2 3 y 2 1 6 y

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49. A circular hedge surrounds a sculpture on a square base. Te

radius o the hedge is 6 x . Te side length o the square sculpture

base is 4 x . What is the area o the hedge? Write your answer inactored orm.

50. Suppose you are making a giant chocolate chip cookie or a raf e. You roll out

a square slab o cookie dough. Ten you use a circular plate that touches the

edges o the square slab o cookie dough and cut the cookie out o the dough.

What is the area o the extra dough? Write your answer in actored orm.

Simplify. Write in standard form.

51. 23 x (4 x 22 6 x 1 12) 52. 27 y

2(24 y

31 6 y ) 53. 9a(23a

21 a 2 5)

54. p(p 1 4)22p(p 2 8) 55. t (t 1 4)2 t (4t 2 2 2) 56. 6c (2c 2 2 4)2 c (8c )

57. 25m(2m32 7m2

1 m) 58. 2q(q 1 1)2q(q 2 1) 59. 2n2(26n2 1 2n)


60. 15 xy 4 1 60 x 2 y 3 61. 8m3n4 1 32mn2 62. 26a5b2 1 51a4

63. 36 j 2k 4 1 24 j 4k 2 64. 12w 4 x 3 2 42wx 2 65. 54c 2d 3 2 36c 3d 2

66. 12st 4 1 46s3t 4 67. 9v 6w 3 1 33v 4w 5 68. 11e 3 f 3 1 132e 2 f 4

69. Error Analysis A student actored the polynomial at the

right. Describe and correct the error made in actoring.

70. Reasoning Te GCF o two numbers j and k is 8. What is the GCF o 2 j and

2k ? Justiy your answer.

71. A cylinder has a radius o 3m2n and a height o 7mn. Te ormula or the

volume o a cylinder is V 5 pr 2h, where r is the radius and h is the height.

What is the volume o the cylinder? Simpliy your answer.



6 x

4 x

63x4 – 14x3 + 35x2

= 7x(9x3 – 2x2 + 5x)

4 x 2(9π 2 4)

r 2(4 2 π)

212 x 3 1 18 x 2 2 36 x

2 p21 20 p

210m41 35m3

2 5m2

15 xy 3( y 1 4 x )

12 j 2k 2(3k 2 1 2 j 2)

2 st 4(6 1 23 s2)

The student did not find the correct GCF. 7 x 2(9 x 2 2 2 x 1 5)

16 The GCF will be the product of 2 and 8.

63πm5n3

28 y 5 2 42 y 3

24t 3 1 t 2 1 6t

q21 3q

8mn2(m2n2 1 4)

6wx 2(2w 3 x 2 7)

3v 4w 3(3v 2 1 11w 2)

227a31 9a2

2 45a

12c 3 2 8c 2 2 24c

6n42 2n3

a4(26ab2 1 51)

18c 2d 2(3d 2 2c )

11e2f 3(e 1 12f )

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8-2Practice Form K



1. 3w (w 1 2) 2. (z 1 5)2z 3. 3m2(4 1 m)

4. 2p(p2 2 6p 1 1) 5. 2 y (5 y 3 2 3 y 2 1 2 y ) 6. 3a(23a2 1 2a 2 7)

7. 6 x 3(3 x 2 2 x 1 10) 8. 24h(2h3 2 8h2 1 2h) 9. 4n(n2 1 5n 1 6)


10. 16q 1 32 11. 4t 3 2 24t 12. 32 y 2 24

13. x 3 1 3 x 2 1 5 x 14. 5d 3 1 20d 2 35 15. 2m31 10m2

1 12m

16. 7 g 4 1 21 g 3 2 14 g 2 17. 15z 3 1 3z 2 2 27z 18. 33w 7 1 55w 5 2 22w 3


19. 9t 2 3 20. 12 j 3 1 28 21. 72 x 2 2 63 x

22. 12k 3 2 9k 2 1 6 23. 30n3 1 18n2 1 54n 24. 32z 4 2 80z 3 1 112z 2

25. 12n4 1 16n3 1 20n2 26. 24 y 6 1 36 y 4 1 42 y 2 27. 7q5 1 21q3 2 49q

16 4t 8

x 5 2m

7g2 3 z 11w 3

3(3t 2 1) 4(3 j 3 1 7) 9 x (8 x 2 7)

3(4k 3 2 3k 2 1 2) 6n(5n21 3n 1 9) 16 z 2(2 z 2 2 5 z 1 7)

4n2(3n21 4n 1 5) 6 y 2(4 y 4 1 6 y 2 1 7) 7q(q4

1 3q22 7)

3w 2 1 6w 2 z 2 1 10 z 12m21 3m3

2 p32 12 p2

1 2 p 25 y 4 1 3 y 3 2 2 y 2 29a31 6a2

2 21a

18 x 5 2 6 x 4 1 60 x 3 4h41 32h3

2 8h2 4n31 20n2

1 24n

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28. You are painting a rectangular wall with length 5 x 2 t and width 12 x t. Tere is

a rectangular door that measures x t by 2 x t that will not be painted. What is

the area o the wall that is to be painted? Write your answer in actored orm.

Simplify. Write in standard form.

29. 23m(2m22 5m 1 10) 30. 25t 2(26t 3 1 12t ) 31. 10 x (24 x 2 1 x 2 3)

32.2

2v (3v 32

6v 21

2v ) 33. 5 y ( y 1

2)2

y ( y 2

3) 34.2

2b2

(2

4b21

3b)


35. 13cd 3 1 39c 2d 2 36. 5 x 3 y 4 2 25 xy 2 37. 42m5n 1 28m4

38. 36 f g 2 1 54 f 2 g 4 39. 8s8t 4 1 20s4t 3 40. 12a2b51 156a2b3

41. Open-Ended Write a quadratic monomial and a cubic trinomial. Ten fnd

their product and write it in standard orm.

42. A rectangle has a length o 6 x 3 y 2 2 1 and a width o 3 xy 1 2. Te ormula or

the perimeter o a rectangle is P 5 2l 1 2w , where l is the length and w is the

width. What is the perimeter o the rectangle? Simpliy your answer.

2 x 2(30 x 2 1)

26m31 15m2

2 30m 30t 5 2 60t 3 240 x 3 1 10 x 2 2 30 x

26v 4 1 12v 3 2 4v 2 4 y 2 1 13 y 8b42 6b3

13cd 2(d 1 3c ) 5 xy 2( x 2 y 2 2 5) 14m4(3mn 1 2)

18fg2(2 1 3fg2) 4 s4t 3(2 s4t 1 5) 12a2b3(b21 13)

Answers may vary. Sample: x 2 and 2 x 3 1 x 2 1 x ; 2 x 5 1 x 4 1 x 3

12 x 3 y 2 1 6 xy 1 2

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Multiple Choice


1. How can this product be simplifed?

5 x 2(2 x 2 3)

A. 5 x 2 1 2 x 2 3 B. 10 x 3 2 15 x 2 C. 25 x 2 D. 7 x 3 2 15 x 2

2. What is the GCF o the terms o 8c 3 1 12c 2 1 10c ?

F. 2 G. 4 H. 2c I. 4c

3. How can the polynomial 6d 4 1 9d 3 2 12d 2 be actored?

A. 3d 2(2d 2 1 3d 2 4)

B. 3d 2(3d 2 1 6d 2 9)

C. 3d (d 3 1 3d 2 2 4)

D. 6d 2(d 2 1 3d 3 2 6)

4. Tere is a circular garden in the middle o a square yard. Te radius o the circle is 4 x .

Te side length o the yard is 20 x . What is the area o the part o the yard that is not

covered by the circle?

F. 4 x (5) G. 8 x 2(5 2 p) H. 16 x (25 1 p) I. 16 x 2(25 2 p)

5. What is the simplifed orm o 23z 2(z 1 2)24(z 2 1 1)?

A. 27z 2 1 1

B. 23z 3 2 4z 2 2 6z 2 4

C. 23z 3 2 2z 2 2 4

D. 23z 3 2 10z 2 2 4

Short Response

6. A rectangular blacktop with a length o 5 x and a width o 3 x has been erected

inside a rectangular feld that has a length o 12 x and a width o 7 x .

a. What is the area o the part o the feld that is not blacktop?

b. Tere is a circular ountain in the rectangular feld that has a radius o 3 x .

What is the area o the part o the feld that does not include the blacktop or

the ountain? Factor your answer.



H

A

B

I

D

[2] Both parts answered correctly

[1] One part answered correctly


69 x 2

3 x 2(23 2 3π)

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o fnd the area o irregular fgures, split the fgure into simple fgures and then

add the areas o each fgure.

1. What is the area o the fgure to the right?

2. What is the perimeter o the fgure?

A circle is inscribed in a square as shown.

3. What is the area o the circle?

4. What is the area o the square?

5. What is the area o the shaded region?

6. Te area o a right triangle is 10 y 3 1 5 y 2 1 37.5 y . Te length o base o the

triangle is a monomial with a whole number coe cient. Te length o the

height is a trinomial. Factor the polynomial to fnd the base and height o the

triangle. (Remember to multiply the area by 2 frst because Atriangle 5b 3 h

2 .)

Base 5

Height 5

8-2Enrichment


24 x 1 5

24 x

23 x

23 x

10 z

37 x 2 2 35 x

228 x 1 10

25π z 2

100 z 2

100 z 2 2 25π z 2, or 25 z 2(4 2 π)

5 y

4 y 2 1 2 y 1 15

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You can multiply a monomial and a trinomial by solving simpler problems. You

can use the Distributive Property to make three simpler multiplication problems.

Problem

What is the simplifed orm o 3 x (2 x 2 1 4 x 2 1)?

Use the Distributive Property to rewrite the problem as three separate

multiplication problems.

3 x (2 x 2 1 4 x 2 1) 5 (3 x ? 2 x 2) 1 (3 x ? 4 x ) 1 (3 x ? (21))

Remember that when you multiply same-base terms containing exponents, you

add the exponents.

Solve 3 x ? 2 x 2 5 6 x 3 Multiply inside the first pair of parentheses.

3 x ? 4 x 5 12 x 2 Multiply inside the second pair of parentheses.

3 x ? (21) 5 23 x Multiply inside the third pair of parentheses.

6 x 3 1 12 x 2 2 3 x Add the products.

Check 6 x 34 2 x

25 3 x Check your solution using division. 12 x 2 4 4 x 5 3 x

23 x 4 (21) 5 3 x

Solution: 3 x (2 x 2 1 4 x 2 1) 5 6 x 3 1 12 x 2 2 3 x

Exercises


1. 4 x (2 x 2 7) 2. 3 y (3 y 1 4) 3. 2z 2(2z 2 3)

4. 3a(24a 2 6) 5. 6b(3b2 1 2b 2 4) 6. 3c 2(2c 2 2 4c 1 3)

7. 22d (4d 2 1 3d 2 2) 8. 5e 2(23e 2 2 2e 2 3) 9. 4 f (23 f 3 1 2 f 2 1 6)

8-2Reteaching


8 x 2 2 28 x

212a22 18a

28d 3 2 6d 2 1 4d

9 y 2 1 12 y

18b31 12b2

2 24b

215e42 10e3

2 15e2

4 z 3 2 6 z 2

6c 4 2 12c 3 1 9c 2

212f 4 1 8f 31 24f

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o actor a polynomial, fnd the greatest common actor (GCF) o the coe cients and

constants and also the GCF o the variables.

Problem

What is the actored orm o 8 x 4 1 12 x 2 2 16 x ?

Solve Find the GCF o the coe cients. Use prime actorization.

8 5 2 ? 2 ? 2

12 5 2 ? 2 ? 3

16 5 2 ? 2 ? 2 ? 2

Te GCF o the numbers is 4.Each term has a variable. Remember, x 5 x 1.

Te GCF is the least exponent.

Te GCF o the variables is x .

Te GCF is 4 x . Combine the GCFs.

Factor out the GCF o each term.

4(2 1 3 2 4) Factor the coefficients.

4 x (2 x 3 1 3 x 2 4) Insert the variables.

Check 4 x (2 x 3 1 3 x 2 4) 5 8 x 4 1 12 x 2 2 16 x Check by multiplying.

Solution: Te actored orm o 8 x 4 1 12 x 2 2 16 x is 4 x (2 x 3 1 3 x 2 4).

Exercises


10. 12 x 2 2 6 x 11. 4 y 2 1 12 y 1 8 12. 6z 3 1 15z 2 2 9z


13. 8a 1 10 14. 12b22 18b 15. 9c 3 1 12c 2

16. 5d 3 2 10d 2 1 20d 17. 6e 2 1 10e 2 8 18. 8 g 3 2 24 g 2 1 16 g



6 x

2(4a 1 5)

5d (d 2 2 2d 1 4)

4

6b(2b 2 3)

2(3e21 5e 2 4)

3 z

3c 2(3c 1 4)

8g(g22 3g 1 2)

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8-3ELL Support

Multiplying Binomials

Use the Distributive Property to fnd the simplifed orm o (3 x 1 2)(4 x 2 3).

(3 x 1 2)(4 x 2 3) Write the problem.

3 x (4 x 2 3) 1 2(4 x 2 3) Distribute the second factor, 4 x 2 3.

12 x 22 9 x 1 2(4 x 2 3) Distribute 3x.

12 x 22 9 x 1 8 x 2 6 Distribute 2.

12 x 22 x 2 6 Combine like terms.

Exercises


(5 x 1 6)(2 x 2 4)

5 x (2 x 2 4) 1 6(2 x 2 4)

10 x 22 20 x 1 6(2 x 2 4)

10 x 22 20 x 1 12 x 2 24

10 x 22 8 x 2 24


(7 x 2 3)(4 x 1 6)

7 x (4 x 1 6) 2 3(4 x 1 6)

28 x 21 42 x 2 3(4 x 1 6)

28 x 21 42 x 2 12 x 2 18

Write the problem.

Write the problem.

Distribute the second factor, 2 x 2 4.

Distribute the second factor, 4 x 1 6.

Distribute 5 x .

Distribute 7 x .

Distribute 6.

Distribute 23.

Combine like terms.

Combine like terms.28 x 21 30 x 2 18

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Geometry Te dimensions o a rectangular prism are n, n 1 7, and n 1 8. Use

the ormula V 5 lwh to write a polynomial in standard orm or the volume o

the prism.

Know

1. What are the dimensions o the rectangular prism? , ,

2. What is the ormula or the volume o a rectangular prism?

3. In the volume ormula, what do l , w , and h represent? , ,

4. Explain how to write a polynomial in standard orm.

Need

5. o solve the problem you need to fnd

.

Plan

6. Draw a diagram o the rectangular prism and

label the inormation you know.

7. Write an expression or the volume o the rectangular prism.

8. Write the volume o the rectangular prism as a polynomial in standard orm.



n n 1 7 n 1 8

V 5 lwh

length

a polynomial in standard form that represents

The terms are arranged in order of

width height

degree, with the highest degree first.

n 1 7

n

n 1 8

V 5 n(n 1 7)(n 1 8)

n31 15n2

1 56n

the volume of the rectangular prism

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Simplify each product using the Distributive Property.

1. ( x 1 3)( x 1 8) 2. ( y 2 4)( y 1 7) 3. (m 1 9)(m 2 3)

4. (c 2 6)(c 2 4) 5. (2r 2 5)(r 1 3) 6. (3 x 1 1)(5 x 2 3)

7. (d 1 2)(4d 2 3) 8. (5t 2 1)(3t 2 2) 9. (a 1 11)(11a 1 1)

Simplify each product using a table.

10. ( x 1 3)( x 2 5) 11. (a 2 2)(a 2 13) 12. (w 2 4)(w 1 8)

13. (5h 2 3)(h 1 7) 14. ( x 2 3)(2 x 1 3) 15. (2p 1 1)(6p 1 4)

Simplify each product using the FOIL method.

16. (2 x 2 6)( x 1 3) 17. (n 2 5)(3n 2 4) 18. (4p2 1 2)(3p 2 1)

19. (a 1 7)(a 2 3) 20. ( x 1 3)(3 x 2 2) 21. (k 2 9)(k 1 5)

22. (b 2 5)(b 2 11) 23. (4m 2 1)(m 1 4) 24. (7z 1 3)(4z 2 6)

25. (2h 1 6)(5h 2 3) 26. (3w 1 12)(w 1 3) 27. (6c 2 2)(9c 2 8)

8-3Practice Form G


x 2 2 2 x 2 15

5h21 32h 2 21

x 2 1 11 x 1 24

c 2 2 10c 1 24

4d 2 1 5d 2 6

2 x 2 2 18

a21 4a 2 21

b22 16b 1 55

10h21 24h 2 18

a22 15a 1 26

2 x 2 2 3 x 2 9

y 2 1 3 y 2 28

2r 2 1 r 2 15

15t 2 2 13t 1 2

3n22 19n 1 20

3 x 2 1 7 x 2 6

4m21 15m 2 4

3w 2 1 21w 1 36

w 2 1 4w 2 32

12 p21 14 p 1 4

m21 6m 2 27

15 x 2 2 4 x 2 3

11a21 122a 1 11

12 p32 4 p2

1 6 p 2 2

k 2 2 4k 2 45

28 z 2 2 30 z 2 18

54c 2 2 66c 1 16

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28. What is the surace area o the cylinder at the right? Write your

answer in simplifed orm.

29. Te radius o a cylindrical popcorn tin is (3 x 1 1) in. Te height

o the tin is three times the radius. What is the surace area o the

cylinder? Write your answer in simplifed orm.

30. Te radius o a cylindrical tennis ball can is (2 x 1 1) cm. Te height o the

tennis ball can is six times the radius. What is the surace area o the cylinder?

Write your answer in simplifed orm.


31. ( x 1 3)( x 2 2 2 x 1 4) 32. (k 2 2 5k 1 2)(k 2 5)

33. (3a2 1 a 1 4)(2a 2 6) 34. (2 x 2 1 2 x 2 6)(3 x 2 4)

35. (4 g 1 5)(2 g 2 2 7 g 1 3) 36. (m22 2m 1 7)(3m 1 6)

37. (2c 1 8)(2c 2 2 4c 2 1) 38. (t 1 8)(3t 2 1 4t 1 5)

39. A medical center’s rectangular parking lot currently has a length o 30 metersand a width o 20 meters. Te center plans to expand both the length and

the width o the parking lot by 2 x meters. What polynomial in standard orm

represents the area o the expanded parking lot?

40. Error Analysis Describe and correct the error made in

fnding the product.

41. Multi Step Te height o a painting is twice its width x . You want

a 3 inch wide wooden rame or the painting. Te area o the rame

alone is 216 square inches.

a. Draw a diagram that represents this situation.

b. Write a variable expression or the area o the rame alone.

c. What are the dimensions o the rame?



x 4

x 7

(2x – 3)(x + 7)

2x2+ 17x + 21

2x

3x3

2x2 14x

21

7x

4π x 21 38π x 1 88π

72π x 2 1 48π x 1 8π

56π x 2 1 56π x 1 14π

x 3 1 x 2 2 2 x 1 12

6a32 16a2

1 2a 2 24

8g32 18g2

2 23g 1 15 3m31 9m 1 42

4c 3 1 8c 2 2 34c 2 8

4 x 2 1 100 x 1 600

length is 26; width is 16

18 x 1 36

In the table, the 3 should be 23. Therefore, 3 x should

be 23 x and 21 should be 221. The answer is

2 x 2 1 11 x 2 21.

k 3 2 10k 2 1 27k 2 10

6 x 3 2 2 x 2 2 26 x 1 24

3t 3 1 28t 2 1 37t 1 40

x 1 6

x

2 x 1 62 x

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8-3Practice Form K


Simplify each product using the Distributive Property.

1. (b 2 2)(b 1 1) 2. ( x 1 6)( x 1 5)

3. (3n 1 1)(n 2 8) 4. (2t 2 7)(t 2 5)

5. ( y 1 3)( y 1 7) 6. (b 2 6)(b 1 3)

Simplify each product using a table.

7. ( x 1 1)( x 2 11) 8. (h 2 2)(3h 1 5) 9. (8w 2 3)(4w 2 7)

10. (3c 1 13)(13c 1 3) 11. (3a 1 2)(a 2 2) 12. (t 1 7)(2t 2 4)

13. (3q2 1 6)(2q 2 5) 14. ( x 1 6)( x 2 7) 15. (p 2 10)(2p 1 5)

16. ( j 2 12)( j 2 11) 17. (3z 2 4)(7z 2 5) 18. (2m 1 11)(6m 2 1)

19. (7h 1 6)(7h 2 6) 20. (23z 1 7)(4z 2 8) 21. (23t 1 5)(23t 2 2)

b22 b 2 2 x 2 1 11 x 1 30

3n22 23n 2 8 2t 2 2 17t 1 35

y 2 1 10 y 1 21 b22 3b 2 18

x 2 2 10 x 2 11 3h22 h 2 10 32w 2 2 68w 1 21

39c 2 1 178c 1 39 3a22 4a 2 4 2t 2 1 10t 2 28

6q32 15q2

1 12q 2 30 x 2 2 x 2 42 2 p22 15 p 2 50

j 2 2 23 j 1 132 21 z 2 2 43 z 1 20 12m21 64m 2 11

49h22 36 212 z 2 1 52 z 2 56 9t 2 2 9t 2 10

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22. Te radius o a circle is (7 x 1 3) cm. Write an expression to represent the area

o the circle in simplifed orm.

23. A rectangle has a length o ( x 1 2) in. and a width o (2 x 1 3) in. Find an

expression that represents the area o the rectangle. Write the expression in

simplifed orm.

Simplify each product using the FOIL method.

24. ( x 1

4)( x 1

6) 25. (a2

5)(2a2

6) 26. (6d 21

4)(8d 2

3)

27. (t 2 4)(t 2 9) 28. (n 1 8)(2n 2 7) 29. ( f 2 7)( f 1 3)


30. (c 1 4)(c 2 2 3c 1 5) 31. (p22 2p 1 5)(p 2 7)

32. (4 x 2 1 2 x 1 3)(3 x 2 8) 33. (5t 2 1 3t 2 11)(6t 2 1)

34. A community center is expanding the size o its rectangular meeting hall. Te

hall is currently 300 t long and 150 t wide. Te center plans to expand both

the length and the width o the meeting hall by 3 x t. What polynomial in

standard orm represents the area o the expanded meeting hall?

35. Open-Ended Write a cubic monomial and a ourth-degree trinomial. Ten

fnd their product and write it in standard orm.

49π x 2 1 42π x 1 9π cm2

2 x 2 1 7 x 1 6 in.2

x 2 1 10 x 1 24 2a22 16a 1 30 48d 3 2 18d 2 1 32d 2 12

t 2 2 13t 1 36 2n21 9n 2 56 f 2 2 4f 2 21

c 31 c

22 7c 1 20 p

32 9 p

21 19 p 2 35

12 x 3 2 26 x 2 2 7 x 2 24 30t 3 1 13t 2 2 69t 1 11

9 x 2 1 1350 x 1 45,000 ft2

Answers may vary. Sample: 2 x 3 and x 4 1 2 x 1 3; 2 x 7 1 4 x 4 1 6 x 3

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Multiple Choice


1. What is the simplifed orm o ( x 2 2)(2 x 1 3)? Use the Distributive Property.

A. 2 x 2 2 x 2 6 B. 2 x 2 2 6 C. 2 x 2 2 7 x 2 6 D. 2 x 2 1 x 2 6

2. What is the simplifed orm o (3 x 1 2)(4 x 2 3)? Use a table.

F. 12 x 2 1 18 x 1 6 G. 12 x 2 1 x 2 6 H. 12 x 2 1 18 x 2 6 I. 12 x 2 2 x 2 6

3. What is the simplifed orm o (4p 2 2)(p 2 4)?

A. 4p21 6p 2 16 B. 4p2

2 18p 1 8 C. 4p22 14p 2 6 D. 4p2

2 6p 1 16

4. Te radius o a cylinder is 3 x 2 2 cm. Te height o the cylinder is x 1 3 cm. What is the surace area o the cylinder?

F. 2p(3 x 2 1 10 x 2 8)

G. 2p(12 x 2 1 7 x 2 2)

H. 2p(12 x 2 2 2 x 1 13)

I. 2p(12 x 2 2 5 x 2 2)

5. What is the simplifed orm o (2 x 2 1 4 x 2 3)(3 x 1 1)?

A. 6 x 3 1 10 x 2 2 5 x 1 3

B. 6 x 3 1 14 x 2 1 5 x 2 3C. 6 x 3 1 14 x 2 2 5 x 2 3

D. 6 x 3 2 10 x 2 2 5 x 2 3

Short Response

6. A soup can that is a cylinder has a radius o 2 x 2 1 and a height o 3 x . What is

the surace area o the soup can? Show your work.



A

I

B

I

C

[2] Correct polynomial written with all work shown

[1] Polynomial written with minor calculation error or inadequate work shown

[0] No correct work shown

20π x 22 14π x 1 2π

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You can fnd the volume o irregular fgures by dividing

the fgure into smaller rectangular prisms, fnding the

volume o each separate fgure, and then adding themtogether. Te fgure to the right can be divided into two

rectangular prisms.

V 1 5 ( x 1 1)( x 1 1)( x 2 1)

5 ( x 2 1 2 x 1 1)( x 2 1)

5 x 31 x

22 x 2 1

Subtract to fnd the length o Prism 2.

(2 x 1 3) 2 ( x 1 1) 5 x 1 2

V 2 5 ( x 1 2)( x 2 1)(2 x 2 2)

5 ( x 2 1 x 2 2)(2 x 2 2)

5 2 x 3 2 6 x 1 4

V otal 5 ( x 3 1 x 22 x 2 1) 1 (2 x 3 2 6 x 1 4)

5 3 x 3 1 x 22 7 x 1 3

You can also fnd the volume o an irregular fgure by fnding the volume o the

whole fgure, as i no pieces were cut away. Next, fnd the volume o the cut away

piece, and then subtract that volume rom the whole. Prism 2 is x 2 3 taller than

Prism 1.

V Whole 5 ( x 2 1)(2 x 1 3)(2 x 2 2) 5 (2 x 2 1 x 2 3)(2 x 2 2) 5 4 x 3 2 2 x 2 2 8 x 1 6

V Piece 5 ( x 2 1)( x 1 1)( x 2 3) 5 ( x 2 2 1)( x 2 3) 5 x 32 3 x 2 2 x 1 3

V otal 5 (4 x 3 2 2 x 2 2 8 x 1 6) 2 ( x 3 2 3 x 2 2 x 1 3) 5 3 x 3 1 x 22 7 x 1 3

What is the volume o each fgure? Write your answer as a polynomial in

standard orm.

1. 2.

8-3Enrichment


2 x 2

2 x 3

x 1

x 1

x 1

PRISM 1 PRISM 2

?

2 x 2

2 x 3

x 1

x 1

x 1

3 x 2

5 x 2

x 2

x 2

x 3

x 3 x 1

x 4

4 x 1

3 x 5

2 x 7

x 1

10 x 31 54 x

21 54 x 2 8 10 x

31 25 x

22 14 x 2 48

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You can multiply binomials by using the FOIL method. FOIL stands or First,

Outer, Inner, and Last.

Problem

What is the simplifed orm o (4 x 1 3)(2 x 1 6)?

Use the FOIL method to simpliy the binomial.

Solve 4 x ? 2 x 5 8 x 2 Multiply the First terms.

4 x ? 6 5 24 x Multiply the Outer terms.

3 ? 2 x 5 6 x Multiply the Inner terms.

3 ? 6 5 18 Multiply the Last terms.

8 x 2 1 24 x 1 6 x 1 18 Add the products.

8 x 2 1 30 x 1 18 Add the like terms.

Check Substitute any number or x . ry x 5 2. I the two sides o the

equation are equal the simplifcation may be correct.

(4 x 1 3)(2 x 1 6) 0 8 x 2 1 30 x 1 18

(4 ? 2 1 3)(2 ? 2 1 6) 0 (8 ? 22) 1 (30 ? 2) 1 18

(11)(10) 0 32 1 60 1 18

110 5 110 3

Solution: Te simplifed orm o (4 x 1 3)(2 x 1 6) is 8 x 2 1 30 x 1 18.

Exercises


1. (a 1 6)(a 2 3) 2. (b 2 4)(b 1 5) 3. (c 1 3)(c 1 7)

4. (2d 1 4)(3d 2 2) 5. (4e 2 5)(3e 1 3) 6. (3 f 2 2)(2 f 2 4)

7. (5 g 1 3)( g 2 3) 8. (4h 1 4)(2h 1 5) 9. (3 j 2 5)(4 j 2 3)

8-3Reteaching


a2 1 3a 2 18

6d 2 1 8d 2 8

5g22 12g 2 9

b2 1 b 2 20

12e22 3e 2 15

8h21 28h 1 20

c 2 1 10c 1 21

6f 22 16f 1 8

12 j 2 2 29 j 1 15

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o multiply a trinomial by a binomial, use the same steps as you would to multiply

a 3-digit number by a 2-digit number. Find the partial products or each term o the

binomial and then add the like terms o the partial products.

Problem

What is the simplifed orm o (2 x 2 1 3 x 2 4)(3 x 1 2)?

Solve Start by arranging the polynomials vertically.

Multiply each part o the trinomial by 2.

2 x 21 3 x 2 4

2 x 2 1 3 x 1 2

4 x 21 6 x 2 8

Multiply each part o the trinomial by 3 x .

6 x 3 1 2 x 21 3 x 2 4

6 x 2 1 4 x 2 1 3 x 1 2

6 x 3 1 4 x 2 1 6 x 2 8

6 x 31 9 x

22 12 x 2 8

2 x 2 ? 3 x 5 6 x 3

3 x ? 3 x 5 9 x 2

24 ? 3 x 5 212 x

Add the partial products.

6 x 3 1 4 x 2 1 6 x 2 8

6 x 3 1 9 x 2 2 12 x 2 8

6 x 3 1 13 x 2 2 6 x 2 8

Check Substitute any number or x . ry x 5 2. I the two sides o the

equation are equal, the simplifcation may be correct.

(2 x 2 1 3 x 2 4)(3 x 1 2) 0 6 x 3 1 13 x 2 2 6 x 2 8

(8 1 6 2 4)(6 1 2) 0 48 1 52 2 12 2 8

80 5 80 3Solution: Te simplifed orm o (2 x 2 1 3 x 2 4)(3 x 1 2) is 6 x 3 1 13 x 2 2 6 x 2 8.

Exercises


10. (w 21 3w 2 4)(2w 1 3) 11. ( x 2 2 8 x 1 6)(3 x 2 4)

12. (2 y 2 1 4 y 2 5)(4 y 1 2) 13. (3z 2 2 6z 1 4)(4z 1 1)



2 x 2 ? 2 5 4 x 2

3 x ? 2 5 6 x

24 ? 2 5 28

2w 3 1 9w 2 1 w 2 12

8 y 3 1 20 y 2 2 12 y 2 10

3 x 3 2 28 x 2 1 50 x 2 24

12 z 3 2 21 z 2 1 10 z 1 4

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8-4ELL Support

Multiplying Special Cases

Use the list below to complete the diagram.

Te square o a binomial isthe square o the frst term

plus twice the product o the

two terms plus the square o

the last term.

( x 1

3)( x 2

3)5

x 2 2 325 x 2 2 9 Te product o the sum anddierence o the same two

terms is the dierence o

their squares.

(a 1 b)25 a2

1 2ab 1 b2 (a 2 b)25 a2

2 2ab 1 b2 (a 1 b)(a 2 b) 5 a22 b2

TheSquare of a Binomial

The

Product of a Sum andDifference

The square of thebinomial is thesquare of the firstterm plus twice theproduct of the twoterms plus the squareof the last term.

(a 1 b)2 5

a21 2ab 1 b2

(a 1 b)(a 2 b) 5

a22 b2

( x 1 3)( x 2 3) 5

x 2 2 325 x 2 2 9

(a 2 b)2 5

a22 2ab 1 b2

The product of thesum and differenceof the sametwo terms is thedifference of theirsquares.

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Name Class Date

Construction A square deck has a side length o x 1 5. You are expanding the

deck so that each side is our times as long as the side length o the original deck.

What is the area o the new deck? Write your answer in standard orm.

Understanding the Problem

1. What is the shape o the deck?

2. How long is each side o the deck?

3. Te new deck has sides that are times longer than the original sides.

4. What is the problem asking you to fnd?

Planning the Solution

5. Write an expression or the new side length o the deck.

6. Write an expression or the area o the new deck.

Getting an Answer

7. What is the standard orm o the expression or the area o the new deck?



square

4

area of new deck

4( x 1 5), or 4 x 1 20

(4 x 1 20)2

16 x 21 160 x 1 400

x 1 5

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Name Class Date



Simpliy each expression.

1. ( x 1 7)2 2. (w 1 9)2 3. (h 1 3)2

4. (2s 1 4)2 5. (3s 1 1)2 6. (5s 1 2)2

7. (a 2 5)2 8. (k 2 10)2 9. (n 2 4)2

10. (3m 2 4)2 11. (6m 2 2)2 12. (4m 2 2)2

Te fgures below are squares. Find an expression or the area o each shaded

region. Write your answers in standard orm.

13. 14.

15. 16.

17. A square brown tarp has a square green patch green in the corner. Te side

length of the tarp is ( x 1 8) and the side length of the patch is x . What is the

area of the brown part of the tarp?

18. A square red placemat has a gold square in the center. Te side length of the

gold square is ( x 2 2) inches and the width of the red region is 4 inches. What

is the area of the red part of the placemat?

8-4Practice Form G


x 1 2

x 2 1

x 2 1

x 1 2

x 1 5

x 1 1

x 1 1

x 1 5

x 1 6

x

x

x 1 6

x 1 7

x 2 2

x 2 2 x 1 7

x 2 1 14 x 1 49

4 s21 16 s 1 16

a22 10a 1 25

9m22 24m 1 16

6 x 1 3

8 x 1 24

16 x 1 64

2 x 2 1 4 x 1 12 square inches

12 x 1 36

18 x 1 45

w 2 1 18w 1 81

9 s21 6 s 1 1

k 2 2 20k 1 100

36m22 24m 1 4

h21 6h 1 9

25 s21 20 s 1 4

n22 8n 1 16

16m22 16m 1 4

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Name Class Date



Mental Math Simplify each product.

19. 482 20. 312 21. 292

22. 522 23. 632 24. 412

25. 892 26. 1992 27. 3022


28. (v 1 7)(v 2 7) 29. (b 1 2)(b 2 2) 30. (z 2 9)(z 1 9)

31. ( x 1 12)( x 2 12) 32. (8 1 y )(8 2 y ) 33. (t 2 15)(t 1 15)

34. (m 1 1)(m 2 1) 35. (a 1 4)(a 2 4) 36. (5 1 g )(5 2 g )

37. (p 1 20)(p 2 20) 38. ( f 2 18)( f 1 18) 39. (2c 1 3)(2c 2 3)


40. 61 ? 59 41. 27 ? 33 42. 202 ? 198

43. 74 ? 66 44. 597 ? 603 45. 85 ? 75


46. (m 1 4n)2 47. (3a 1 b)2 48. (6s 2 t )2

49. (s 1 7t 2)2 50. (p5 2 8q3)2 51. (e 4 1 f 2)2

52. (r 2 1 5s)(r 2 2 5s) 53. (6p2 1 2q)(6p2 2 2q) 54. (3w 4 2 z 3)(3w 4 1 z 3)

55. Error Analysis Describe and correct the error

made in simplifying the product.

56. Te formula V 543p

r 3 gives the volume of a sphere

with radius r . Find the volume of a sphere with

radius x 1 9. Write your answer in standard form.



(2x + 7)(2x – 7)= 4x2 – 28x – 49

2304

2704

7921

v 2 2 49

x 2 2 144

m22 1

p22 400

3599

4884

m21 8mn 1 16n2

s21 14 st 2 1 49t 4 e8

1 2e4f 21 f 4

b22 4

64 2 y 2

a22 16

f 22 324

891

359,991

9a21 6ab 1 b2

p102 16 p5q3

1 64q6

36 p4 2 4q2

z 2 2 81

t 2 2 225

25 2 g2

4c 2 2 9

39,996

6375

36 s22 12 st 1 t 2

r 4 2 25 s2

The x terms should have a sum of zero; 4 x 2 2 49

V 5 43π x

31 36π x 2 1 324π x 1 972π

9w 8 2 z 6

39,601 91,204

3969 1681

961 841

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Name Class Date



8-4Practice Form K


Simpliy each expression.

1. ( y 1 1)2 2. (n 1 11)2 3. (t 1 7)2

4. (3m 1 6)2 5. (4 x 1 1)2 6. (3n 1 2)2

7. (t 2 3)2 8. (7v 2 3)2 9. (6p 2 5)2

Te fgures below are squares. Find an expression or the area o each shaded

region. Write your answers in standard orm.

10. 11.

12. A fat, square roo needs a square patch in the corner to seal a leak. Te side

length o the roo is ( x 1 12) t and the side length o the patch is x t. What is

the area o the good part o the roo?

13. A white, square quilt has a purple square in the center. Te side length o thepurple square is ( x 2 5) inches and the width o the quilt is 60 inches. What is

the area o the white part o the quilt?

x à8

x à8

x

x x à5

x à5

x

x

y 2 1 2 y 1 1 n21 22n 1 121 t 2 1 14t 1 49

9m21 36m 1 36 16 x 2 1 8 x 1 1 9n2

1 12n 1 4

t 2 2 6t 1 9 49v 2 2 42v 1 9

16 x 1 64 10 x 1 25

(24 x 1 144) ft

(2 x 2 1 10 x 1 3575) in.2

36 p22 60 p 1 25

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Name Class Date






14. 522 15. 182 16. 1192

17. 4952 18. 722 19. 1512


20. ( x 1 1)( x 2 1) 21. (m 1 5)(m 2 5) 22. (a 2 4)(a 1 4)

23. (s 2 13)(s 1 13) 24. (2z 2 3)(2z 1 3) 25. (4d 1 6)(4d 2 6)


26. 99 ? 101 27. 48 ? 52 28. 178 ? 182


29. (s 1 3t )2 30. (2 x 1 y )2 31. (4a 2 b)2

32. (m21 3n)(m2

2 3n) 33. (9 f 21 4 g )(9 f 22 4 g ) 34. (6m42 n3)(6m4

1 n3)

35. Te formula V 5 pr 2h gives the volume of a cylinder with radius r and height

h. Find the volume of a cylinder with radius ( x 1 4) cm and height 5 cm.

Write your answer in standard form.

2704 324 14,161

245,025 5184 22,801

x 2 2 1 m22 25 a2

2 16

s22 169 4 z 2 2 9 16d 2 2 36

9999 2496 32,396

s21 6 st 1 9t 2 4 x 2 1 4 xy 1 y 2 16a2

2 8ab 1 b2

m42

9n2

81f 42

16g2

36m82

n6

(5π x 2 1 40π x 1 80π) cm3

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Name Class Date

Gridded Response

Solve each exercise and enter your answer on the grid provided.

1. What is coef cient o the x -term in the simplied orm o (2 x 1 4)2?

2. What is 272? Use mental math.

3. What is constant in the simplied orm o ( x 2 6)2?

4. What is the product o 38 and 42? Use mental math.

`5. How much greater is the product o 73 and 67 than the product o 74 and 66?



16

729

36

1596

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9

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2

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Name Class Date

Find the volume of each cube.

1. 2.

3. Find the volume of the

rectangular prism.

4. How much greater is the volume of Cube B than the volume of Cube A?

8-4Enrichment


x à5

x à1

x Ľ4

x Ľ4

x à6

Cube A

x

2

Ľ3

x à9

Cube B

x 31 15 x

21 75 x 1 125 x

62 9 x

41 27 x

22 27

x 32 7 x

21 8 x 1 16

9 x 2 1 135 x 1 513

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Name Class Date

A binomial is squared when it is multiplied by itsel. Te square o a binomial is

the square o the frst term plus the twice the product o the two terms plus the

square o the last term. Tis can be expressed as (a 1 b)2 5 a2 1 2ab 1 b2.

Problem

What is the simplifed orm o ( x 1 5)2?

Use the rules or squaring a binomial.

Solve x ? x 5 x 2 Square the first term.

2(5 ? x ) 5 10 x Multiply the product of the two terms by 2.

5 ? 5 5 25 Square the last term.

So, ( x 1 5)25 x 2 1 10 x 1 25.

Check ( x 1 5)25 ( x 1 5)( x 1 5) Rewrite the binomials.

x ? x 5 x 2 Multiply the First addends.

x ? 5 5 5 x Multiply the Outer addends.

5 ? x 5 5 x Multiply the Inner addends.

5 ? 5 5 25 Multiply the Last addends.

x 2 1 5 x 1 5 x 1 25 Add the products.

x 2 1 10 x 1 25 Combine the like terms.

Solution: Te simplifed orm o ( x 1 5)2 is x 2 1 10 x 1 25.

Exercises

Simplify each expression.

1. (a 1 7)2 2. (b 2 4)2 3. (2c 1 3)2 4. (3d 2 5)2

5. (4e 1 1)2 6. (2 f 2 6)2 7. ( g 2 10)2 8. (5h 1 8)2

9. (3 j 2 3)2 10. (2k 1 4)2 11. (4m 2 2)2 12. (3n 1 6)2

8-4Reteaching


a21 14a 1 49

16e21 8e 1 1

9 j 2 2 18 j 1 9

b22 8b 1 16

4f 2 2 24f 1 36

4k 2 1 16k 1 16

4c 2 1 12c 1 9

g22 20g 1 100

16m22 16m 1 4

9d 2 2 30d 1 25

25h21 80h 1 64

9n21 36n 1 36

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Name Class Date

Te product o the sum and the diference o the same two terms produces a pattern

that can be expanded algebraically as (a 1 b)(a 2 b) 5 a22 ab 1 ab 2 b2. Te

sum o the two ab- terms is 0. Tereore, (a 1 b)(a 2 b) 5 a2 2 b2. Te product isthe square o the rst term minus the square o the last term.

Problem

What is the simplied orm o (2 x 2 3)(2 x 1 3)?

Use the rules or nding the product o the sum and the diference o the same two terms.

Solve 2 x ? 2 x 5 4 x 2 Square the first term.

3?

35

9 Square the last term. Remember, the product is the diference o the two squares.

Te product is 4 x 2 2 9.

Check Multiply the binomials using the FOIL Method.

2 x ? 2 x 5 4 x 2 Multiply the First addends.

2 x ? 3 5 6 x Multiply the Outer addends.

23 ? 2 x 5 26 x Multiply the Inner addends.

23 ? 3 5 29 Multiply the Last addends.

4 x 2 1 6 x 2 6 x 2 9 Add the products.

4 x 2 2 9 Combine the like terms.

Solution: Te simplied orm o (2 x 2 3)(2 x 1 3) is 4 x 2 2 9.

Exercises


13. (p 2 4)(p 1 4) 14. (q 1 5)(q 2 5) 15. (3r 1 2)(3r 2 2)

16. (4s 2 6)(4s 1 6) 17. (2t 2 1)(2t 1 1) 18. (5u 2 3)(5u 1 3)

19. (6v 2 4)(6v 1 4) 20. (3w 2 8)(3w 1 8) 21. (7 x 2 9)(7 x 1 9)



p22 16

16 s22 36

36v 2 2 16

q22 25

4t 2 2 1

9w 2 2 64

9r 2 2 4

25u22 9

49 x 2 2 81

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Name Class Date

8-5ELL Support

Factoring x 2 1 bx 1 c

For Exercises 1–5, draw a line rom each term in Column A to its defnition in

Column B. Te frst one is done or you.

( x 1 4)( x 1 8) 5 x 2 1 (8 1 4) x 1 4 ? 8 5 x 2 1 12 x 1 32

Column A Column B

1. 1 coef cient o trinomial’s x 2 term

2. 12 binomial

3. 32 coef cient o trinomial’s x term

4. x 1 4 product o ( x 1 4) and ( x 1 8)

5. x 21 12 x 1 32 trinomial’s constant term

For Exercises 6–9, match the expression in Column A with its defnition in

Column B.

n22 9n 2 36 5 (n 2 12)(n 1 3)

Column A Column B6. (n 2 12)(n 1 3) actors o 236

7. n22 9n 2 36 sum o 212 and 3

8. 212 and 3 trinomial

9. 29 actored orm o n22 9n 2 36

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Name Class Date

Recreation A rectangular skateboard park has an area o x 21 15 x 1 54. What

are possible dimensions o the park? Use actoring.

Know

1. Te area o the skateboard park is .

2. Te dimensions o a rectangle are its and .

3. Te o the area polynomial are possible dimensions o the skateboard park.

Need

4 o solve the problem I need to fnd

Plan

5. Complete the table. List the pairs o actors o u.

Identiy the pair that has a sum o u.

6. Write the actored polynomial.

7. What are possible dimensions o the skateboard park?

8. Justiy your answer.



Factors

of 54

Sum of

Factors

x 21 15 x 1 54

length

factors

the factors of x 21 15 x 1 54

54

15

( x 1 6)( x 1 9)

length: x 1 9; width x 1 6

If the length is x 1 9 and the width is x 1 6, then the area is

( x 1 9)( x 1 6) 5 x 21 15 x 1 54.

width

15

Factors Sum

55

29

21

1 and 54

2 and 27

3 and 18

6 and 9

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Name Class Date



Complete.

1. k 21

11k 1

305

(k 1

5)(k 1

u) 2. x 21

6 x 1

95

( x 1

3)( x 1

u)

3. t 2 1 7t 1 10 5 (t 1 2)(t 1u) 4. n2 1 9n 1 14 5 (n 1 7)(n 1u)

5. w 2 1 13w 1 36 5 (w 1 4)(w 1 ) 6. y 2 1 18 y 1 65 5 ( y 1 13)( y 1 )

7. s2 2 12s 1 32 5 (s 2 8)(s 2 ) 8. g 2 2 14 g 1 45 5 ( g 2 9)( g 2 )

9. v 2 2 17v 1 60 5 (v 2 12)(v 2 ) 10. q2 2 13q 1 42 5 (q 2 6)(q 2 )

11. d 2 2 9d 1 8 5 (d 2 8)(d 2u) 12. r 2 2 9r 1 20 5 (r 2 5)(r 2u)

Factor each expression. Check your answer.

13. y 2 1 5 y 1 6 14. t 2 1 9t 1 18 15. x 2 1 16 x 1 63

16. n2 2 12n 1 35 17. r 2 2 12r 1 27 18. q2 2 12q 1 20

19. w 2 1 19w 1 60 20. b2 2 11b 1 24 21. z 2 2 13z 1 12

Complete.

22. q2 1 q 2 56 5 (q 2 7)(q 1 ) 23. z 2 2 3z 2 18 5 (z 2 6)(z 1 )

24. n2 2 6n 2 40 5 (n 1 4)(n 2u) 25. y 2 1 3 y 2 4 5 ( y 1 4)( y 2u)

26. v 2 2 5v 2 36 5 (v 2 9)(v 1u) 27. d 2 1 2d 2 15 5 (d 2 3)(d 1u)

28. m22 5m 2 14 5 (m 1 2)(m 2 ) 29. p2 2 6p 2 16 5 (p 2 8)(p 1 )

8-5Practice Form G


6

5

3

2

5

5

4

7

1

5

4

9

( y 1 3)( y 1 2)

(n 2 7)(n 2 5)

(w 1 15)(w 1 4)

(t 1 6)(t 1 3)

(r 2 9)(r 2 3)

(b 2 8)(b 2 3)

( x 1 9)( x 1 7)

(q 2 10)(q 2 2)

( z 2 12)( z 2 1)

8

10

4

7 2

5

1

3

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30. r 2 1 3r 2 10 31. w 2 1 2w 2 8 32. z 2 1 3z 2 40

33. d 2 2 4d 2 12 34. p22 7p 2 8 35. s2

2 5s 2 24

36. x 2 1 5 x 2 6 37. v 2 1 3v 2 28 38. n21 2n 2 63

39. t 2 2 2t 2 24 40 a22 7a 2 18 41. c 2 2 c 2 30

42. Te area of a rectangular door is given by the trinomial x 2 2 14 x 1 45. Te

door’s width is ( x 2

9). What is the door’s length?

43. Te area of a rectangular painting is given by the trinomial a22 6a 2 16. Te

painting’s length is (a 1 2). What is the painting’s width?

Write the correct factored form for each expression.

44. k 2 1 4kn 2 96n2 45. g 2 2 13 gh 1 42h2 46. m22 4mn 2 32n2

47. x 2 1 5 xy 2 14 y 2 48. s2 1 17st 1 72t 2 49. h2 1 3hj 2 88 j 2

50. Error Analysis Describe and correct the

error made in factoring the trinomial.

51. A rectangular pool cover has an area of p21 9p 2 36. What are possible

dimensions of the pool cover? Use factoring.



x2 + 2x – 80= (x + 8)(x – 10)

(r 1 5)(r 2 2)

(d 2 6)(d 1 2)

( x 1 6)( x 2 1)

(t 2 6)(t 1 4)

(w 1 4)(w 2 2)

( p 2 8)( p 1 1)

(v 1 7)(v 2 4)

(a 2 9)(a 1 2)

x 2

5

a 2 8

(k 1 12n)(k 2 8n)

( x 1 7 y )( x 2 2 y )

(g 2 6h)(g 2 7h)

( s 1 8t )( s 1 9t )

The operation signs are wrong. The answershould be ( x 2 8)( x 1 10).

( p 1 12) and ( p 2 3)

(m 2 8n)(m 1 4n)

(h 1 11 j )(h 2 8 j )

( z 1 8)( z 2 5)

( s 2 8)( s 1 3)

(n 1 9)(n 2 7)

(c 2 6)(c 1 5)

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8-5Practice Form K


Complete.

1. n2 1 9n 1 18 5 (n 1 3)(n 1u) 2. t 2 1 9t 1 14 5 (t 1 2)(t 1u)

3. d 2 1 11d 1 30 5 (d 1 5)(d 1 ) 4. v 2 1 2v 1 1 5 (v 1 1)(v 1 )

5. m22 8m 1 15 5 (m 2 5)(m 2u) 6. a2 2 13a 1 22 5 (a 2 2)(a 2u)

7. z 2 2 17z 1 72 5 (z 2 8)(z 2 ) 8. w 2 2 7w 1 12 5 (w 2 3)(w 2 )


9. g 2 1 6 g 1 8 10. y 2 1 10 y 1 24 11. r 2 1 12r 1 35

12. k 21

9k 1

8 13. x 22

16 x 1

60 14. h22

19h1

78

Complete.

15. g 2 1 5 g 2 24 5 ( g 2 3)( g 1 ) 16. b2 2 6b 2 7 5 (b 2 7)(b 1 )

17. y 2 1 4 y 2 45 5 ( y 1 9)( y 2 ) 18. k 2 1 4k 2 12 5 (k 1 6)(k 2 )

19. p2 2 7p 2 60 5 (p 1 5)(p 2 ) 20. n2 2 6n 2 40 5 (n 2 10)(n 1 )

8 1

5 2

12 4

(g 1 2)(g 1 4) ( y 1 6)( y 1 4) (r 1 5)(r 1 7)

(k 1 1)(k 1 8) ( x 2 10)( x 2 6) (h 2 13)(h 2 6)

7

1

3 11

9 4

6

6

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21. x 2 2 4 x 2 5 22. t 2 1 t 2 20 23. z 2 2 z 2 72

24. m22 6m 2 27 25. a2

1 4a 2 21 26. v 2 2 4v 2 12

27. c 2 2 7c 2 44 28. r 2 1 6r 2 16 29. f 21 f 2 6

30. j 2 2 6 j 2 55 31. y 2 1 3 y 2 54 32. n22 10n 2 11

33. Te area of a rectangular window is given by the trinomial x 2 2 14 x 1 48. Te

window’s length is ( x 2 8). What is the window’s width?

34. Te area of a rectangular area rug is given by the trinomial f 2 2 4 f 2 77. Telength of the rug is ( f 1 7). What is the width of the rug?

35. Reasoning Write possible expressions for the length and the width of a

rectangle with area x 2 1 13 x 1 42.

36. A rectangular tabletop has an area of t 21 2t 2 99. What are possible

dimensions of the tabletop? Use factoring.

( x 1 1)( x 2 5) (t 1 5)(t 2 4) ( z 1 8)( z 2 9)

(m 1 3)(m 2 9) (a 1 7)(a 2 3) (v 1 2)(v 2 6)

(c 1 4)(c 2 11) (r 1 8)(r 2 2) (f 1 3)(f 2 2)

( j 1 5)( j 2 11) ( y 1 9)( y 2 6) (n 1 1)(n 2 11)

( x 2 6)

(f 2 11)

( x 1 6); ( x 1 7)

t 1 11 and t 2 9

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Name Class Date

Multiple Choice


1. Which number makes this equation true?

v 2 1 10v 1 16 5 (v 1 8)(v 1 u )

A. 2 B. 4 C. 6 D. 8

2. What is the factored form of x 2 1 6 x 1 8?

F. ( x 1 5)( x 1 3) G. ( x 1 4)( x 1 2) H. ( x 1 7)( x 1 1) I. ( x 1 3)( x 1 3)


A. ( x 2 5)( x 2 3) B. ( x 2 6)( x 2 1) C. ( x 2 2)( x 2 5) D. ( x 2 4)( x 2 3)

4. Which number makes this equation true?

q21 3q 2 18 5 (q 1 6)(q 2 u )

F. 1 G. 2 H. 3 I. 12


A. ( x 1 5)( x 2 2) C. ( x 2 2)( x 2 5)

B. ( x 2 5)( x 1 2) D. ( x 1 5)( x 1 2)

6. Te area of a garden is given by the trinomial g 2 2 2 g 2 24. Te garden’s

length is g 1 4. What is the garden’s width?F. g 2 2 G. g 2 6 H. g 2 8 I. g 1 2

7. What is the factored form of x 2 1 3 xy 2 28 y 2?

A. ( x 1 14 y )( x 2 2 y )B. ( x 1 2 y )( x 2 14 y ) C. ( x 1 4 y )( x 2 7 y ) D. ( x 2 4 y )( x 1 7 y )

Short Response

8. Te area of a rectangular backyard is given by the trinomial b21 5b 2 24. What

are possible dimensions of the backyard? Show why your answer is correct.



A

G

D

H

A

G

D

[2] Both length and width calculated correctly with all work shown

[1] Correct answer with minor calculation error or inadequate work shown


length: (b 1 8); width: (b 2 3); (b 1 8)(b 2 3) 5 b21 5b 2 24

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o actor a trinomial o the orm x 2 1 bx 1 c as the product o binomials, you

must fnd actor pairs that have a sum o b and a product o c . Examine what

happens to c as you increase b when c is greater than zero.I b = 2, the actor pair is 1, 1 and the product is 1. ( x 1 1)( x 1 1) 5 x 2 1 2 x 1 1

I b = 3, the actor pair is 1, 2 and the product is 2. ( x 1 1)( x 1 2) 5 x 2 1 3 x 1 2

I b = 4, the actor pairs are 1, 3 and 2, 2. Te products are 3 and 4.( x 1 1)( x 1 3) 5 x 2 1 4 x 1 3 ( x 1 2)( x 1 2) 5 x 2 1 4 x 1 4

I b = 5, the actor pairs are 1, 4 and 2, 3. Te products are 4 and 6.

( x 1 1)( x 1 4) 5 x 2 1 5 x 1 4 ( x 1 2)( x 1 3) 5 x 2 1 5 x 1 6

1. What are the actor pairs and products (values o c ) or the ollowing values

o b, or x 2 1 bx 1 c i c . 0?

b 5 6 b 5 7 b 5 8 b 5 9 b 5 10

2. What pattern do you see in the number o actor pairs (and thus values or c )

as you increase the value o b?

3. Describe at least one pattern you see in the value o c in terms o b.

Now examine what happens to the value o b when the value o c changes, when c . 0.

I c = 1, the actor pair is 1, 1, and the sum is 2. ( x 1 1)( x 1 1) 5 x 2 1 2 x 1 1

I c = 2, the actor pair is 1, 2 and the sum is 3. ( x 1 1)( x 1 2) 5 x 2 1 3 x 1 2

I c = 3, the actor pair is 1, 3 and the sum is 4. ( x 1 1)( x 1 3) 5 x 2 1 4 x 1 3

I c = 4, the actors pairs are 1, 4 and 2, 2. Te sums are 5 and 4.

( x 1 1)( x 1 4) 5 x 2 1 5 x 1 4 ( x 1 2)( x 1 2) 5 x 2 1 4 x 1 4

4. What are the actor pairs and sums (values o b) or the ollowing values o c ,

or x 2 1 bx 1 c i c . 0?

c 5 5 c 5 6 c 5 7 c 5 8

5. Describe at least one pattern you see in the value o b in terms o c . Explain

why this might be.

8-5Enrichment


b 5 6; pairs: 1,5; 2, 4; 3, 3; products: 5, 8, 9 b 5 7; pairs: 1, 6; 2, 5; 3, 4; products: 6, 10, 12

b 5 8; pairs: 1,7; 2, 6; 3, 5; 4, 4; products: 7, 12, 15, 16 b 5 9; pairs: 1, 8; 2, 7; 3, 6; 4, 5;products: 8, 14, 18, 20 b 5 10; pairs: 1, 9; 2, 8; 3, 7; 4, 6; 5, 5; products: 9, 16, 21, 24, 25

The number of factor pairs increases as b increases

Answers may vary. Sample: If b is even then b

2 is the number of c values. If b is

odd, then(b 2 1)

2

is the number of c values.

c 5 5; pairs: 1, 5; sums: 6 c 5 6; pairs: 1, 6; 2, 3; sums: 7, 5

c 5 7; pairs: 1, 7; sums: 8 c 5 8; pairs: 1, 8; 2, 4; sums: 9, 6

Answers may vary. Sample: Prime numbers have only one pair of factors becausethe factors of a prime number are the number and 1.

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8-5Reteaching


I a trinomial o the orm x 2 1 bx 1 c can be written as the product o two

binomials, then:

• Te coef cient o the x -term in the trinomial is the sum o the constants in the

binomials.

• Te trinomial’s constant term is the product o the constants in the binomials.

Problem

What is the actored orm o x 2 1 12 x 1 32?

o write the actored orm, you are looking or two actors o 32 that have a sum o 12.

Solve Make a table showing the actors o 32.

x 2 1 12 x 1 32 5 ( x 1 4)( x 1 8)

Check ( x 1 4)( x 1 8)

x

21 8

x 1 4

x 1 32 Use FOIL Method.

x 2 1 12 x 1 32 Combine the like terms.

Solution: Te actored orm o x 2 1 12 x 1 32 is ( x 1 4)( x 1 8).

Exercises

Factor each expression.

1. x 2 1 9 x 1 20 2. y 2 1 12 y 1 35 3. z 2 1 8z 1 15

4. a2 1 11a 1 28 5. b2 1 10b 1 16 6. c 2 1 12c 1 27

7. d 2 1 6d 1 5 8. e 2 1 15e 1 54 9. f 2 1 11 f 1 24

Factors of 32 Sum of Factors

33

18

12

1 and 32

2 and 16

4 and 8

( x 1 5)( x 1 4) ( y 1 7)( y 1 5) ( z 1 5)( z 1 3)

(c 1 9)(c 1 3)

(f 1 8)(f 1 3)

(b 1 8)(b 1 2)

(e 1 9)(e 1 6)

(a 1 4)(a 1 7)

(d 1 5)(d 1 1)

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Some actorable trinomials in the orm o x 2 1 bx 1 c will have negative

coef cients. Te rules or actoring are the same as when the x -term and the

constant are positive.• Te coef cient o the x -term o the trinomial is the sum o the constants in the

binomials.

• Te trinomial’s constant term is the product o the constants in the binomials.

However, one or both constants in the binomial actors will be negative.

Problem

What is the actored orm o x 2 2 3 x 2 40?

o write the actored orm, you are looking or two actors o 240 that have a sumo 23. Te negative constant will have a greater absolute value than the positive

constant.

Solve Make a table showing the actors o 240.

x 2 2 3 x 2 40 5 ( x 2 8)( x 1 5)

Check ( x 2 8)( x 1 5)

x 2 1 5 x 2 8 x 2 40 Use FOIL Method.

x 2 1 (23 x ) 2 40 Combine the like terms.

Solution: Te actored orm o x 2 2 3 x 2 40 is ( x 2 8)( x 1 5).

Exercises


10. s2 1 2s 2 35 11. t 2 2 4t 2 32 12. u2 1 6u 2 27

13. v 2 2 2v 1 48 14. w 2 2 8w 2 9 15. x 2 1 3 x 2 18

Ľ3

Factors of ] 40 Sum of Factors

Ľ39

Ľ18

Ľ6

1 and Ľ40

2 and Ľ20

4 and Ľ10

5 and Ľ8

( s 1 7)( s 2 5)

(v 2 8)(v 1 6)

(t 2 8)(t 1 4)

(w 2 9)(w 1 1)

(u 1 9)(u 2 3)

( x 1 6)( x 2 3)

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8-6ELL Support

Factoring ax 2 1 bx 1 c

A student is trying to factor 3 x 21 13 x 1 4. She wrote these steps to solve the

problem on note cards, but they got mixed up.

Use the note cards to complete the steps below.

1. First,

2. Second,

3. Third,

4. Then,

5. Finally,

Find factors of ac that havesum b .

Since ac 5 12 and b 5 13, findpositive factors of 12 thathave sum 13.(3x 1 1)(x 1 4)

To factor the trinomial, usethe factors you found torewrite bx as 1x 1 12x.

Make a table.

Factorsof 12

2, 6 3, 4 1, 12

Sum offactors 8 7 133

find factors of ac that have sum b.

Factors of 12 2, 6 3, 4 1, 12

Sum of factors 8 7 13 3

since ac 5 12 and b 5 13, find positive factors of 12 that have sum 13.

(3 x 1 1) ( x 1 4)

to factor the trinomial, use the factors you found to rewrite bx .

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Carpentry Te top o a rectangular table has an area o 18 x 21 69 x 1 60. Te

width o the table is 3 x 1 4. What is the length o the table?

Know

1. Te area o the table top is .

2. Te width o the table top is .

3. Some quadratic trinomials can be written as the product o two .

4. One o the actors o the polynomial 18 x 21 69 x 1 60 is .

Need

5. o solve the problem I need to fnd

.

Plan

6. Find the missing actor.

What can you multiply by 3 x to get 18 x 2? 3 x ? 5 18 x

2

What can you multiply by 4 to get 60? 4 ?u 5 60

7. What is the actored orm o 18 x 21 69 x 1 60?

8. What is the length o the table? Check your answer.



18 x 21 69 x 1 60

3 x 1 4

binomials

3 x 1 4

the other factor

6 x

15

(3 x 1 4)(6 x 1 15)

length: (6 x 1 15)

Check: (3 x 1 4)(6 x 1 15) 5 18 x 2 1 69 x 1 60

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1. 2w 2 1 13w 1 15 2. 3d 2 1 20d 1 12 3. 4n2 1 62n 2 32

4. 3p2 2 7p 2 40 5. 6r 2 2 10r 2 24 6. 5z 2 2 17z 1 14

7. 14k 2 2 67k 1 63 8. 2m22 m 2 15 9. 3 x 2 1 9 x 2 84

10. 4 y 2 1 26 y 1 30 11. 5t 2 2 24t 2 5 12. 7c 2 2 2c 2 9

13. 8k 2 2 42k 1 27 14. 6 g 2 2 2 g 2 20 15. 2c 2 2 23c 1 11

16. Te area of a rectangular computer screen is 4 x 2 1 20 x 1 16. Te width of

the screen is 2 x 1 8. What is the length of the screen?

17. Te area of a rectangular granite countertop is 12 x 2 1 10 x 2 12. Te width of

the countertop is 2 x 1 3. What is the length of the countertop?

18. Te area of a rectangular book cover is 4 x 2 2 6 x 2 40. Te width of the book

cover is 2 x 2 8. What is the length of the book cover?

19. Te area of a rectangular parking lot is 21 x 2 2 44 x 1 15. Te width of the

parking lot is 3 x 2 5. What is the length of the parking lot?

Factor each expression completely.

20. 6 x 2 2 10 x 2 4 21. 6d 2 1 21d 1 15 22. 8n2 1 68n 1 84

23. 20p2

2 115p 2 30 24. 15r 2

1 141r 2 90 25. 12z 2

2 14z 1 4

26. 20k 2 1 110k 1 120 27. 9m22 66m 1 21 28. 40 x 2 2 136 x 2 96

29. 42 y 2 1 28 y 2 14 30. 8t 2 2 16t 2 90 31. 24c 2 1 96c 1 90

8-6Practice Form G


(2w 1 3)(w 1 5)

(3 p 1 8)( p 2 5)

(2k 2 7)(7k 2 9)

(5t 1 1)(t 2 5)

(4k 2 3)(2k 2 9)

(3d 1 2)(d 1 6)

2(3r 1 4)(r 2 3)

(2m 1 5)(m 2 3)

2(2 y 1 3)( y 1 5)

2(3g 1 5)(g 2 2)

2 x 1 2

6 x 2 4

2 x 1 5

7 x 2 3

2(3 x 1 1)( x 2 2)

5(4 p 1 1)( p 2 6)

10(2k 1 3)(k 1 4)

14(3 y 2 1)( y 1 1)

3(2d 1 5)(d 1 1)

3(5r 2 3)(r 1 10)

3(3m 2 1)(m 2 7)

2(2t 1 5)(2t 2 9)

4(2n 1 3)(n 1 7)

2(2 z 2 1)(3 z 2 2)

8(5 x 1 3)( x 2 4)

6(2c 1 5)(2c 1 3)

2(2n 2 1)(n 1 16)

(5 z 2 7)( z 2 2)

3( x 1 7)( x 2 4)

(7c 2 9)(c 1 1)

(2c 2 1)(c 2 11)

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Open-Ended Find two diferent values that complete each expression so that

the trinomial can be actored into the product o two binomials. Factor your

trinomials.

32. 4 x 2 1u x 1 12 33. 6t 2 2ut 2 4 34. 9m22 m 1 8

35. 8n2 1 n 2 10 36. 12v 2 2 v 1 15 37. 5w 2 2 w 2 24

38. Error Analysis Describe and correct the

error made in factoring the expression at

the right.

39. A parallelogram has an area of 4 x 2 1 7 x 2 15. Te base of the parallelogram

is x 1 3

. What is the height of the parallelogram?a. Write the formula for the area of a parallelogram.

b. Writing Explain how factoring the trinomial helps you solve the problem.

40. A rectangular window pane has an area of 15 x 2 2 19 x 1 6. Te width of the

window pane is 3 x 2 2. What is the length of the window pane?


41. 28 y 2 1 43 y 2 48 42. 16z 2 2 54z 1 35 43. 27n2 2 54n 1 15

44. 36p2 1 63p 1 20 45. 28r 2 2 20r 2 33 46. 30z 2 2 53z 1 12

47. 32 x 3 1 28 x 2 1 5 x 48. 25p2 1 20pq 2 12q2 49. 72 g 2h 2 43 gh 1 6h



(6x2 + 3x – 9) = 3(2x2 + x – 3)= 3(2x2 - 3x + 2x – 3)

= 3(2x2 - 3x + (2x – 3)= 3[ x (2x - 3) + 1 (2x – 3) ]

= 3(x + 1) (2x – 3)

Answers may vary. Sample:19, 16: (4 x 1 3)( x 1 4);(4 x 1 4)( x 1 3)

Answers may vary. Sample:11, 211; (8n 2 5)(n 1 2);

(n 2 2)(8n 1 5)

Answers may vary. Sample:23, 25; (6t 1 1)(t 2 4);

(3t 1 4)(2t 2 1)

Answers may vary. Sample:29, 27; (4v 2 3)(3v 2 5);

(4v 2 5)(3v 2 3)

Answers may vary. Sample:73, 27; (9m 2 1)(m 2 8);

(3m 2 8)(3m 2 1)

Answers may vary. Sample.26, 14; (5w 1 4)(w 2 6);

(5w 1 6)(w 2 4)

In the second step, the student wrote21 x instead of 1 x . x should be writtenas 3 x 2 2 x . Answer: 3(2 x 1 3)( x 2 1)

A 5 bh

Factor to find h: ( x 1 3)(4 x 2 5) 5 4 x 2 1 7 x 2 15; h 5 4 x 2 5

5 x 2 3

(4 y 2 3)(7 y 1 16)

(3 p 1 4)(12 p 1 5)

x (4 x 1 1)(8 x 1 5)

(8 z 2 7)(2 z 2 5)

(2r 2 3)(14r 1 11)

(5 p 2 2q)(5 p 1 6q)

3(3n 2 1)(3n 2 5)

(2 z 2 3)(15 z 2 4)

h(9g 2 2)(8g 2 3)

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8-6Practice Form K



1. 3n2 2 8n 2 3 2. 5a2 2 22a 1 8 3. 2s2 1 13s 1 6

4. 6t 2 1 21t 2 12 5. 9b2 2 65b 1 14 6. 5z 2 1 11z 1 6

7. 7r 2 2 9r 2 10 8. 2m21 m 2 21 9. 3 g 2 1 20 g 1 32

10. Te area o a rectangular driveway is 2 x 2 1 15 x 1 25. Te width o the

driveway is x 1 5. What is the length o the driveway?

11. Te area o a rectangular foor is 8 x 2 1 6 x 2 20. Te width o the foor is

2 x 1 4. What is the length o the foor?

12. Te area o a rectangular desktop is 6 x 2 2 3 x 2 3. Te width o the desktop is

2 x 1 1. What is the length o the desktop?


13. 24n2 1 2n 2 12 14. 72q2 2 12q 2 40 15. 30 j 2 2 27 j 2 21

16. 60h2 1 280h 1 45 17. 40a2 1 126a 1 44 18. 45 f 2 1 24 f 2 189

2(4n 1 3)(3n 2 2) 4(3q 1 2)(6q 2 5) 3(2 j 1 1)(5 j 2 7)

5(6h 1 1)(2h 1 9) 2(4a 1 11)(5a 1 2) 3(5f 2 9)(3f 1 7)

2 x 1 5

4 x 2 5

3 x 2 3

(3n 1 1)(n 2 3) (5a 2 2)(a 2 4) (2 s 1 1)( s 1 6)

3(2t 2 1)(t 1 4) (9b 2 2)(b 2 7) (5 z 1 6)( z 1 1)

(7r 1 5)(r 2 2) (2m 1 7)(m 2 3) (3g 1 8)(g 1 4)

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Open-Ended Find two diferent values that complete each expression so that

the trinomial can be actored into the product o two binomials. Factor your

trinomials.

19. 4n2 1un 2 3 20. 12r 2 1u 1 6

21. 24a2 1 a 2 15 22. 18b2 1 b 1 8

23. A parallelogram has an area o 8 x 2 2 2 x 2 45. Te height o the

parallelogram is 4 x 1 9.

a. Write the ormula or the area o a parallelogram.

b. What is the length o the base o the parallelogram?

c. Writing Explain how you solved the problem.

24. A rectangular athletic feld has an area o 40 x 2 1 190 x 2 50. Te width o the

athletic feld is 8 x 2 2. What is the length o the athletic feld?


25. 96d 2 2 76d 2 77 26. 48h2 2 86h 1 35

27. 24m21 18m 2 15 28. 36c 2 1 27c 2 55

2 x 2 5

A 5 bh

Sample: You know that the product of 4 x 1 9 and another factor is 8 x 2 2 2 x 2 45.

4 x times 2 x is 8 x 2 and 9 times 25 is 245. So, 8 x 2 2 2 x 2 45 5 (4 x 1 9)(2 x 2 5). Thenuse FOIL to check.

5 x 1 25

(12d 1 7)(8d 2 11) (8h 2 5)(6h 2 7)

3(2m 2 1)(4m 1 5) (3c 1 5)(12c 2 11)

Answers may vary. Sample:

24, 11; 4n22 4n 2 3 5 (2n 2 3)(2n 1 1);

4n21 11n 2 3 5 (n 1 3)(4n 2 1)

Answers may vary. Sample:

17, 38; 12r 2 1 17r 1 6 5 (3r 1 2)(4r 1 3);

12r 2 1 38r 1 6 5 (2r 1 6)(6r 1 1)

Answers may vary. Sample: 218, 37;

24a22 18a 2 15 5 (6a 1 3)(4a 2 5);

24a21 37a 2 15 5 (8a 1 15)(3a 2 1)

Answers may vary. Sample: 24, 74;

18b21 24b 1 8 5 (3b 1 2)(6b 1 4);

18b21 74b 1 8 5 (9b 1 1)(2b 1 8)

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Multiple Choice


1. What is the factored form of 4 x 21 12 x 1 5?

A. (2 x 1 4)(2 x 1 3) B. (4 x 1 5)( x 1 1) C. (2 x 1 1)(2 x 1 5) D. (4 x 1 1)( x 1 5)

2. What is the factored form of 2 x 21 x 2 3 ?

F. (2 x 1 3)( x 2 1) G. (2 x 1 1)( x 2 3) H. (2 x 2 3)( x 1 1) I. (2 x 2 1)( x 1 3)

3. Te area of a rectangular swimming pool is 10 x 22 19 x 2 15. Te length of

the pool is 5 x 1 3. What is the width of the pool?

A. 2 x 2 18 B. 2 x 2 5 C. 5 x 2 5 D. 5 x 2 22


F. 4(2 x 2 2)(2 x 1 2)

G. 4(4 x 2 6)( x 1 2)

H. 4(2 x 2 2)(2 x 1 3)

I. 4(2 x 2 3)(2 x 1 1)


A. 3( x 1 8)( x 2 1)

B. 3( x 1 6)( x 1 1)

C. 3( x 1 5)( x 2 3)

D. 3( x 1 7)( x 2 3)

Short Response

6. Te perimeter around a dog’s running space is 20 x 21 28 x 1 8. Te length

of the dog’s running space is 10 x 1 4. What is the width of the dog’s running

space? Show why your answer is correct.



C

F

B

I

A

[2] Correct expression written with all work shown

[1] Expression written with minor calculation error or inadequatework shown


2 x 1 2

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You can use a unction to estimate the volume o an adult body based on the

length o one part, such as the length o an index fnger, x . Start by using x to

calculate the volume o an index fnger. Assume the ratio o the length to height to width o an average index fnger is 7 : 1 : 2. Tereore, the volume is

249 x

3.

You can then estimate that approximately 10 index fngers make up one hand.

Multiply the volume o one index fnger by 10 to fnd the volume in one hand:2049 x

3. Use this more convenient hand measure to fgure out how many hands

make up each large body area.

Hand = 1 hand Arm ≈ 12 hands Head ≈ 12 hands Neck ≈ 8 hands

orso ≈ 100 hands Leg ≈ 45 hands Foot ≈ 3 hands

Add up all the parts, making sure to double the hands, arms, and legs:

2 1 24 1 12 1 8 1 100 1 90 1 6 5 242 hands

Now multiply the number o hands by the volume in one hand:

242Q2049R x 3 5

484049 x

35 98

3849 x 3

Now that you have a unction or the volume o a human body, you can use it to

fnd expressions or other body parts without measuring.

1. Use the unction V 5 983849 x 3 , the volume o an adult body to write an

expression or the length o the oot in an adult body. where the ratio o thelength to height to width o the oot is 6 : 1 : 1.

2. Use the same unction to write an expression or the length o an arm in

an adult body where the ratio o the length to height to width o the arm is 10 : 1 : 1.

3. Measure the lengths o three people’s index fngers, eet, and arms. How do

the results compare to your estimates?

8-6Enrichment


6049 x

24049 x

Check students’ work.

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You can use your knowledge o prime numbers to help you actor some trinomials as two

binomials. A prime number has only 1 and itsel as actors. For trinomials o the orm

ax 2 1 bx 1 c , i a is a prime number then you already know the frst term o eachbinomial: ax and 1 x . Ten list the actors that will multiply to produce c . Use guess and check

to fnd the actor pair that will add to b.

Problem

What is the actored orm o 7 x 2 1 31 x 1 12?

7 x 2 1 31 x 1 12 5 (7 x )(1 x ) a is 7, which is prime, so the factors are 7 and 1.

5 (7 x )( x ) You don’t need the 1 in front of the variable, so drop it.

7 x 2 1 31 x 1 12 5 (7 x 1 )( x 1 ) The trinomial has two plus signs, so the

binomials also have plus signs.

Because c is 12, fnd actor pairs that multiply to 12: (1 and 12), (2 and 6), (3 and 4).

ry each pair in the expression to see i the INNER and OUER products add to b, or 31.

(7 x 1 1)( x 1 12) 5 7 x 2 1 x 1 84 x 5 7 x 2 1 85 x 1 12 (NO)

(7 x 1 2)( x 1 6) 5 7 x 2 1 2 x 1 42 x 5 7 x 2 1 44 x 1 12 (NO)

(7 x 1 3)( x 1 4) 5 7 x 2 1 3 x 1 28 x 5 7 x 2 1 31 x 1 12 (YES)

Te actored orm o 7 x 2 1 31 x 1 12 is (7 x 1 3)( x 1 4).

Exercises


1. 3 x 2 1 14 x 1 8 2. 5 y 2 1 43 y 1 24 3. 2z 2 1 19z 1 42

4. 11a21 39a 1 18 5. 13b2

1 58b 1 24 6. 23c 2 1 56c 1 20

7. 7d 2 1 d 2 8 8. 3e 2 1 20e 2 32 9. 19 f 21 10 f 2 9

10. 5s22 18s 1 16 11. 17t 2 2 12t 2 5 12. 29u2

1 48u 2 20

8-6Reteaching


(3 x 1 2)( x 1 4)

(11a 1 6)(a 1 3)

(7d 1 8)(d 2 1)

(5 s 2 8)( s 2 2)

(5 y 1 3)( y 1 8)

(13b 1 6)(b 1 4)

(3e 2 4)(e 1 8)

(17t 1 5)(t 2 1)

(2 z 1 7)( z 1 6)

(23c 1 10)(c 1 2)

(19f 2 9)(f 1 1)

(29u 2 10)(u 1 2)

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I you are given the area and one side o a rectangle, you can fnd the second side by

actoring the trinomial. One binomial is the width and the other binomial is the length.

Problem

Te area o a rectangular swimming pool is 6 x 21 11 x 1 3. Te width o the pool

is 2 x 1 3. What is the length o the pool?

You are given the area and length o the pool. Set up an equation with what you

are given and solve or length.

6 x 21 11 x 1 3 5 (2 x 1 3)(uuu ) Area = length × width.

6 x 21 11 x 1 3 5 (2 x 1 3)(3 x uu ) 6 x

25 (2 x )(3 x ), so the first term of the second

binomial is 3 x .

6 x 21 11 x 1 3 5 (2 x 1 3)(3 x 1 u ) The trinomial has two plus signs, so the sign for

the second binomial must also be plus.

6 x 21 11 x 1 3 5 (2 x 1 3)(3 x 1 1) The value of c is 3. Since 3 5 3 3 1, the second

term must be 1.

Multiply to check your answer. Use FOIL.

(2 x 1 3)(3 x 1 1) 5 6 x 21 2 x 1 9 x 1 3 5 6 x

21 11 x 1 3 3

Te length o the swimming pool is 3 x 1 1.

Exercises

13. Te area o a rectangular cookie sheet is 8 x 21 26 x 1 15. Te width o the

cookie sheet is 2 x 1 5. What is the length o the cookie sheet?

14. Te area o a rectangular lobby oor in the new o ce building is

15 x 21 47 x 1 28. Te length o one side o the lobby is 5 x 1 4. What is the

width?

15. Te area o a rectangular school banner is 12 x 21 13 x 2 90. Te width o the

banner is 3 x 1 10. What is the length o the banner?

16. Te distance a train has traveled is 6 x 22 23 x 1 20. Te train’s average speed

is 3 x 2 4. How long has the train been traveling?



4 x 1 3

3 x 1

7

4 x 2 9

2 x 2 5

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8-7 ELL Support

Factoring Special Cases

Complete the vocabulary chart by flling in the missing inormation.

Word or Word Phrase Definition Picture or Example

difference of twosquares

A binomial in which a perfect square

monomial is subtracted from another

perfect square monomial

x 2 2 16

factoring adifference of two

squares

1. x 2 2 25 5 ( x 1 5)( x 2 5)

factoringperfect-squaretrinomials

For every real number a and b:

a21 2ab 1 b2

5

(a 1 b)(a 1 b) 5 (a 1 b)2 or

a22 2ab 1 b2

5

(a 2 b)(a 2 b) 5 (a 2 b)2.

2.

perfect-squaretrinomial

3. 9 x 2 1 24 x 1 16Any trinomial of the form

a21 2ab 1 b2 or

a22 2ab 1 b2 is a

perfect-square trinomial becauseit is the result of squaring abinomial.

To factor the difference of two

squares a2 and b2, multiply thesum of the two factors a and b bythe difference of the two factorsa and b.

4 x 2 2 20 x 1 25 5

(2 x 2 5)(2 x 2 5)

a22 b2

5 (a 1 b)(a 2 b)

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8-7 Think About a Plan


Interior Design A square rug has an area o 49 x 22 56 x 1 16. A second square

rug has an area o 16 x 21 24 x 1 9. What is an expression that represents the

diference o the areas o the rugs? Show two diferent ways to nd the solution.

1. What are two methods you could use to solve this problem?

2. How would you nd the diference without actoring?

3. What polynomial do you get when you use this method?

4. Can you actor that polynomial?

5. How could you use actoring to solve the problem?

6. What do the shape o the rug and the polynomials tell you about how to actor

the polynomials or the area o the rugs?

7. Factor each trinomial.

49 x 22 56 x 1 16 5 (u 2 u ) (u 2 u ) 5 (u u u )2

16 x 21 24 x 1 9 5 (u 1 u ) (u 1 u ) 5 (u u u )2

8. Use your results rom Exercise 7 to write an expression or the diference in the areas.

9. Factor the expression rom Exercise 8 using the diference o two squares.

Simpliy the expressions within each set o parentheses.

10. Do the two methods give you the same result?

subtraction; factoring before subtracting

subtract the polynomials

33 x 2 2 80 x 1 7

yes ; (11 x 2 1)(3 x 2 7)

(7 x 2 4)(7 x 2 4) 5 (7 x 2 4)2

(4 x 1 3)(4 x 1 3) 5 (4 x 2 3)2

(7 x 2 4)2 2 (4 x 1 3)2

f(7 x 2 4) 1 (4 x 1 3)g f(7 x 2 4) 2 (4 x 1 3)g 5 (11 x 2 1)(3 x 2 7)

yes

Factoring gives you a second way to

find the difference. You can represent the difference in the form a22 b2.

The factors of each square polynomial will be

the same.

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8-7 Practice Form G



1. h2 1 10h 1 25 2. v 2 2 14v 1 49 3. d 2 2 22d 1 121

4. m21 4m 1 4 5. q2 1 6q 1 9 6. p2 2 24p 1 144

7. 36 x 2 1 60 x 1 25 8. 64 x 2 1 48 x 1 9 9. 49n2 1 14n 1 1

10. 16s2 2 72s 1 81 11. 25r 2 2 80r 1 64 12. 9 g 2 2 24 g 1 16

13. 81w 2 1 144w 1 64 14. 16e 2 2 88e 1 121 15. 25 j 2 1 100 j 1 100

16. 144 f 2 2 24 f 1 1 17. 4a2 2 36a 1 81 18. 49d 2 2 84d 1 36

Te given expression represents the area. Find the side length of the square.

19. 20. 21.

22. 23. 24.


made in factoring the expression at the right.

64 x 2 + 80 x + 25 9 y 2 - 24 y + 16 4t 2 + 36t + 81

36n2 + 84n + 49 100w 2 + 20w + 1 16 s2 + 104 s + 169

175x2 - 28 = 7(25x2 - 4)

= 7(5x - 2)(5x - 2)

= 7(5x - 2)2

(h 1 5)2

(m 1 2)2

(6 x 1 5)2

(4 s 2 9)2

(9w 1 8)2

(12f 2 1)2

8 x 1 5

6n 1 7

(25 x 2 2 4) factors to (5 x 2 2)(5 x 1 2), not

(5 x 2 2)2

(v 2 7)2

(q 1 3)2

(8 x 1 3)2

(5r 2 8)2

(4e 2 11)2

(2a 2 9)2

3 y 2 4

10w 1 1

(d 2 11)2

( p 2 12)2

(7n 1 1)2

(3g 2 4)2

(5 j 1 10)2

(7d 2 6)2

2t 1 9

4 s 1 13

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8-7 Practice (continued) Form G



26. m22 49 27. c 2 2 100 28. p2 2 16

29. 4a2 2 25 30. 64n2 2 1 31. 25 x 2 2 144

32. 50 g 2 2 8 33. 8d 2 2 8 34. 27 x 2 2 48

35. 24e 2 2 54 36. 245k 2 2 20 37. 112h2 2 63

38. 48 x 2 1 72 x 1 27 39. 8b2 1 80b 1 200 40. 48w 2 1 48w 1 12

41. 45s2 2 210s 1 245 42. 45t 2 2 72t 1 24 43. 100z 2 2 120z 1 36

44. Writing Explain how to recognize a perect-square trinomial.

45. a. Open-Ended Write an expression that shows the actored orm o a

diference o two squares.

b. Explain how you know that your expression is a diference o two squares.


46. 36s8 2 60s4 1 25 47. c 10 2 30c 5d 2 1 225d 4 48. 25n6 1 40n3 1 16

Mental Math For Exercises 49–51, fnd a pair o actors or each number by

using the dierence o two squares.

49. 24 50. 28 51. 72

52. Reasoning Explain how reversing the rules or multiplying squares o

binomials can help you actor a perect-square trinomial.

53. Writing Te area o a square parking lot is 49p4 2 84p2 1 36. Explain how

you would nd the length o the parking lot.

(m 1 7)(m 2 7)

(2a 1 5)(2a 2 5)

2(5g 1 2)(5g 2 2)

6(2e 1 3)(2e 2 3)

3(4 x 1 3)2

5(3 s 2 7)2

(c 5 2 15d 2)2 (5n31 4)2(6 s4

2 5)2

24 5 522 12

5 (5 1 1)(5 2 1) 5 (6)(4)

28 5 822 62

5 (8 2 6)(8 1 6) 5 (2)(14)

72 5 922 32

5 (9 1 3)(9 2 3) 5 (12)(6)

When the b term in a trinomial is exactly twice the product of a and c , you can

factor it as (a 1 b)2 or as (a 2 b)2.

Factor 49 p42 84 p2

1 36 to find the length. You get (7 p22 6)2 so each side has a

length of (7 p22 6).

(c 1 10)(c 2 10)

(8n 1 1)(8n 2 1)

8(d 1 1)(d 2 1)

5(7k 1 2)(7k 2 2)

8(b 1 5)2

3(15t 2 2 24t 1 8)

( p 1 4)( p 2 4)

(5 x 1 12)(5 x 2 12)

3(3 x 1 4)(3 x 2 4)

7(4h 1 3)(4h 2 3)

12(2w 1 1)2

4(5 z 2 3)2

The coefficient of the squared term and the constant will be perfect squares. Twice

the product of these numbers is the coeffiecient of the middle term.The sign before

the constant will be positive.

Answers may vary. Sample: (2 x 1 3)(2 x 2 3)

Answers may vary. Sample: 4 x 2 2 9; 4 x 2 and 9 are squares and they are

separated by a subtraction.

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8-7 Practice Form K



1. c 2 1 2c 1 1 2. d 2 2 10d 1 25 3. p2 2 24p 1 144

4. w 2 1 14w 1 49 5. s2 1 16s 1 64 6. 9 g 2 1 24 g 1 16

7. 25m22 60m 1 36 8. 4q2 2 32q 1 64 9. 49 y 2 2 84 y 1 36

10. 121n2 2 66n 1 9 11. 81 x 2 2 18 x 1 1 12. 100t 2 2 100t 1 25

Te given expression represents the area. Find the side length of the square.

13. 14.

15. 16.

17. Writing How can you tell that x

22 19

x 1 90

is not a perfect squaretrinomial?

36w 2 à 12w à 1 81w

2 Ľ 72w à 16

9w 2 Ľ 48w à 64 121w

2 Ľ 66w à 9

(c 1 1)2 (d 2 5)2 ( p 2 12)2

(w 1 7)2 ( s 1 8)2 (3g 1 4)2

(5m 2 6)2 4(q 2 4)2 (7 y 2 6)2

(11n 2 3)2 (9 x 2 1)2

6w 1 1 9w 2 4

3w 2 8 11w 2 3

Sample: 90 is not a perfect square.

25(2t 2 1)2

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8-7 Practice (continued) Form K



18. b22 121 19. d 2 2 81 20. f 22 625

21. 108 x 2 2 3 22. 50n22 8 23. 405z 2 2 245

24. 216h22 150 25. 28 y 2 2 28 26. 50t 2 1 40t 1 8

27. 12n22 36n 1 27 28. 180a2

2 300a 1 125 29. 250k 2 2 200k 1 40

30. Writing Explain how to recognize a diference o two squares.

31. a. Open-Ended Write an expression that shows the actored orm o a

perect-square trinomial.

b. Explain how you know your expression is a perect-square trinomial when

expanded.

Mental Math For Exercises 32–34, fnd a pair o actors or each number by

using the dierence o two squares.

32. 84 33. 55 34. 80

35. Writing Te area o a square painting is 225 x 4 1 240 x 2 1 64. Explain how

you would nd a possible length o one side o the painting.

(14)(6) (11)(5) (20)(4)

Answers may vary. Sample: (5 x 1 3)(5 x 1 3) or (5 x 1 3)2

It is in the form a21 2ab 1 b2.

The expression is the difference of two terms that are both perfect squares.

Since the trinomial is a perfect-square trinomial, the length of theside could be a factor of the trinomial.

(b 1 11)(b 2 11) (d 1 9)(d 2 9) (f 1 25)(f 2 25)

3(6 x 1 1)(6 x 2 1) 2(5n 1 2)(5n 2 2) 5(9 z 1 7)(9 z 2 7)

6(6h 1 5)(6h 2 5) 28( y 1 1)( y 2 1) 2(5t 1 2)(5t 1 2)

3(2n 2 3)(2n 2 3) 5(6a 2 5)(6a 2 5) 10(5k 2 2)(5k 2 2)

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Multiple Choice


1. What is the actored orm o q22 12q 1 36?

A. (q 1 6)(q 2 6) B. (q 2 6)(q 2 6) C. (q 2 9)(q 1 4) D. (q 1 4)(q 1 9)

2. What is the actored orm o 9 x 2 1 12 x 1 4?

F. (3 x 1 2)2 G. (3 x 1 3)2 H. (3 x 2 2)2 I. (3 x 2 3)2

3. What is the actored orm o x 2 2 196?

A. ( x 2 14)2 B. ( x 1 14)2 C. ( x 2 28)(4 x 1 7) D. ( x 2 14)( x 1 14)

4. What is the actored orm o 9 x 2 2 64?

F. (3 x 2 8)2 G. (3 x 1 8)2 H. (3 x 2 8)(3 x 1 8) I. (9 x 2 8)( x 1 8)

5. What is the actored orm o 12m22 75?

A. 3(2m 2 5)2 B. 3(2m 1 5)(2m 2 5) C. 3(2m 1 5)2 D. (6m 2 25)(2m 1 3)

6. What is the actored orm o 49 x 2 2 56 x 1 16?

F. (7 x 2 4)2 G. (7 x 1 4)(7 x 2 4) H. (7 x 1 4)2 I. (7 x 2 8)2

Extended Response

7. A our-sided building has an area o 36 x 2 1 48 x 1 16. Explain how to fnd a

possible length and width o the building. What is a possible shape o the

building?

8-7 Standardized Test Prep


B

F

D

H

B

F

(6 x 1 4)2 ; The length and width could be the same, so the shape is a square.

[4] Answer correctly factors the polynomial and indicates the building could bea square with sides 6 x 1 4. Complete explanation is provided.

[3] Minor calculation error in the answer or incomplete explanation

[2] Polynomial correctly factored but not related to length and width of thebuilding

[1] Some steps in solution of problem completed correctly


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8-7 Enrichment


Te surface area of a cube is determined by the formula SA 5 6s2, where s is the length of a

side of the cube. You can use this formula to analyze a polynomial that represents the surface

area of a cube.

Start by dividing the polynomial by 6. Tis will leave an expression for of the area of one face

of the cube. You can see that the area is a perfect-square trinomial. Reverse the rules for

multiplying squares of binomials to factor the trinomial.

For example, a cube with a surface area of 24 x 2 1 24 x 1 6 has a side measure of 2 x 1 1.

6s2 5 24 x 2 1 24 x 1 6

s2 524 x 2 1 24 x 1 6

6 5 4 x 2 1 4 x 1 1

s2 5 (2 x 1 1)(2 x 1 1)

s 5 2 x 1 1

Te surface area of a rectangular prism with two square faces is determined by the formula

SA 5 4ls 1 2s2, where l is the length and s is the measure of the side of the square face. If

you are given the surface area and the area of the square face, you can determine the

dimensions of the rectangular prism.

Suppose a rectangular prism has a surface area of 24 x 1 30 and each square face measures

9 cm2.

24 x 1 30 2 18 5 24 x 1 12 Subtract the area of the square faces.

24 x 1 12

4 5 6 x 1 3 Divide by 4 to get the area of each remaining side.

6 x 1 3

3 5 2 x 1 1 Divide by the side length of the square base, or the square

root of the base’s area.

1. Te surface area of a cube is 96 x 2 1 144 x 1 54. What is the measure of each side?



4. Te surface area of a rectangular prism is 100 x 1 90. Te areas of the two square faces of

the prism are 25 m2 each. What are the dimensions of the rectangular prism?

5. Te surface area of a rectangular prism is 2 x 2 1 48 x 1 88. Te areas of the two square

faces of the prism are x 2 1 4 x 1 4 each. What are the dimensions of the rectangular

prism?

4 x 1 3

3 x 2 1

12 x 1 5

5, 5, and 5 x 1 2

10, x 1 2, and x 1 2

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8-7 Reteaching


Te area of a square is given by A 5 s2, where s is a side length. When

the side length is a binomial, the area can be written as a perfect-square

trinomial. If you are given the area of such a square, you can usefactoring to write an expression for a side length.

Problem

A mosaic is made of small square tiles called tesserae. Suppose the area of one

tessera is 9 x 2 1 12 x 1 4. What is the length of one side of a tessera?

Because the tile is a square, you know the side lengths must be equal. Terefore,

the binomial factors of the trinomial must be equal.

9 x 2 1 12 x 1 4 5 ( u u u )2 This is a perfect square trinomial and can be factored as the

square of a binomial.9 x 2 5 (3 x )2 9 x

2 and 4 are perfect squares. Write them as squares.

4 5 22

2(3 x )(2) 5 12 x Check that 12 x is twice the product of the first and last terms. It

is, so you are sure that you have a perfect-square trinomial.

9 x 2 1 12 x 1 4 5 (3 x 1 2)2 Rewrite the equation as the square of a binomial.

Multiply to check your answer.

(3 x 1 2)(3 x 1 2) 5 9 x 2 1 6 x 1 6 x 1 4 5 9 x 2 1 12 x 1 43

Te length of one side of the square is 3 x 1 2.

ExercisesFactor each expression to fnd the side length.

1. Te area of a square oil painting is 4 x 2 1 28 x 1 49. What is the length of one

side of the painting?

2. You are installing linoleum squares in your kitchen. Te area of each linoleum

square is 16 x 2 2 24 x 1 9. What is the length of one side of a linoleum square?

3. You are building a table with a circular top. Te area of the tabletop is

(25 x 2 2 40 x 1 16)π. What is the radius of the tabletop?

4. A fabric designer is making a checked pattern. Each square in the pattern has

an area of x 2 2 16 x 1 64. What is the length of one side of a check?

s

s A 5 s 2

2 x 1 7

4 x 2 3

5 x 2 4

x 2 8

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8-7 Reteaching (continued)


Some binomials are a diference o two squares. o actor these expressions, write

the actors so the x -terms cancel and you are let with two perect squares.

Problem

What is the actored orm o 4 x 22 9?

4 x 22 9 5 ( u 1 u )( u 2 u ) Both 4 x

2 and 9 are perfect squares. You know the signs of

the factors will be opposite, so the x -terms will cancel out.

" 4 x 25 2 x Find the square root of each term.

! 9 5 3

(2 x 1 3)(2 x 2 3) Write each term as a binomial with opposite signs, so the

x -terms will cancel out.


(2 x 1 3)(2 x 2 3) 5 4 x 21 6 x 2 6 x 2 9

5 4 x 22 93

Te actored orm o 4 x 22 9 is (2 x 1 3)(2 x 2 3).

ExercisesFactor each expression.

5. 9 x

22

4 6. 25 x

22

49 7. 144 x

22

1

8. 64 x 22 25 9. 49 x

22 16 10. 36 x

22 49

11. 81 x 22 16 12. 16 x

22 121 13. 25 x

22 144

14. 16 x 2 2 9 15. x 2 2 81 16. 4 x 2 2 49

(3 x 1 2)(3 x 2 2)

(8 x 1 5)(8 x 2 5)

(9 x 1 4)(9 x 2 4)

(4 x 1 3)(4 x 2 3)

(5 x 1 7)(5 x 2 7)

(7 x 1 4)(7 x 2 4)

(4 x 1 11)(4 x 2 11)

( x 1 9)( x 2 9)

(12 x 1 1)(12 x 2 1)

(6 x 1 7)(6 x 2 7)

(5 x 1 12)(5 x 2 12)

(2 x 1 7)(2 x 2 7)

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8-8ELL Support

Factoring by Grouping

Use the list to complete the diagram.

common factors factor GCF pair of binomial factors squares

Steps for Factoring aPolynomial Completely

1. Factor out the .

2. If the polynomial has two or three terms,

look for a difference of two ______________ ,

a perfect-square trinomial, or a

____________________.

3. If the polynomial has four or more terms,

group terms and __________ to find commonbinomial factors.

4. Make sure there are no ______________ other

than 1.

GCF

squares

common factors

factor

pair of binomial factors

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Art Te pedestal o a sculpture is a rectangular prism with a volume o

63 x 32 28 x . What expressions can represent the dimensions o the pedestal? Use

actoring.

KNOW

1. Te pedestal o the sculpture is shaped like a .

2. Te volume o the pedestal is .

3. Te ormula you can use to fnd the dimensions o the pedestal is .

NEED

4. o solve the problem you need to fnd

PLAN

5. Factor out the GCF rom the volume o the pedestal.

6. What type o expression is o the remaining expression?

7. Factor the expression completely.

8. What expressions represent possible dimensions o the pedestal?

rectangular prism

63 x 3 2 28 x

V 5 lwh

3 factors

7 x (9 x 2 2 4)

difference of two squares

7 x (3 x 2 2)(3 x 1 2)

7 x , (3 x 2 2), and (3 x 1 2)

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8-8Practice Form G


Find the GCF o the frst two terms and the GCF o the last two terms or each

polynomial.

1. 12 x 3 1 3 x 2 1 20 x 1 5 2. 6v 3 1 42v 2 1 5v 1 35

3. 8t 3 1 36t 2 1 2t 1 9 4. 10s3 1 35s2 1 6s 1 21

5. 9m32 6m2

1 12m 2 8 6. 8w 3 1 6w 2 2 28w 2 21

7. 7r 3 1 16r 2 2 9r 2 72 8. 21 x 3 2 28 x 2 2 6 x 1 8


9. 8 j 3 1 4 j 2 1 10 j 1 5 10. 2m31 8m2

1 9m 1 36

11. 10s3 1 25s2 1 8s 1 20 12. 6 x 3 1 9 x 2 1 2 x 1 3

13. 21 x 3

1 6 x 2

2 28 x 2 8 14. 8w 3

1 12w 2

1 10w 1 15

15. 18r 3 2 12r 2 1 21r 2 14 16. 36n3 2 27n2 2 8n 1 6

17. 110b3 1 77b2 2 60b 2 42 18. 64d 3 2 40d 2 2 24d 1 15

19. 10s3 1 80s2 2 7s 2 56 20. 25 j 3 1 15 j 2 2 5 j 2 3

21. 24c 3 2 84c 2 1 10c 2 35 22. 27 f 31 9 f 22 24 f 2 8

3 x 2, 5

4t 2, 1

r 2, 29

(4 j 2 1 5)(2 j 1 1)

(5 s21 4)(2 s 1 5)

(3 x 2 2 4)(7 x 1 2)

(6r 2 1 7)(3r 2 2)

(11b22 6)(10b 1 7)

(10 s22 7)( s 1 8)

(12c 2 1 5)(2c 2 7)

6v 2, 5

5 s2, 3

3m2, 4 2w 2, 27

7 x 2, 22

(2m21 9)(m 1 4)

(3 x 2 1 1)(2 x 1 3)

(4w 2 1 5)(2w 1 3)

(9n22 2)(4n 2 3)

(8d 2 2 3)(8d 2 5)

(5 j 2 2 1)(5 j 1 3)

(9f 22 8)(3f 1 1)

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Factor completely.

23. 32 x 3 1 8 x 2 1 48 x 1 12 24. 45w 4 2 36w 3 1 15w 2 2 12w

25. 32k 4 2 16k 3 1 12k 2 2 6k 26. 6 g 3 1 18 g 2 1 60 g 1 180

27. 30b4 2 45b3 2 10b2 1 15b 28. 32m31 72m2

2 80m 2 180

29. 63 j 4 1 84 j 3 2 18 j 2 2 24 j 30. 96n3 2 240n2 2 168n 1 420

31. 12e 4 1 18e 3 1 36e 2 1 54e 32. 60a5 2 72a4 2 210a3 1 252a2

Find linear expressions for the possible dimensions of each rectangular prism.

33. 34.

35. 36.

37. A shipping box in the shape of a rectangular prism has a volume of

12 x 3 1 32 x 2 1 20 x . What linear expressions can represent possible

dimensions of the box?

38. Error Analysis Describe and correct the error made in factoring completely.

39. Open-Ended Write a 3-term expression for the volume of a rectangular

prism that you can factor by grouping. Factor your polynomial.

16x4 + 24x3 + 64x2 + 96x = 4x(4x3 + 6x2 + 16x + 24)

= 4x[2x2 (2x + 3) + 8(2x + 3)]= 4x(2x2 + 8)(2x + 3)

V = 32 p3 - 224 p2 + 360 p

V = 18d 3 + 84d 2 + 48d

V = 24 y 3 + 54 y 2 -15 y

V = 15 x 3 + 52 x 2 + 32 x

4(2 x 2 1 3)(4 x 1 1)

2k (8k 2 1 3)(2k 2 1)

5b(3b22 1)(2b 2 3)

3 j (7 j 2 2 2)(3 j 1 4)

6e(e21 3)(2e 1 3)

x , 5 x 1 4, 3 x 1 8

3 y , 4 y 2 1, 2 y 1 5

4 x , 3 x 1 5, x 1 1

In the first step, the GCF is 8 x , not 4 x .

Answers may vary. Sample: x 5 1 4 x 4 1 3 x 3 5 x 3( x 1 3)( x 1 1)

3w (3w 2 1 1)(5w 2 4)

6(g21 10)(g 1 3)

4(2m22 5)(4m 1 9)

12(4n22 7)(2n 2 5)

6a2(2a22 7)(5a 2 6)

6d , 3d 1 2, d 1 4

8 p, 2 p 2 5, 2 p 2 9

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8-8Practice Form K


Find the GCF o the frst two terms and the GCF o the last two terms or each

polynomial.

1. 6n3 1 3n2 1 10n 1 5 2. 12z 3 1 36z 2 1 4z 1 12

3. 9k 3 1 45k 2 1 2k 1 10 4. 11a3 1 33a2 1 8a 1 24

5. 2 f 3 1 5 f 2 2 4 f 2 10 6. 16d 3 2 24d 2 2 6d 1 9


7. 6 x 3 2 4 x 2 1 15 x 2 10 8. 5q3 2 40q2 2 4q 1 32

9. 28m31 7m2

2 8m 2 2 10. 3p3 1 5p2 1 9p 1 15

11. 18 y 3 2 6 y 2 2 63 y 1 21 12. 3t 3 2 18t 2 1 5t 2 30

13. 250c 3 2 250c 2 1 100c 2 100 14. 18 g 3 2 33 g 2 1 30 g 2 55

15. 88n3

1 77n2

2 72n 2 63 16. 50h3

2 40h2

1 60h 2 48

17. 24b3 2 96b2 2 14b 1 56 18. 54r 3 1 9r 2 2 6r 2 1

(2 x 2 1 5)(3 x 2 2) (5q22 4)(q 2 8)

(7m22 2)(4m 1 1) ( p2

1 3)(3 p 1 5)

3(2 y 2 2 7)(3 y 2 1) (3t 2 1 5)(t 2 6)

50(5c 2 1 2)(c 2 1) (3g21 5)(6g 2 11)

(11n22 9)(8n 1 7) 2(5h2

1 6)(5h 2 4)

2(12b22 7)(b 2 4) (9r 2 2 1)(6r 1 1)

3n2; 5 12 z 2; 4

9k 2; 2 11a2; 8

f 2; 22 8d 2; 23

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Factor completely.

19. 49s3 1 14s2 1 14s 1 4 20. 32h4 1 72h3 1 36h2 1 81h

21. 42z 4 2 48z 3 2 7z 2 1 8z 22. 60p3 1 48p2 1 25p 1 20

23. 26n4 2 14n3 1 91n2 2 49n 24. 40t 3 1 28t 2 2 30t 2 21

25. 45k 4 2 9k 3 1 10k 2 2 2k 26. 18b5 2 3b4 1 30b3 2 5b2


27. 28.

29. A storage bin in the shape o a rectangular prism has a volume o

10 x 3 1 9 x 2 1 2 x . What linear expressions can represent possible

dimensions o the bin?

30. Writing Describe the frst step to look or in actoring a cubic expression

containing our terms.

31. Open-Ended Write a 4-term expression that you can actor by grouping.

Factor your polynomial.

V â x 3 à x 2 Ľ6 x V â12a 3 à13a 2à3a

x ; (5 x 1 2) ; (2 x 1 1)

Check to see if you can factor a GCF from all four terms.

Answers may vary. Sample: 4 x 3 1 36 x 2 1 7 x 1 63 5 (4 x 2 1 7)( x 1 9)

(7 s21 2)(7 s 1 2) h(8h2

1 9)(4h 1 9)

z (6 z 2 2 1)(7 z 2 8) (12 p21 5)(5 p 1 4)

n(2n21 7)(13n 2 7) (4t 2 2 3)(10t 1 7)

k (9k 2 1 2)(5k 2 1) b2(3b21 5)(6b 2 1)

a by (3a 1 1) by (4a 1 3) x by ( x 1 3) by ( x 2 2)

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Multiple Choice


1. What is the GCF o the frst two terms o the polynomial 4 y 3 1 8 y 2 1 5 y 1 10?

A. 4 y B. 4 y 2 C. 4 y 3 D. 4

2. What is the actored orm o 4 x 3 1 3 x 2 1 8 x 1 6?

F. (2 x 2 1 3)(2 x 1 3)

G. (2 x 2 1 2)(2 x 1 3)

H. ( x 2 1 2)(2 x 1 3)

I. ( x 2 1 2)(4 x 1 3)

3. What is the actored orm o 9 x 4 2 6 x 3 1 18 x 2 2 12 x ?

A. 3 x ( x 2 2 2 x )( x 2 4)

B. 3 x ( x 2 2 2)(3 x 1 2)

C. 3 x ( x 2 1 2)(3 x 2 2)

D. 3 x (3 x 2 2 2 x )(6 x 2 4)

4. What is the actored orm o 20p31 40p2

1 15p 1 30?

F. 5(2p21 3)(p 1 2)

G. 5(2p21 6)(p 1 4)

H. 5(4p21 3)(p 1 2)

I. 5(4p21 8p)(3p 1 6)

5. A box in the shape o a rectangular prism has a volume o 9 x 3 1 24 x 2 1 12 x .

Which is not one o the possible dimensions? (Its dimensions are all linear

expressions with integer coe cients.)

A. 2 x 1 3 B. 3 x 1 2 C. 3 x D. x 1 2

Short Response

6. Te polynomial 3π x 31

24π x 21

48π x represents the volume o a cylinder.Te ormula or the volume o a cylinder with radius r and height h is

V 5 πr 2h.

a. Factor 3π x 3 1 24π x 2 1 48π x .

b. Write a linear expression or a possible radius o the cylinder. Explain.

B

I

C

H

A

3π x ( x 1 4)2

x 1 4 because that is the term that is squared

[2] Both parts answered correctly with full explanation

[1] One part answered correctly or both parts answered correctlywith incomplete explanation


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8-8Enrichment


Pascal’s triangle is named ater French

mathematician Blaise Pascal, but this special

number pattern had been studied in India,China, Persia, and Italy long beore Pascal.

o generate Pascal’s triangle, start with the

number 1 in Row 0. Each successive row has

a 1 at both ends. Add the numbers directly

above-let and above-right to fnd the new

value.

You can use Pascal’s triangle to quickly expand a binomial expression. Te exponent tells you

the row number to choose. Te numbers in the correct row are the coe cients to use in the

expansion.

o expand (a 1 b)4, look to Row 4. Te coe cients are 1, 4, 6, 4, 1. Expand the variables,

raising the frst variable to 4 and decreasing by one or each term. Raise the second variable

to 0 and increase by 1 at each new term. Multiply each term by the coe cients:

a4 1 4a3b 1 6a2b2 1 4ab3 1 b4.

You can also use Pascal’s triangle to actor polynomials that are expansions o binomial

expressions. Arrange the polynomial in standard orm. Check to see i the coe cients

correspond to a row in Pascal’s triangle. Work backwards to actor.

o actor 15 xy 2 2 y 3 1 125 x 3 2 75 x 2 y , frst rearrange the terms in standard orm:

125 x 3 2 75 x 2 y 1 15 xy 2 2 y 3. In expansions o binomial expressions the x -exponentsdecrease by one in every term and y -exponents increase by one. Since the frst and last terms

have exponents o 3, the binomial is raised to the third power. Find the cube root o 125 to fnd

the coe cient o x : 5. Te fnal term is negative and has a coe cient o one, so the expression

is (5 x 2 y )3. Expand the binomial to check your answer:

(5 x 2 y )3 5 1(5 x )3 1 3(5 x )2(2 y ) 1 3(5 x )(2 y )2 1 1(2 y )3 5 125 x 3 2 75 x 2 y 1 15 xy 2 2 y 3

Expand the binomial using Pascal’s triangle.

1. (4k 1 j )4

2. (7 x 2 y )7

Factor the polynomial using Pascal’s triangle. Ten expand the binomial to check

your answer.

3. 8a3 1 12a2b 1 6ab2 1 b3

4. 40 x 2 y 3 1 32 x 5 1 10 xy 4 1 80 x 4 y 1 y 5 1 80 x 3 y 2

5. 1215 x 4 y 2 1 135 x 2 y 4 1 729 x 6 1 18 xy 5 1 y 6 1 1458 x 5 y 1 540 x 3 y 3

Row 0 1

Row 1 1 1

Row 2 1 2 1

Row 3 1 3 3 1

Row 4 1 4 6 4 1

Row 5 1 5 10 10 5 1

Row 6 1 6 15 20 15 6 1

Row 7 1 7 21 35 35 21 7 1

Row 8 1 8 28 56 70 56 28 8 1

256k 4 1 256k 3 j 1 96k 2 j 2 1 16kj 3 1 j 4

(2a 1 b)3

(2 x 1 y )5

823,543 x 7 2 823,543 x 6 y 1 352,947 x 5 y 2 2 84,035 x 4 y 3 1 12,005 x 3 y 4 2 1029 x 2 y 5 1 49 xy 6 2 y 7

(3 x 1 y )6

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8-8Reteaching


You can actor some higher-degree polynomials by grouping terms and actoring

out the GCF to fnd the common binomial actor. Make sure to actor out a

common GCF rom all terms frst beore grouping.

Problem

What is the actored orm o 2b4 2 8b3 1 10b2 2 40b?

2b4 2 8b3 1 10b2 2 40b 5 2b(b3 2 4b2 1 5b 2 20) 2b is the GCF of all four terms. Factor out 2b

from each term.

5 2bfb2(b 2 4) 1 5(b 2 4)g Group terms into pairs and look for the GCF of

each pair. b2 is the GCF of the first pair, and 5is the GCF of the second pair.

5 2b(b2 1 5)(b 2 4) b 2 4 is the common binomial factor. Use theDistributive Property to rewrite the expression.


2b(b2 1 5)(b 2 4) 5 2b(b3 1 5b 2 4b2 2 20) Multiply b21 5 and b 2 4.

5 2b4 1 10b2 2 8b3 2 40b Multiply by 2b.

5 2b4 2 8b3 1 10b2 2 40b 3 Reorder the terms by degree.

Te actored orm o 2b4 2 8b3 1 10b2 2 40b is 2b(b2 1 5)(b 2 4).

Exercises

Factor completely. Show your steps.

1. 4 x 4 1 8 x 3 1 12 x 2 1 24 x 2. 24 y 4 1 6 y 3 1 36 y 2 1 9 y

3. 72z 4 1 48z 3 1 126z 2 1 84z 4. 2e 4 2 8e 3 1 18e 2 2 72e

5. 12 f 32 36 f 2 1 60 f 2 180 6. 16 g 4 2 56 g 3 1 64 g 2 2 224 g

7. 56m32 28m2

2 42m 1 21 8. 40n4 2 60n3 2 50n2 1 75n

9. 60 x 3 2 90 x 2 2 30 x 1 45 10. 12p5 1 8p4 1 18p3 1 12p2

11. 6r 3 1 9r 2 2 60r 12. 20s6 2 50s5 2 30s4

4 x ( x 2 1 3)( x 1 2)

6 z (4 z 2 1 7)(3 z 1 2)

12(f 21 5)(f 2 3)

7(4m22 3)(2m 2 1)

15(2 x 2 2 1)(2 x 2 3)

3r (2r 2 5)(r 1 4)

3 y (2 y 2 1 3)(4 y 1 1)

2e(e21 9)(e 2 4)

8g(g21 4)(2g 2 7)

5n(4n22 5)(2n 2 3)

2 p2(2 p21 3)(3 p 1 2)

10 s4(2 s 1 1)( s 2 3)

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Polynomials can be used to express the volume of a rectangular prism. Tey can

sometimes be factored into 3 expressions to represent possible dimensions of the

prism. Te three factors are the length, width, and height.

Problem

Te plastic storage container to the right has a volume of

12 x 31 8 x

22 15 x . What linear expressions could

represent possible dimensions of the storage container?

12 x 31 8 x

22 15 x 5 x (12 x

21 8 x 2 15) Factor out x , the GCF for all three terms.

5 x (12 x 21 18 x 2 10 x 2 15) ac is –180 and b is 8. Break 8 x into two terms

that have a sum of 8 x and a product of 2180x2.

5 x f6 x (2 x 1 3) 2 5(2 x 1 3)g Group the terms into pairs and factor out the GCF

from each pair. The GCF of the first pair is 6 x . TheGCF of the second pair is 25.

5 x (6 x 2 5)(2 x 1 3) 2 x 1 3 is the common binomial term. Use the

Distributive Property to reorganize the factors.


x (6 x 2 5)(2 x 1 3) 5 x (12 x 21 18 x 2 10 x 2 15) Multiply 6 x 2 5 and 2 x 1 3.

5 x (12 x 21 8 x 2 15) Combine like terms.

5 12 x 31 8 x

22 15 x 3 Multiply by x .

Possible dimensions of the storage container are x , 6 x 2 5, and 2 x 1 3.

Exercises


13. 14.

15. 16.

V = 12 x 3 + 8 x 2 -15 x

V = 60 x 3 - 68 x 2 -16 x V = 12 x 3 - 15 x 2 -18 x

V = 10 x 3 + 65 x 2 +105 x V = 12 x 3 + 34 x 2 +14 x

2 x , 3 x 1 7, 2 x 1 1

4 x , 5 x 1 1, 3 x 2 4

5 x , 2 x 1 7, x 1 3

3 x , 4 x 1 3, x 2 2

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Chapter 8 Quiz 1 Form G

Lessons 8-1 through 8-4

Do you know HOW?


1. 8 x 3 2. 57 3. 6p3q2 4. 81 x 6 y 3

Simplify.

5. (7t 2 1 9) 1 (6t 2 1 8) 6. 5 x 3 y 2 2 7 x 3 y 2

7. (3m21 2m 2 8) 1 (4m2

2 5m 1 6)


8. 3n(4n2 1 5n) 9. 4k 2(3 2 4k ) 10. 27 y 3(4 y 2 1 y 2 3)


11. 18s 2 63 12. 30b2 1 48b 2 24 13. w 5 1 4w 4 1 10w 3 1 40w 2


14. ( x 1 7)( x 1 5) 15. ( j 1 3)( j 2 4) 16. (3 x 2 1)( x 2 6)

17. (d 1 4)(d 1 4) 18. (3a 1 7)(3a 2 7) 19. (2z 2 3)2

20. A rectangle has length x 1 9 and width 2 x 2 1. What is the area o the rectangle?

21. A square has side length (5 x 2 3) cm. What is the area o the square?

Do you UNDERSTAND?

22. Vocabulary What are the parts that make up a polynomial?

23. Open-Ended Write a trinomial with 3 x as the GCF o its terms.

24. Writing Explain how to use the Distributive Property to fnd the product o

two binomials.

3

13t 2 1 17 22 x 3 y 2

7m22 3m 2 2

12n31 15n2

9(2 s 2 7)

x 2 1 12 x 1 35

d 2 1 8d 1 16

2 x 2 1 17 x 2 9

(25 x 2 2 30 x 1 9) sq cm

1 or more monomials

Answers may vary. Sample 3 x 3 1 6 x 2 1 3 x

Answers may vary. Sample (a 1 b)(c 1 d ) 5 a(c 1 d ) 1 b(c 1 d )

216k 3 1 12k 2

6(5b 2 2)(b 1 2)

j 2 2 j 2 12

9a22 49

228 y 5 2 7 y 4 1 21 y 3

w 2(w 2 1 10)(w 1 4)

3 x 2 2 19 x 1 6

4 z 2 2 12 z 1 9

0 5 9

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Chapter 8 Quiz 2 Form G


Do you know HOW?


1. x 2 1 11 x 1 24 2. s2 2 7s 1 12 3. 2m21 27m 1 70

4. 4z 2 2 16z 1 15 5. 8 y 2 2 22 y 2 21 6. 9 x 2 1 48 x 1 64

7. g 2 2 64 8. 4s2 2 25 9. 49t 2 2 9

10. 6r 3 1 15r 2 1 8r 1 20 11. 10c 3 2 12c 2 1 15c 2 18 12. 16w 3 1 8w 2 1 28w 1 14

13. Te area o a rectangular feld is given by the trinomial t 2 2 4t 2 45. Te

length o the rectangle is t 1 5. What is the expression or the width o the

feld?

14. Te area o a rectangle is given by the trinomial 10 x 2 2 31 x 2 14. Te length

o the rectangle is 5 x 1 2. What is the expression or the width o the

rectangle?

15. Te area o a square room is 16 x 2 1 72 x 1 81. What is the length o one side

o the room?

16. A rectangular prism has a volume o 4 x 3 1 30 x 2 1 36 x . What linearexpressions can represent possible dimensions o the prism?

Do you UNDERSTAND?

Describe how you would factor each expression.

17. 81m22 25 18. 4 x 2 2 16 x 1 16 19. 9 x 2 1 42 x 1 49

20. Reasoning In ax 2 1 bx 1 c , i ac is negative and b is positive, what do you

know about the actors o ac ?

21. Writing Describe how to fnd linear expressions or the possible dimensions

o a rectangular prism with a volume o 8k 3 1 26k 2 1 6k .

22. Open-Ended Write two trinomials that you can actor into two binomials.

Factor each trinomial. Ten write one trinomial that you cannot actor and

explain why.

( x 1 8)( x 1 3)

(2 z 2 3)(2 z 2 5)

(g 1 8)(g 2 8)

(3r 2 1 4)(2r 1 5)

t 2 9

2 x 2 7

4 x 1 9

difference of two squares:(9m 1 5)(9m 2 5)

The factors have different signs.


Factor out the GCF and then factor the other factor. 2k (4k 1 1)(k 1 3)

factor out GCF, perfectsquare: 4( x 2 2)2

perfect square: (3 x 1 7)2

2 x , 2 x 1 3, x 1 6

( s 2 3)( s 2 4)

(4 y 1 3)(2 y 2 7)

(2 s 1 5)(2 s 2 5)

(2c 2 1 3)(5c 2 6)

(2m 1 7)(m 1 10)

(3 x 1 8)2

(7t 1 3)(7t 2 3)

2(4w 2 1 7)(2w 1 1)

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Chapter 8 Chapter Test Form G

Do you know HOW?

Find the degree o each monomial.

1. 6 xy 2. 23b2c 4 3. 12m7n

Simpliy each sum or diference.

4. 6r 3 1 7r 3 5. 23u2v 2 19u2v 6. (5 g 2 2 g ) 1 (2 g 2 1 6 g )

7. Te perimeter o a pentagon is 20t 1 7. Four sides have the ollowing lengths:

6t , 2t , 4t 2 5, and 5t 1 1. What is the length o the fth side?

Simpliy each product.

8. 3 x ( x 1 6) 9. 2z 2(z 2 9) 10. 2 x (4 x 2 2 7 x 1 6)


11. 12 x 2 9 12. 24n32 40n2

1 72n 13. 14b2c 3 1 21bc 5

14. An artist is making a square stained glass window in which a green

glass circle is surrounded by blue glass. Te side length o the

window is shown, and the area o the green piece is 64π x 2. What is

the area o the blue glass? Write your answer in actored orm.

Simpliy each product using the stated method.

15. ( x 2 2)(3 x 2 4); table

16. (3 x 1 2)( x 1 7); Distributive Property 17. (4 x 2 1)(2 x 1 5); FOIL Method

18. What is the surace area o a cylinder with radius x 1 3 and height x 1 11?


19. ( x 1 6)2 20. (2s 1 7)2 21. (3 x 2 8)2

Complete.

22. k 2 1 9k 1 18 5 (k 1 3)(k 1 ) 23. x 2 2 11 x 1 28 5 ( x 2 4)( x 2 )


24. (v 1 7)(v 2 7) 25. (5s 2 t )2 26. (3p21 10q)(3p2

2 10q)

20 x

2

3t 1 11

3 x 2 1 18 x 2 z 3 1 9 z 2

3(4 x 2 3) 8n(3n22 5n 1 9)

16 x 2(25 2 4π)

3 x 2 2 10 x 1 8

3 x 2 1 23 x 1 14

4π( x 2 1 10 x 1 21)

x 2 1 12 x 1 36

v 2 2 49 25 s22 10 st 1 t 2 9 p4

2 100q2

6 7

4 s2 1 28 s 1 49 9 x 2 2 48 x 1 64

8 x 2 1 18 x 2 5

7bc 3(2b 1 3c 2)

8 x 3 2 14 x 2 1 12 x

13r 3 4u2v

2g21 9g

6 8

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Chapter 8 Chapter Test (continued) Form G

Find an expression for the area of each shaded region.

27. 28.

29. Te area o a rectangular cofee table is given by the trinomial t 2 1 7t 2 8.

Te table’s length is t 1 8. What is the table’s width?


30. r 2 1 12r 1 27 31. g 2 2 8 g 2 48 32. m21 2m 2 35

33. 3d 2 2 13d 1 12 34. 8 y 2 1 60 y 1 72 35. 9w 2 2 75w 2 54

Factor completely.

36. 6n3 2 24n2 1 n 2 4 37. 2p4 1 6p3 2 8p2 2 4p 38. 8h2 1 36h 1 16

39. A cereal box in the shape o a rectangular prism has a volume o 18 x 3 2 3 x 2 2 6 x .

What are three possible linear expressions or the dimensions o the cereal box?

40. Te area o a rectangular serving tray is 3 x 2 1 17 x 2 56. Te width o the tray

is x 1 8. What is the length o the tray?

Do You UNDERSTAND?

41. Writing Write a binomial with 2 x 3 y 2 as the GCF o its terms. Explain how you

ound your answer.


made in simpliying the product.

43. Reasoning Let x 2 1 7 x 2 18 5 ( x 1 p)( x 1 q) and z 2 2 7z 2 18 5 (z 1 r )(z 1 s).

a. What do you know about the signs o p and q? r and s?

b. Suppose u p u . u q u and u r u . u s u . What is the value o p 2 r ?

x 2 2

x 2 2

x 1 5

x 1 5 2 x

2 x

3 x 1 4

3 x 1 4

(r2 + 7s)(r2 - 7s) = r4 + 14r2s - 49s2

14 x 1 21

t 2 1

(r 1 9)(r 1 3)

(3d 2 4)(d 2 3)

(g 1 4)(g 2 12)

4(2 y 1 3)( y 1 6)

2 p( p31 3 p2

2 4 p 2 2)

(m 1 7)(m 2 5)

3(3w 1 2)(w 2 9)

4(2h 1 1)(h 1 4)(6n21 1)(n 2 4)

3 x , 3 x 2 2, 2 x 1 1

3 x 2 7

18

they are opposite; they areopposite

Answers may vary. Sample: 2 x 4 y 2 1 2 x 3 y 2 ; mental math: multiplied by ( x 1 1)

The student did not notice that the r 2 s terms

should cancel. Correct answer is r 4 2 49 s2.

(5 x 1 4)( x 1 4)

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Chapter 8 Part A Test Form K


Do You Know HOW?

Find the degree o each monomial.

1. 24t 5 2. 22a4b7

3. 28mn2 4. 10a2b3

Simpliy each sum or diference.

5. (4n

31

12)1

(10n

31

1) 6. (5 y

21

3 y 2

6)2

(2 y

22

5 y 1

3)

7. t 4 1 9t 4 8. (3n2 2 4n 1 8) 2 (24n2 1 5n)

9. Te perimeter of a triangle is 12b 1 5. wo sides have the following lengths:

4b and 3b 1 15. What is the length of the third side?


10. 5w (2w 2 1 6w ) 11. 23p2(5p2 1 p 2 7)

12. 2w (w 2 7) 13. 4n(3n2 1 6n 2 9)

14. (t 1 1)(t 1 6) 15. (2n 2 5)(n 2 3)

16. A square has side length (7 x 2 2) in. What is the area of the square?

14n31 13 3 y 2 1 8 y 2 9

10t 4 7n22 9n 1 8

5b 2 10

10w 3 1 30w 2 215 p42 3 p3

1 21 p2

2w 2 2 14w 12n31 24n2

2 36n

t 2 1 7t 1 6 2n22 11n 1 15

(49 x 2 2 28 x 1 4) in.2

5 11

3 5

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Chapter 8 Part A Test (continued) Form K


17. A rectangle has a length o (2 x 1 1) t and a width o (3 x 1 5) t. What is the

area o the rectangle?

18. A carpenter is cutting a circle out o a square piece o plywood. Te side length

o the plywood is (2 x 1 5), and the area o the circle is 49p x 2. What is the

area o the plywood ater the circle has been cut out?

Simplify each product using the stated method.

19. (4 x 1 5)( x 1 3) ; Distributive Property 20. (2 x 2 9)(5 x 1 4) ; FOIL Method


21. (n 1 5)2 22. (5 x 2 3)2

Write the missing value in each trinomial.

23. s21 14s 1 33 5 (s 1 3)(s 1 ) 24. c 2 2 9c 1 14 5 (c 2 2)(c 2 )


25. ( j 1 9)( j 2 9) 26. (6t 2 1 11)(6t 2 2 11)

Do You UNDERSTAND?

27. Open-Ended Write two trinomials whose diference is 5 x 2 2 6 x 1 9.

28. Writing Explain how to use the FOIL Method to nd the product o two

binomials.

29. Writing Explain how you know which terms in diferent polynomials can be

added or subtracted. What do you add or subtract?

11 7

(6 x 2 1 13 x 1 5) ft2

(4 2 49π) x 2 1 20 x 1 25

4 x

21

17 x 1

15 10 x 22

37 x 2

36

n21 10n 1 25 25 x 2 2 30 x 1 9

j 2 2 81 36t 4 2 121

Answers may vary. Sample: (7 x 2 1 x 1 11) 2 (2 x 2 1 7 x 1 2).

Multiply the first terms of the binomials. Multiply the outside terms of thebinomials. Multiply the inside terms of the binomials. Multiply the last termsof the binomials. Add the products and combine like terms.

Like terms in polynomials can be added or subtracted. Like terms are added orsubtracted by adding or subtracting the coefficients only of the like terms.

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Chapter 8 Part B Test Form K


Do You Know HOW?


1. v 2 1 11v 1 18 2. 3h21 32h 1 20

3. 12 y 2 1 16 y 2 3 4. 25a21 70a 1 49

5. p22 100 6. 81 x 2 2 1

7. 6k 3 2 2k 2 1 15k 2 5 8. 80n31 30n2

2 56n 2 21

9. 16k 2 40 10. 6m41 12m3

1 3m21 21m

11. 15r 1 6 12. 18 x 2 y 2 27 x 3 y 2

13. Te area of a rectangular classroom is given by the trinomial a22 4a 2 21.

Te length of the rectangle is a 1 3. What is the expression for the width of

the classroom?

14. Te area of a square mural is 144 x 2 2 72 x 1 9. What is the length of one side

of the mural?

15. A box shaped like a rectangular prism has a volume of 6 x 3 1 7 x 2 2 24 x . What

expressions can represent possible dimensions of the box?

16. Te area of a rectangular blanket is given by the trinomial w 2 1 5w 2 6. Te

length of the blanket is w 1 6. What is the width of the blanket?

17. Te area of a rectangular curtain is 5 x 2 1 42 x 1 16. Te width of the curtain

is 5 x 1 2. What is the length of the curtain?

(v 1 9)(v 1 2) (3h 1 2)(h 1 10)

(6 y 2 1)(2 y 1 3) (5a 1 7)2

( p 1 10)( p 2 10) (9 x 1 1)(9 x 2 1)

(2k 2 1 5)(3k 2 1) (10n2 2 7)(8n 1 3)

8(2k 2 5) 3m(2m31 4m2

1 m 1 7)

3(5r 1 2) 9 x 2 y (2 2 3 xy )

a 2 7

12 x 2 3

x by (3 x 1 8) by (2 x 2 3)

w 2 1

x 1 8

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Chapter 8 Part B Test (continued) Form K



18. m21 11m 1 24 19. a2 2 6a 2 7

20. 7 y 2 1 11 y 2 6 21. 10 x 2 2 53 x 2 11

22. g 2 1 17 g 1 72 23. 112n2 2 252

Factor completely.

24. 12n3 2 3n2 1 16n 2 4 25. 9z 2 1 42z 1 49

26. A shipping box in the shape o a rectangular prism has a volume o

18 x 3 1 5 x 2 2 2 x . What are three expressions that can represent possible

dimensions o the shipping box?

Do You UNDERSTAND?

Writing Describe how you would factor each expression.

27. 9t 2 2 49

28. 36n2 1 60n 1 25

29. 25t 3 2 20t 2 1 4t

30. Open-Ended Find two diferent values that complete the expression

8n2 1 u n 1 4 so that the trinomial can be actored into the product o two

binomials. Factor your trinomials.

31. Open-Ended Find two diferent values that complete the expression

24 x 2 1 u x 2 18 so that the trinomial can be actored into the product o two

binomials. Factor your trinomials.

First, factor out the GCF, t , giving you t (25t 2 2 20t 1 4). What is left is inside

the parentheses is in the form a21 2ab 1 b2 where a 5 5t and b 5 22. So,

the factored form of the trinomial is t (5t 2 2)2.

Answers may vary. Sample: 33 and 12. 8n21 33n 1 4 5 (8n 1 1)(n 1 4) and

8n21 12n 1 4 5 (2n 1 2)(4n 1 2).

Answers may vary. Sample: 66 and 24. 24 x 2 1 66 x 2 18 5 (3 x 1 9)(8 x 2 2) and

24 x 2 1 24 x 2 18 5 (4 x 1 6)(6 x 2 3).

(m 1 8)(m 1 3) (a 1 1)(a 2 7)

(7 y 2 3)( y 1 2) (5 x 1 1)(2 x 2 11)

(g 1 8)(g 1 9) 28(2n 1 3)(2n 2 3)

(3n21 4)(4n 2 1) (3 z 1 7)

2

x , 9 x 2 2, 2 x 1 1

This expression is the difference of two squares. You would use

a22 b2

5 (a 2 b)(a 1 b) which gives you (3t 1 7)(3t 2 7).

This trinomial is in the form a21 2ab 1 b2 where a 5 6n and b 5 5. So, the

factored form of the trinomial is (6n 1 5)2.

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Chapter 8 Performance Task

Give complete answers. Show all your work.

TASK 1

How would you describe a polynomial to some new algebra students? What

advice would you give to help them learn how to add and subtract polynomials?

TASK 2

In your own words, explain what is accomplished by factoring. As part of your

explanation, write a sample problem in which you factor a monomial from a

polynomial. Ten write and factor another problem that is a quadratic expression.

Check your solutions by multiplying after you factor.


[4] Complete explanation of polynomials and polynomial addition andsubtraction given

[3] Explanation includes minor errors or omits small details

[2] Explanation covers many important points and is generally correct

[1] Some parts of the explanation are correct

[0] No correct information given


[4] Factoring correctly explained and two correct examples given and checkedcorrectly

[3] Answer contains minor errors or omits one part of the task

[2] At least two of the four parts of the task completed correctly

[1] At least one of the four parts of the task completed correctly

[0] No part of the task completed correctly

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Chapter 8 Performance Task (continued)

TASK 3

A classmate is having trouble factoring special-case polynomials. Point out

the errors she has made, and write a brief suggestion to help her with similar

problems.

a. 4a2 2 100

5 (2a 2 10)(2a 2 10)

5 (2a 2 10)2

b. 81m41 72m2n 1 16n2

5 (9m21 8n)(9m2

1 8n)

5 (9m2 1 8n)2

c. a10b4 2 16

5 (a5b2 2 16)

5 (a5b2 2 4)(a5b2 1 4)

d. 4d 2 1 36bd 1 81

5 (2d 1 9)(2d 2 9)

TASK 4

Explain how to factor a polynomial with four terms by grouping.

Use grouping to factor 20ay 2 10ax 1 42by 2 21bx . Explain each step.

The first step should be to factor the GCD, 4.In Step 2, signs should be different. Remember

that a22 b2

5 (a 1 b)(a 2 b).

The square root of 16 is 4, not 8. Learn the common

squares and square roots.

In Step 2, the parentheses are around the square

of a5b2 and the exponent 2 should be used.

No parentheses are needed after 16.

The expression is a perfect square, so the answer

should be (2d 1 9)2. Watch the signs when you

work.

[4] Correct explanation of grouping and demonstration with each step fully explained

[3] Correct explanation of grouping and demonstration with some steps inadequately

explained

[2] Inadequate explanation of grouping or minor errors in demonstration

[1] Some steps in grouping explained or demonstrated correctly

[0] No part of task explained or demonstrated correctly

[4] All four errors correctly identified with helpful suggestions

[3] Three errors correctly identified with helpful suggestions

[2] Two errors correctly identified with helpful suggestions

[1] One error correctly identified with a helpful suggestion

[0] No errors correctly identified

First look for the GCF of all four terms. Look for the GCF of two terms at a time.

10a(2 y 2 x ) 1 21b(2 y 2 x ) Found GCFs.

(10a 1 21b)(2 y 2 x ) Simplified with Distributive Property

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Cumulative Review Chapters 1–8

Multiple Choice

For Exercises 1–11 choose the correct letter.

1. What are the next three terms in the sequence 6, 12, 24, 48,c?

A. 72, 96, 120 B. 86, 162, 240 C. 96, 192, 384 D. 50, 52, 54

2. Solve 8 y 5 2100.

F. 2800 G. 212.5 H. 800 I. 12.5

3. Find the equation of the line passing through (2,21) and parallel to

y 5 23 x 2 1.

A. y 5 23 x 1 5 B. y 5 23 x

22 1 C. y 5

x

31 5 D. y 5 3 x 1 1

4. Solve3 x 1 7 y 5 22

4 x 2 3 y 5 22.

F. (24,22) G. (24, 2) H. (4, 2) I. (4,22)

5. Simplify 10 x 5 y 3

2 x 6 y .

A. 5 xy 2 B.5 y 2

x C.

5 x

y 2 D.

x

5 y 2

6. Simplify (3 x 2 1)( x 1 4).

F. 3 x 2 2 4 G. 3 x 2 2 11 x 2 4 H. 3 x 2 1 11 x 2 4 I. 3 x 2 1 13 x 2 4

7. A scuba diver at a depth of 80 ft begins her ascent to the ocean surface. Her

rate of change in depth is 2ft/s. Which expression represents her depth in feet

t seconds after she begins her ascent?

A. 2t 2 80 B. 80 2 2t C. 280 2 2t D. 80 1 2t

8. Factor 4 x 2 2 x 2 14.

F. (4 x 1 7)( x 2 2) G. (2 x 2 7)(2 x 1 2) H. (4 x 2 7)( x 1 2) I. (2 x 1 7)(2 x 2 2)

9. What is the GCF of the terms of 3 x 3 1 6 x 2 2 9 x ?

A. x B. 3 C. 3 x D. 3 x 2

10. Which number isnot a solution of the compound inequality 7 2 4 x # 3 and

2 x 2 5 . 210?

F. 5 G. 4 H. 2 I. 1

11. Which of the following is a cubic binomial?

A. w 3 2 6w 2 1 9 B. 7a31 4a22 C. 2 y 3 1 3 y 5 D. x 2 2 2 x 3

C

G

A

I

B

H

B

F

C

F

D

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12. A city is growing at a rate o 8 percent per year. What multiplier is used to fnd

the new population each year?

13. Simpliy 624 4 1 2(7 2 3) ? 4.

14. What is the slope o a line that passes through the origin and the point (6, 3)?

15. Evaluate x 2 1 3 y or x 5 4 and y 5 0.5.

16. A weight o 6 lb stretches a spring a distance o 12 in. Find the constant k or

the spring.

17. Solve18

x 5

21

14.

18. What is the x -intercept o the line with equation 5 x 1 4 y 5 30?

19. How many positive solutions are there to the equation Z 2 x 2 5 Z 5 4?

20. Write an equation in standard orm passing through the points (22, 0) and

(23,21).

21. Te product o two negative integers is 36. Te second integer is 5 more than

the frst. Find the integers.

22. Te length o a rectangular pizza is 4 in. less than twice its width. Te area o

the pizza is 160 in.2. Find the dimensions o the pizza.

23. Write a polynomial that is a dierence o two squares using the variablem.

Write the polynomial in actored and standard orms.

24. Solve the ollowing system o inequalities by graphing:

2 x 2 4 y # 4

23 x 2 6 y . 6

Cumulative Review (continued) Chapters 1–8

6

2

1.08

41

17.5

12

y 2 x 5 2

29, 24

10 in. by 16 in.

Sample :4m22 9; (2m 1 3)(2m 2 3)

graph of y L 12 x 2 1 and y R 2

12 x 2 1

12

12

x

O

y 4

2

2

Ź4

Ź2

Ź4 4Ź2

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T E A C H E R I N S T R U C T I O N S



About the Project

In this project students will learn about the uses o trees. Tey will use ormulas toanalyze data and predict the production o wood and ruit.

Introducing the Project

• Ask students to think o something fat that is made o wood, such as a table

top or door.

• Instruct them to estimate the number o pieces o wood, each 1 t2, that make

up their objects.

• Ask students to compare results with partners.

• Direct student attention to Activity 1. Explain that they will research types o

wood and the tools carpenters use to work with wood.

Activity 1: Researching

Students research lumber and tool requirements or the construction o a house.

Activity 2: Calculating

Students evaluate the given expression to calculate the useable board eet o a log.


Students use the given expression to determine the diameter o a tree.

Activity 4: Graphing

Students use the given unction to calculate and graph the number o bushels o

walnuts produced on an acre o land.

Finishing the Project

You may wish to plan a project day on which students share their completed

projects. Encourage groups to explain their processes as well as their results. Have

students review the project work and update their olders.

• Have students review the research, equations, graphs, and explanations

needed or the project.

• Ask groups to share their insights that resulted rom completing the project,such as any shortcuts they ound or doing research, solving equations, or

making graphs.

Chapter 8 Project Teacher Notes: Trees Are Us

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Beginning the Chapter Project

Many schools celebrate Arbor Day by planting young trees to replenish ourecosystem. rees use carbon dioxide that humans and animals exhale to make

oxygen. rees anchor the soil and prevent erosion. Tey also produce ruit. Wood

rom trees is used or the construction o everything rom pencils to houses.

As you work through the activities, you will learn more about the uses o trees. You

will use ormulas to analyze data and predict the production o wood and ruit.

Ten you will decide how to organize and display your results.

List of Materials

• Calculator

• Graph paper

Activities

Activity 1: Researching

A board oot is a cubic measure o lumber equal to a square oot o wood 1 in.

thick.

• What can you make rom 10 board eet? 100 board eet? 1000 board eet? How

is the size o a house related to the amount o wood used to build it?

• What diferent types o wood are needed or cabinets, oors, and roos? What

tools do carpenters use to make these items?


You can use the expression 0.0655l (1 2 p)(d 2 s)2 to

nd the number o useable board eet in a log.

• Estimate the useable board eet in a 35-t log i

its diameter is 20 in. Assume the log loses 10% o

its volume rom the saw cuts and a total o 2 in. is

trimmed of the log.

• Te diameter o a log is 25 in. A total o 2 in. will be

trimmed of the log. Te estimated volume loss dueto saw cuts is 10%. How long must the log be to yield

600 board eet o lumber?

Chapter 8 Project: Trees Are Us

i lengthin feet

P percent lossdue to cut of saw

d diameter in inches

s inches trimmed off thelog to make boards

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With aerial photography, you can study a orest o

ponderosa pines without ever walking through it. o fnd the

diameter in inches o trees in the orest, use the expression:

3.76 1 (1.35 3 1022)hv 2 (2.45 3 1026)hv 2 1 (2.44 3 10210)hv 3.

Te variable h is the height o the tree in eet, and v is the crown

diameter visible in eet (rom a photograph). Determine the diameter

o a 100-t tree that has a visible crown diameter o 20 t.

Activity 4: Graphing

You can use the unction b 5 20.01t 2 1 0.8t to fnd the number o bushels b

o walnuts produced on an acre o land. Te variable t represents the number o

walnut trees per acre.• Use your graphing calculator to graph this unction. Include an accurate graph

in your notebook. You may wish to investigate the ABLE eature on your

calculator. Use the maximum eature under the CALC menu to determine the

number o trees per acre that gives the greatest yield.

• How many walnut trees would you advise a armer to plant on 5 acres o land

to produce the most walnuts possible? Explain your reasoning.

Finishing the Project

Te answers to the our activities should help you complete your project. Assemble all the parts o your project in a older. Add a summary telling what you

have learned about the uses o trees.

Reflect and Revise

Ask a classmate to review your project older with you. ogether, check that your

graph is clearly labeled and accurate. Check that you have used ormulas correctly

and that your calculations are accurate. Make any revisions necessary to improve

your work.

Extending the Project

rees have many uses that you can investigate. You can begin your research by

contacting the United States Department o Agriculture Forest Service or a local,

state, or national park. You can also get more inormation by using the Internet.

Chapter 8 Project: Trees Are Us (continued)

20 ft

Aerial View o

Ponderosa Pine

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Getting StartedRead the project. As you work on the project, you will need a calculator, materials

on which you will record your calculations, and materials to make accurate and

attractive graphs. Keep all o your work or the project in a older.

Checklist Suggestions

☐ Activity 1: researching lumber ☐ Measure wooden objects to help you estimate.

☐ Activity 2: calculating ☐ Use the given ormula.

☐ Activity 3: calculating ☐ Have someone check your solution.

☐ Activity 4: graphing the

unction

☐ Select an appropriate viewing window.

☐ project report ☐ What have you learned about trees and lumber

while working on this project? o whom might the

ormulas in the activities be most useul, and why?

Scoring Rubric3 Torough research techniques are demonstrated with many diferent sources

o inormation accessed. Calculations are correct. Te graph is neat, accurate,

and has an appropriate scale. Explanations are complete and well thought-

out.

2 Good research techniques are evident. Calculations are mostly correct, but

have minor errors. Te graph is neat and mostly accurate with minor errors

in scale.

1 Needed inormation is located with some help. Calculations contain both

minor and major errors. Te graph could be more accurate.

0 Major elements o the project are incomplete or missing.

Chapter 8 Project Manager: Trees Are Us

a1ao08a

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