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Unit Unit Unit Unit 9999: : : : FactoringFactoringFactoringFactoring

Lesson 1: Introduction to Factoring

Try These on Your Own

1. List the factors of 72. 4. ______ is a common factor of 32 & 40.

2. 36 is divisible by _____.

3. _____ is a multiple of 12.

Factors and Multiples

6 ● 5 = 30 6 ● 8 = 48

_________ _______

___ and ____ are _________ of _____.

_________ is a multiple of ________.

_________ is a multiple of ________.

Common Factors

If two numbers are divisible by the same number, then they are called common factors

___ is a common factor of ______ and ________.

___ and ____ are _________ of _____.

_________ is a multiple of ________.

_________ is a multiple of ________.



Prime and Composite Numbers

Greatest Common Factor

Example 1 Example 2

Example 3 Example 4

Prime Numbers Composite Numbers

What is the greatest common factor for 18

and 27?

18 27

The Greatest Common Factor is the greatest integer that is a common factor

of two or more numbers.

What is the greatest common factor of

6x2 y and 9x3 y2 ?

6x2 y 9x3 y2

What is the greatest common factor for

3x and 9x2

3x 9x2

What is the greatest common factor of

2x and 3y?

2x 3y

If: 2x + 3y

This polynomial is ____________ because



Lesson 1: Introduction to Factoring Practice

Part 1: Read each question and answer completely.

1. 35 is divisible by _______ (give 2 solutions)

2. Name 6 factors of 12.

3. ________ is a multiple of 13. (Give 3 solutions)

4. _______ is a common factor of 36 and 48.

5. ______ is a common factor of 144 and 120.

6. ______ is the greatest common factor of 48 and 64.

7. ______ is the greatest common factor of 48 and 60.

8. Is 9 a prime number? Explain why or why not.

9. 15 is a composite number. Explain why.

10. Do you think that the number 1 is prime, composite, or neither? Explain.

Part 2: Find the greatest common factor (GCF) for each set of numbers.

1. 12a2 and 14 a 7. 3xy, 2x2y, and 4x3y

2. x2 y3 and 2xy2 8. 3a2b, 27ab2, and 9ab

3. 3ab5 and 4a3b3 9. 2x2y3z, 6xy2z2, and 18xz3

4. 9rs and 12r2s 10. 15rs3t, 7r2st, and 5r2s2t

5. 4a3b2 and 2a3 11. 18xz, 9x2y4z2, and 3xyz

6. 16x4yz2 and 24x3z 12. 7x2y5, 5x5y4, and 8x3y5

1. 13x3yz and 26xy2z2 2. 18r5s8t4, 81r6s5t6, and 27r9s7t5

3. 22abc and 11a2b

Find the greatest common factor (GCF) for each set of numbers.

(2 points each)



Lesson 1: Introduction to Factoring Practice – Answer Key

Part 1: Read each question and answer completely.

1. 35 is divisible by 7 and 5 (give 2 solutions) (7 & 5 are factors of 35)

2. Name 6 factors of 12. (1,2,3,4,6,12) (1 x12) (2 x6) (3x4)

3. 26, 39, 52 is a multiple of 13. (Give 3 solutions) (13 x2 = 26) (13x3 = 39) (13 x4 = 52) (You may

also have had: 65, 78, 91, 104, 117, 130 …)

4. 6 is a common factor of 36 and 48. (6 x6 = 36) (6 x 8 = 48) (You may also have had 4 because

(4 x9 = 36) (4 x 12 = 48). Other answers include: 2 , 3, and 12

5. 12 is a common factor of 144 and 120. (12 x12 = 144) (12 x 10 = 120) (Other answers include:

2, 3, 4, 6, 8, 24)

6. 16 is the greatest common factor of 48 and 64.

7. 12 is the greatest common factor of 48 and 60.

8. Is 9 a prime number? Explain why or why not.

9. 15 is a composite number. Explain why.

Factors of 48: Factors of 64:

1, 2, 3, 4,6, 8, 12, 16, 24 ,48 1, 2, 4, 8, 16, 32, 64

The common factors are: 1,2,4, 8, 16 The greatest common factor would be 16.

Factors of 48: Factors of 60:

1, 2, 3, 4,6, 8, 12, 16, 24 ,48 1,2,3,4,5,6,10,12,15,20,30,60

The common factors are: 1,2,3,4,6, 12 The greatest common factor would be 12.

9 is not a prime number. A prime number only has factors of 1 and itself. 9 has the following factors:

1,3,9. Since 3x3=9, 9 is not a prime number.

15 is a composite number. A composite number has one or more factors other than itself and 1. (It’s the

opposite of a prime number.)

The factors of 15 are: 1,3,5,15. Since 3x5=15, we can classify 15 as a composite number.



10. Do you think that the number 1 is prime, composite, or neither? Explain.

Part 2: Find the greatest common factor (GCF) for each set of numbers.

1. 12a2 and 14 a 7. 3xy, 2x2y, and 4x3y

2. x2 y3 and 2xy2 8. 3a2b, 27ab2, and 9ab

3. 3ab5 and 4a3b3 9. 2x2y3z, 6xy2z2, and 18xz3

4. 9rs and 12r2s 10. 15rs3t, 7r2st, and 5r2s2t

Since the only factor of 1 is 1, it is not considered to be prime or composite.

The prime numbers less than 50 are:

2,3,5,7,11,13,17,19,23,29,31,37,41,43,47.

GCF: 2a

The GCF for 12 and 14 is 2.

The GCF for a2 and a is a

GCF: xy

The GCF for 3,2, and 4 is 1. (We don’t

need to write)

The GCF for x, x2 and x3 is x.

The GCF for y, y, and y is y

GCF: xy2

The GCF for 1 and 2 is 1. (We don’t

need to write.)

The GCF for x2 and x is x.

The GCF for y3 and y2 is y2

GCF: 3ab

The GCF for 3, 27, and 9 is 3.

The GCF for a2 and a is a.

The GCF for b, and b2 is b.

GCF: ab3

The GCF for 3 and 4 is 1. (We don’t

need to write.)


The GCF for b5 and b3 is b3.

GCF: 2xz

The GCF for 2, 6 and 18 is 2.


Since 18xz3 does not have a y term, we

cannot include a y in our GCF.

The GCF for z3, z2 and z is z.

GCF: 3rs


The GCF for r2 and r is r.

The GCF for s is s.

GCF: rst

The GCF for 15, 7 and 5 is 1. (We don’t

need to write.)

The GCF for r2 and r is r.

The GCF for s3,s2, and s is s.

The GCF for t is t.



5. 4a3b2 and 2a3 11. 18xz, 9x2y4z2, and 3xyz

6. 16x4yz2 and 24x3z 12. 7x2y5, 5x5y4, and 8x3y5

1. 13x3yz and 26xy2z2 2. 18r5s8t4, 81r6s5t6, and 27r9s7t5

3. 22abc and 11a2b

GCF: 2a3


The GCF for a3 is a3.

Since 2a3 does not have a b term, we

cannot include a b term in the GCF.

GCF: 3xz



Since 18xz does not have a y term, we

cannot include a y term in the GCF.

The GCF for z2 and z is z.

GCF: 8x3z


The GCF for x4 and x3 is x3.

Since 24x3z does not have a y term, we

cannot include a y term in our GCF.

The GCF for z2 and z is z.

GCF: x2y4

The GCF for 7 and 5, and 8 is 1. (We

don’t need to write.)

The GCF for x2, x3, and x5 is x2.

The GCF for y4 and y5 is y4.

GCF: 13xyz



The GCF for y2 and y is y.

The GCF for z and z2 is z.

GCF: 9r5s5t4


The GCF for r5, r6 and r9 is r5.

The GCF for s8, s5 and s7 is s5.

The GCF for t4, t6, and t5 is t4.

GCF: 11ab



The GCF for b is b.

Since 11a2b does not have a c term, we

cannot include a c term in our GCF.



Lesson 2: Factoring Polynomials Using the Greatest Common Factor (GCF)

Example 1 Example 2

Example 3 Example 4

The Greatest Common Factor (GCF) is the greatest integer that is a common factor

of two or more numbers.

Factor: 3x2 + 27x3 Factor: 2x2y2 - 8xy2

Factor: 3x2y2 + 2x2 + xy2 Factoring can make it easier to simplify some

fractions:

Simplify: ��

�

NOTE: Factoring can always be checked using multiplication!



Lesson 2: Factoring Polynomials by Using the GCF Practice

Part 1: The greatest common factor has been factored out ( in bold). Fill in the blanks to

complete the factorization of each polynomial.

1. 3x2 + 12xy2 = 3x(______ + ________)

2. 2r2s3 + 5rs6 = rs3(________+_________)

3. 8a2bc4 - 6ab4c4 = 2abc4(___________-___________)

4. 2x2y5 + 4xy3 + 6x2y7 = 2xy3(___________+ ____________+ ____________)

5. 3r3s3t5 – 2r2s7t4 + s5t3 = s3t3( ___________ - ___________+______________)

Part 2: Factor Completely.

1. 7a2b5 + 21ab4 6. 9x2y – 27xy3 + 3xy

2. 81x2yz – 27xy2z3 7. 12a4b5c + 2a2c + 8a2bc

3. 3a2b3c + 12a3b2c2 8. 22g2h – 11gh2 + g2h2

4. 5s5t2 – 25s4 9. 2w5xy – 3w2x2y + 2w2x

5. 14x4yz2 + 2x4y3z – 7x3yz 10. 8s2t5u2 – 12st2u2 + 2st4u

Part 3: Simplify each fraction by simplifying the numerator.

1. ��

�� 2.

��

�

1. 4x3y5z – 6x3y5 2. 8r3st + 24r4s2t2

3. Simplify the fraction. ��

�� (2 points)

Factor completely. ( 2 points each)



Lesson 2: Factoring Polynomials by Using the GCF Practice- Answers

Part 1: The greatest common factor has been factored out ( in bold). Fill in the blanks to

complete the factorization of each polynomial.

1. 3x2 + 12xy2 = 3x( x + 4y2)

2. 2r2s3 + 5rs6 = rs3( 2r + 5s3)

3. 8a2bc4 - 6ab4c4 = 2abc4( 4a - 3b3)

4. 2x2y5 + 4xy3 + 6x2y7 = 2xy3( xy2+ 2+ 3xy4 )

5. 3r3s3t5 – 2r2s7t4 + s5t3 = s3t3( 3r3t2 - 2r2s4t + s2)

Part 2: Factor Completely.

1. 7a2b5 + 21ab4 6. 9x2y – 27xy3 + 3xy

2. 81x2yz – 27xy2z3 7. 12a4b5c + 2a2c + 8a2bc

3. 3a2b3c + 12a3b2c2 8. 22g2h – 11gh2 + g2h2

4. 5s5t2 – 25s4 9. 2w5xy – 3w2x2y + 2w2x

Identify the GCF: 7ab4

7ab4(ab + 3)

Identify the GCF: 3xy

3xy(3x – 9y2 + 1)

Identify the GCF: 27xyz

27xyz(3x – yz2)

Identify the GCF: 2a2c

2a2c(6a2b5 + 1 + 4b)

Identify the GCF: 3a2b2c

3a2b2c(b + 4ac)

Identify the GCF: gh

gh(22g – 11h + gh)

Identify the GCF: 5s4

5s4(st2 – 5)

Identify the GCF: w2x

w2x(2w3y – 3xy + 2)



5. 14x4yz2 + 2x4y3z – 7x3yz 10. 8s2t5u2 – 12st2u2 + 2st4u

Part 3: Simplify each fraction by simplifying the numerator.

1. ��

�� 2.

��

�

1. 4x3y5z – 6x3y5 2. 8r3st + 24r4s2t2

3.Simplify the fraction. ��

�� (2 points)

Identify the GCF: x3yz

x3yz(14xz + 2xy2 – 7)

Identify the GCF: 2st2u

2st2u(4st3u – 6u + t2)

2��(2� + �)

2��= 2� + �

Identify the GCF of the numerator: 2xy

��(�� + 1 −�� − �)

��=�� −�� − � + 1

Identify the GCF of the numerator: b3

��(6� + 5�")

��= 6� + 5�"

Identify the GCF of the numerator: ab2

Identify the GCF: 2x3y5

2x3y5(2z – 3)

Identify the GCF: 8r3st

8r3st(1 + 3rst)



Lesson 3: Factoring Trinomials

What is a Trinomial?

Example 1 Example 2

A trinomial is a polynomial that has _____ terms.

x2 –x - 12

Before we start, let’s review multiplication of polynomials.

Multiply: (x+2)(x+3) =

Multiply: (x – 5) (x+6)

When factoring trinomials in the form of: x2 + bx + c, you must find two numbers whose product

is c and whose sum is b. NOTE: (Always take the sign in front of b and c when determining your

numbers.

Factor: x2 + 6x + 8

2#’s: Product = ____ Sum = ____

( ) ( )

Factor: x2 -9x + 20

2#’s: Product = ____ Sum = ____



Example 3 Example 4

Factor: x2 – 5x – 24

2#’s: Product = _____ Sum = _______

Let’s first multiply:

(x – 3) (x + 3)

Factor: x2 - 81



Lesson 3: Factoring Trinomials Practice

Part 1: Factor by filling in the missing parts.

1. x2 + 8x + 12

(x + ____) (x + _____)

2. x2 -7x + 12

(x + ____) (x + _____)

3. x2 -x - 12

(x + ____) (x + _____)

4. y2+ y - 6

(y + ____) (y + _____)

Product: _____ Sum: ______

Product: _____ Sum: ______

Product: _____ Sum: ______

Product: _____ Sum: ______



Part 2: Factor completely.

1. x2 + 15x + 36 5. x2 - 49

2. x2 + 2x – 24 6. x2 - 36

3. x2 – 14x + 48 7. x2 + 14x + 49

4. x2 – 7x – 30 8. x2 – 17x + 72

Part 3: Thinking Questions

1. What polynomial, when factored, gives (x + 5)(x -6)?

2. What polynomial when factored gives (x – 8) (x+8)?

3. One factor of the trinomial: x2 + 85x + 750 is (x+10), what is the other factor?

4. One factor of the trinomial: x2 – 10 x – 375 is (x – 25), what is the other factor?

5. Explain why the trinomial: x2 – 7x + 14 cannot be factored.

Factor Completely. (2 points each)

1. x2 + 16x + 55 2. x2 + 8x + 11



Lesson 3: Factoring Trinomials Practice

Part 1: Factor by filling in the missing parts.

1. x2 + 8x + 12

(x + 2 ) (x + 6 )

2. x2 -7x + 12

(x + -4) (x + -3)

Most commonly written as:

(x-4) (x-3)

3. x2 -x - 12

(x + -4) (x + 3)


(x-4) (x+3)

Product: 12 Sum: 8

1 ● 12 1 + 12 = 13 X

2 ● 6 2 + 6 = 8

4 ● 3 4 + 3 = 7 X

Product: 12 Sum: -7

You know that both factors must be negative since

the sum is negative and the product is positive.

-4 ● -3 -4 + (-3) = -7

-6 ● -2 -6 + -2 = -8 X

-1 ● -12 -1 + -12 = -13 X

Product: -12 Sum: -1

Since the product is negative, you know that one

factor must be positive and one must be negative.

Since the sum is negative also, the larger integer will

be negative.

-12 ● 1 -12 + 1 = -11 X

-6 ● 2 -6+ 2 = -4 X

-4 ● 3 -4 + 3 = -1



4. y2+ y - 6

(y + 3 ) (y + -2)


(y+3)(y-2)


1. x2 + 15x + 36 5. x2 – 49

2. x2 + 2x – 24 6. x2 – 36

Product: -6 Sum: 1

Since the product is negative, we know that one

integer must be positive and one must be

negative. Also, since the sum is positive, the

larger number must be positive.

6 ● (-1) 6 + (-1) = 5 X

3 ● (-2) 3 + (-2) = 1

Need a Sum of 15 and Product of 36.

12 ●3 = 36 and 12+ 3 = 15

Therefore, we can use: 12 and 3

(x+12)(x+3)

Need a Sum of 0 and Product of -49. (There is no

x term, therefore, the sum is 0)

The product is -49, so one integer must be

positive and one must be negative. Since the sum

is 0, the integer should be the same.

7 ● -7 = 49 7 + (-7) = 0

Therefore, we can use: 7 and -7

(x+7)(x-7) This is a Difference of Two Squares

Need a Sum of 2 and Product of -24.

The product is negative, so one integer must be


is positive, the largest integer should be positive.

6 ● (-4) = -24 and 6 + (-4) = 2


(x+6)(x-4)

Need a Sum of 0 and Product of -36. (There is no

x term, therefore, the sum is 0)



is 0, the integer should be the same.

6 ● (-6) = -36 and 6 +(-6) = 0


(x+6)(x-6) This is a Difference of Two Squares



3. x2 – 14x + 48 7. x2 + 14x + 49

4. x2 – 7x – 30 8. x2 – 17x + 72


1. What polynomial, when factored, gives (x + 5)(x -6)?

Need a Sum of -14 and Product of 48.

The product is positive and the sum is negative.

Therefore, both integers must be negative.

-6 ● (-8) = 48 and (-6) + (-8) = -14

Therefore, we can use: -6 and -8

(x-6)(x-8)


Since both the sum and product are positive, both

integers will be positive.

7 ●7 = 49 and 7 + 7 = 14


(x+7)(x+ 7) or (x+7)2

This is a perfect square binomial.

Need a Sum of -7 and Product of -30.



is negative, the largest integer should be negative.

-10 ● 3 = -30 and -10 + 3 = -7

Therefore, we can use: -10 and 3

(x+3)(x-10)

Need a Sum of -17 and Product of 72.

Since the product is positive and the sum is

negative, both integers must be negative.

-9 ● (-8) = 72 and -9 + (-8) = -17

Therefore, we can use: -9 and -8

(x-9)(x-8)

In order to find the original polynomial, we can multiply the two binomials. Use FOIL.

(x+5) (x-6)

x(x) + x(-6) + 5(x) + 5(-6)

x2 – 6x + 5x - 30

x2 – x – 30 is the original trinomial that factors into (x+5) (x-6)



2. What polynomial when factored gives (x – 8) (x+8)?

3. One factor of the trinomial: x2 + 85x + 750 is (x+10), what is the other factor?

4. One factor of the trinomial: x2 – 10 x – 375 is (x – 25), what is the other factor?

5. Explain why the trinomial: x2 – 7x + 14 cannot be factored.

1. x2 + 16x + 55 2. x2 + 8x + 11

This set of binomials is a Difference of Perfect Squares. Notice that they are opposites.

Therefore, the original trinomial is: x2 – 64.

We are given the trinomial and one of the factors. We know that we must find two numbers whose product

is 750 and sum is 85. One of the numbers is 10. So, 10 times what is 750? 10 ● 75 = 750 AND 10 + 75

= 85. Therefore, the other factor is: (x+75)

We are given the trinomial and one of the factors. We know that we must find two numbers whose product

is -375 and sum is -10. One of the numbers is -25. So, -25 times what is -375? Or, -375/-25 = 15

So, -25 ● 15 = -375 and -25 + 15 = 10 Therefore, the other factor is: (x+15)

We are looking for two integers that have a product of 14 and a sum of -7. The two integers must both be

negative since the product is positive and the sum is negative. Let’s try:

-1 ● -14 = 14 but -1 + -14 = -15; -7 ● -2 = 14 but -7 + -2 = -9 There are no other factors of 14 to try.

Therefore, this trinomial cannot be factored.

Need a Sum of 16 and Product of 55

Since both the sum and product are positive, then

both integers must be positive.

11 ● 5 = 55 and 11 + 5 = 16


(x+ 11)(x+ 5)


Since both the sum and product are positive, then

both integers must be positive.

11 is a prime integer. Only 1 ● 11 = 11.

Since 1 + 11 ≠ 8, we can conclude that this

trinomial is prime and cannot be factored.

There is no way to factor this trinomial.




Factoring – Quiz # 1

Part 1: Complete each problem. Express your answer in simplest terms. (Use the GCF)

1. Factor the following polynomial using the GCF. 3x2 – 18x + 33

2. Factor the following polynomial using the GCF. 6a5 – 12a4 + 15a3

3. Factor the following polynomial using the GCF. 8x2y3z + 4xy4 z3 – 12x4y5z2

4. Simplify the following fraction: ��

��

��

��

5. Simplify the following fraction: 6x2y7z5 + 3x2y3z6 + 2xy3 z8

xy3z5

Part 2: Factor each polynomial.

1. x2 + 9x + 18

2. x2 – x – 20

3. y2 – 9y + 14

4. r2 + 2r – 48

5. z2 – 6z + 5

Part 3: Short Answer.

1. Explain the steps used to factor the following polynomial: 3x4 + 6x2 – 3x

2. Is the following polynomial factorable? Explain why or why not.

x2 + 5x + 5



Factoring Quiz #1 – Answer Key

Part 1: Complete each problem. Express your answer in simplest terms. (1 point each)

1. Factor the following polynomial using the GCF. 3x2 – 18x + 33

2. Factor the following polynomial using the GCF. 6a5 – 12a4 + 15a3

3. Factor the following polynomial using the GCF. 8x2y3z + 4xy4 z3 – 12x4y5z2

4. Simplify the following fraction: ��

��

��

��

Step 1: Identify the GCF for 3x2, 18x and 33. The GCF is 3. Factor a 3 out of each term:

3(x2 – 6x + 11)

Final Answer: 3(x2 – 6x + 11)

Step 1: Identify the GCF for 6a5, 12a4 and 16a3. The GCF is 3a3. Factor 3a3 out of each term.

3a3(2a2 – 4a + 5)

Final Answer: 3a3(2a2 – 4a + 5)

Step 1: Identify the GCF for 8x2y3z, 4xy4 z3, and 12x4y5z2 . The GCF is 4xy3z. Factor 4xy3z out of

each term.

4xy3z(2x+yz2 – 3x3y2z)

Final Answer: 4xy3z(2x+yz2 – 3x3y2z)

Step 1: Identify the GCF for the numerator: The GCF is: 2a2b3. Factor this out of each term:

2a2b3 (4+ 3ab)

2a2b3

Step 2: Simplify: 2a2b3 (4+ 3ab) = 4+3ab

2a2b3

Final Answer: 4+3ab



5. Simplify the following fraction: 6x2y7z5 + 3x2y3z6 + 2xy3 z8

xy3z5

Part 2: Factor each polynomial. (1 point each)

1. x2 + 9x + 18

2. x2 – x – 20

3. y2 – 9y + 14

Step 1: Identify the GCF for the numerator: The GCF is: xy3z5 Factor this out of each term:

xy3z5 (6xy4+3xz+2z3)

xy3z5

Step 2: Simplify: xy3z5 (6xy4+3xz+2z3)

xy3z5

Final Answer: 6xy4 + 3xz + 2z3

We need two factors whose product is 18 and sum is 9.

6 & 3 6●3 = 18; 6+3 = 9

Final Answer: (x+6)(x+3)

We need two factors whose product is -20 and sum is -1. (Since the product is negative, we need a

positive number and a negative number. The larger number should be negative since the sum is

negative.

4 & -5 4●(-5) = -20; 4+(-5) = -1

Final Answer: (x+4)(x-5)

We need two factors whose product is 14 and sum is -9. (Since the product is positive and the sum is

negative, we must have two numbers that are negative.

-7& -2 -7●(-2) = 14; -7+(-2) = - 9

Final Answer: (y-7)(y-2)



4. r2 + 2r – 48

5. z2 – 6z + 5

Part 3: Short Answer. (2 points each)

1. Explain the steps used to factor the following polynomial: 3x4 + 6x2 – 3x

2. Is the following polynomial factorable? Explain why or why not.

x2 + 5x + 5

We need two factors whose product is -48 and sum is 2. (Since the product is negative, we need a

positive and a negative number. The larger number will be positive since the sum is positive.

-6 & 8 -6●8 = -48; -6+8 = 2

Final Answer: (r-6)(r+8)

We need two factors whose product is 5 and sum is -6. (Since the product is positive and the sum is

negative, we must have two number that are negative.)

-5 & -1 -5●(-1) = 5; -5+(-1) = -6

Final Answer: (z-5)(z-1)

First you must look to see if there is a greatest common factor. The GCF is 3x

Then factor out the 3x.

3x(x3 + 2x – 1) This is the final answer.

***Look for key words: greatest common factor***

This polynomial is NOT factorable. We would need to find two positive numbers whose product is 5

and whose sum is 5. The only factors of 5 are 5 and 1. Since 5+1 = 6 and not 5, there is no way to

factor this polynomial using integers.

This quiz is worth 14 points.



Lesson 4: Factoring Trinomials: A Special Situation

Using two different factoring methods.

In the previous lesson, we factored trinomials that had a lead coefficient of 1. Of course,

many trinomials have a lead coefficient that is greater than 1. We will explore one method of

factoring these types of trinomials.

Example 1

Factor: 3x2 – 3x – 36

Notice how the lead coefficient is not 1. However, each term is divisible by 3.

Step 1:

Example 2

Factor: 4x3 + 4x2 – 24x

Note: If a trinomial looks difficult to factor, first check to see if a

Greatest Common Factor (GCF) can be factored out.



Lesson 4: Factoring Trinomials (A Special Situation) Practice

Part 1: Factor Completely

1. 2x2 + 10x – 48 7. 2y5 – 16y4 + 32y3

2. 5y2 – 40y + 60 8. 3x2 - 243

3. 6x3 + 54x2 + 120x 9. 2b3 – 2b2 – 24b

4. 3s3 – 108s 10. 4x3 – 20x2 + 24x

5. 5x4 – 50x3 + 125x2 11. 6w4 – 18w3 – 24w2

6. 2a4 + 12a3 + 18a2 12. 3x2 – 9x - 54


1. Give at least one example of a trinomial in the form of: x2 +bx +c that can be factored.

2. Give at least one example of a trinomial without a lead coefficient of one that can be factored.

3. Give an example of a prime trinomial.

1. 3x3 +18x2 – 81x 2. 2x3 – 8x2 – 8x

Factor each trinomial. (3 points each)



Lesson 4: Factoring Trinomials (A Special Situation) Answers

Part 1: Factor Completely

1. 2x2 + 10x – 48 7. 2y5 – 16y4 + 32y3

2. 5y2 – 40y + 60 8. 3x2 – 243

Step 1: Since the lead coefficient is greater than

1, see if there is a GCF for all terms.

2 is the GCF for all terms.

2(x2 + 5x – 24)

Step 2: Factor the expression inside of

parenthesis.

We need a positive and negative number that

have a product of -24 and a sum of 5

(8 & 3, 8 must be positive since we need

a +5 for a sum).

Final Answer: 2(x+ 8) (x-3)



2y3 is the GCF for all terms.

2y3(y2 – 8y + 16)


parenthesis.

We need 2 negative numbers that have a product

of 16 and a sum of -8.

(-4 & -4).

2y3(y-4) (y-4)

Final Answer: 2y3(y-4)2




5(y2 - 8x + 12)


parenthesis.

We need 2 negative numbers that have a product

of 12 and a sum of -8

(-6 & -2).

Final Answer: 5(y-6) (y-2)




3(x2 - 81)


parenthesis.

This is a difference of two squares. So,

what number squared is 81? (9)

Final Answer: 3(x+ 9) (x-9)



3. 6x3 + 54x2 + 120x 9. 2b3 – 2b2 – 24b

4. 3s3 – 108s 10. 4x3 – 20x2 + 24x

5. 5x4 – 50x3 + 125x2 11. 6w4 – 18w3 – 24w2



6x is the GCF for all terms.

6x(x2 + 9x + 20)


parenthesis.

We need 2 positive numbers that have a product

of 20 and a sum of 9

(4 & 5)

Final Answer: 6x(x+ 4) (x+ 5)



2b is the GCF for all terms.

2b(b2 - b – 12)


parenthesis.

We need a positive and negative number that

have a product of -12 and a sum of -1 (-

4 & 3, 4 must be negative since we need

a -1 for a sum).

Final Answer: 2b(b-4) (b+3)



3s is the GCF for all terms.

3s(s2 - 36)


parenthesis.

This is a difference of two squares. What

number when squared equals 36? (6)

Final Answer: 3s(s+ 6) (s-6)




4x(x2 – 5x+ 6)


parenthesis.

We need two negative numbers that have

a product of 6 and a sum of -5. (-3 & -2)

Final Answer: 4x(x-3) (x-2)



5x2 is the GCF for all terms.

5x2(x2 -10x + 25)


parenthesis.

We need two negative numbers whose

product is 25 and sum is -10 (-5 & -5)

This is a perfect square.

Final Answer: 5x2(x - 5) (x-5) or 5x2(x-5)2



6w2 is the GCF for all terms.

6w2(w2 -3w -4)


parenthesis.

We need a positive and negative number

whose product is -4 and whose sum is -3.

(-4 & 1) 4 must be negative since the

sum is negative.

Final Answer: 6w2(w -4) (w +1)



6. 2a4 + 12a3 + 18a2 12. 3x2 – 9x – 54


1. Give at least one example of a trinomial in the form of: x2 +bx +c that can be factored.

2. Give at least one example of a trinomial without a lead coefficient of one that can be factored.



2a2 is the GCF for all terms.

2a2(a2 + 6a + 9)


parenthesis.

This is a perfect square. What number

can you square and get 9, and add to get

6? (3)

Final Answer: 2a2(a+3)2




3(x2 – 3x - 18)


parenthesis.

What positive and negative numbers have

a product of -18 and a sum of -3? (-6 & 3)

6 must be negative since the sum is

negative.

Final Answer: 3(x -6) (x +3)

These answers may vary. The easiest way to find a trinomial that can be factored is to work backwards

and start with two binomials. You can choose any binomials.

For example, I’ll choose (x+4) (x+2) now multiply using foil.

(x)(x) +(2)x+(4)x+4(2)

x2 +6x + 8 is a trinomial that can be factored. We know this because we started with the two factors and

multiplied.

These answers may vary. The easiest way to find a trinomial that can be factored is to work backwards

and start with two binomials. You can choose any binomials. Since you want a lead coefficient that is

greater than 1, make sure that one of your binomials has a coefficient of x that is greater than 1.

For example, I’ll choose (2x+4) (x+2) now multiply using foil.

(2x)(x) +(2x)2+(4)x+4(2)

2x2 +8x + 8 is a trinomial that can be factored. We know this because we started with the two factors and

multiplied.



3. Give an example of a prime trinomial.

1. 3x3 +18x2 – 81x 2. 2x3 – 8x2 – 8x

A prime trinomial cannot be factored. For example:

x2 + 2x + 2 - We would need two positive numbers whose product is 2 and whose sum is 2. The only way

to make a product of 2 is (2 ●1) Since 2+1 = 3, we know that this cannot be factored.

The best way to make up a prime trinomial is to use a prime number as your constant. Then make sure the

coefficient of x is not the sum of the factors of your prime number.




3x(x2 + 6x - 27)


parenthesis.


whose product is -27 and whose sum is

6. (9 & -3) (9 must be positive since the

sum is positive)

Final Answer: 3x(x+9) (x - 3)




2x(x2- 4x - 4)


parenthesis.


whose product is -4 and whose sum is -4.

The inside of the parenthesis cannot be

factored any further because there’s no

positive and negative number whose

product is -4 and whose sum is -4.

Final Answer: 2x(x2 – 4x – 4)



Lesson 5: Factoring Trinomials in the Form of: ax2 + bx + c

If you are factoring a trinomial with a lead coefficient that is greater than 1 and there is NO GCF to

factor out, then we will need to use a guess and check method in order to factor the trinomial.

Example 1

Example 2

Factor: 2x2 +9x + 4

Factor: 3x2 + 7x - 6



Example 3

Factor: 4x2 -4x - 3



Lesson 5: Factoring Trinomials in the Form of: ax2 + bx + c


1. 3x2 + 5x – 2 6. 4x2 + 4x +1

2. 2x2 – 3x – 2 7. 6x2 -5x - 6

3. 5x2 – 14x – 3 8. 4x2 – x - 3

4. 3x2 + 13x + 4 9. 36x2 + 12x + 4

5. 2x2 – 9x + 4 10. 8x2 – 14x + 6

Part 2: Mixed Factoring! Factor each trinomial completely using ANY method (or combination of

methods.)

1. x2 + 9x + 18 9. 24x2 + 80x + 24

2. 3x3 – 33x2 + 84x 10. 6x3 -24x2 -126x

3. 2x2 + 9x – 5 11. 5x3 -35x2 -40x

4. 10x2 + 5x – 15 12. 18x2 -21x -9

5. x2 –x – 56 13. 10x2 -32x +6

6. 6x2 – 10x – 4 14. 12x3 +36x2 + 27x

7. 3x2 – 33x + 54 15. 4x2 - 1

8. 2x3 + 20x2 + 42x 16. 27x2 -12

1. 6x2 + 3x – 3 3. 24x2 +52x - 20

2. 8x2 +14x +6 4. 18x2 -12x + 2




Lesson 5: Factoring Trinomials in the Form of: ax2 + bx + c - Answers


1. 3x2 + 5x – 2 6. 4x2 + 4x +1

2. 2x2 – 3x – 2 7. 6x2 -5x - 6

For all of the problems in Part 1, the lead coefficient is greater than 1. However, there is no GCF, so we must

use the guess and check method.

Using the model (ax + b)(cx+d), we need:

ac=3 (3 &1 )and bd=-2 (-2 &1 or -1 &2)

(3x-2)(x+1) =3x2 +3x -2x -2 or 3x2 +x - 2

(3x+1)(x-2) = 3x2 – 6x +x -2 or 3x2 -5x - 2

(3x -1)(x+2) = 3x2 +6x – x -2 or 3x2 +5x – 2

(3x+2)(x-1) = 3x2 – 3x + 2x – 2 or 3x2 -x – 2

Final Answer: (3x-1)(x+2)


ac=4 (4 &1 or 2 &2 )and bd=1 (1 &1)

(4x+1)(x+1) = 4x2 +4x+x+1 or 4x2 +5x +1

(2x+1)(2x+1) = 4x2 +2x +2x +1 or 4x2 +4x +1

Final Answer: (2x+1)(2x+1) or (2x+1)2


ac=2 (2 &1 )and bd=-2 (-2 &1 or -1 &2)

(2x-2)(x+1) = 2x2 +2x -2x -2 or 2x2 -2

(2x+1)(x-2) = 2x2 -4x +x -2 or 2x2 -3x -2

(2x -1)(x+2) = 2x2 +4x – x -2 or 2x2 +3x - 2

(2x+2)(x-1) = 2x2 -2x +2x - 2 or 2x2 -2

Final Answer: (2x+1)(x-2)


ac=6 (6 &1 or 3 & 2 )and bd=-6 (-6 &1 or -1 &6

or -3 &2 or -2 &3)

(6x-6)(x+1) = 6x2 +6x -6x -6 or 6x2 -6

(6x-1)(x+6) = 6x2 +36x –x -6 or 6x2 +35x - 6

(6x -3)(x+2) = 6x2 +12x -3x -6 or 6x2 +9x -6

(6x-2)(x+3) = 6x2 +18x -2x -6 or 6x2 +16x -6

(3x-6)(2x+1) = 6x2 +3x-12x -6 or 6x2 -9x -6

(3x-1)(2x+6) = 6x2 +18x -2x -6 or 6x2 +16x - 6

(3x -3)(2x+2) = 6x2 +6x -6x -6 or 6x2 -6

(3x-2)(2x+3) = 6x2 +9x -4x -6 or 6x2 +5x – 6

(3x+2)(2x-3) = 6x2 -9x +4x -6 or 6x2 -5x -5

Final Answer: (3x+2)(2x-3)



3. 5x2 – 14x – 3 8. 4x2 – x – 3

4. 3x2 + 13x + 4 9. 36x2 + 12x + 4


ac=5 (5 &1 )and bd=-3 (-3 &1 or -1 &3)

(5x-3)(x+1) = 5x2 +5x – 3x -3 or 5x2 +2x -3

(5x+1)(x-3) = 5x2 -15x +x -3 or 5x2 -14x -3

(5x -1)(x+3) = 5x2 +15x –x -3or 5x2 +14x - 3

(5x+3)(x-1) = 5x2 -5x +3x -3 or 5x2 -2x -3



ac=4 (4 &1 or 2 & 2)and bd=-3 (-3 &1 or -1 &3)

(4x-3)(x+1) = 4x2 +4x – 3x -3 or 4x2 +x - 3

After multiplying this problem, I notice that I need to

make x a negative x and then I would have

the correct factors. Therefore, if I try

(4x +3) (x-1) I may be able to arrive at my

answer and not have to try all of the other

factors.

(4x+3)(x-1) = 4x2 -4x +3x -3 or 4x2 –x -3



ac=3 (3 &1 )and bd= 4 (2 &2 or 4 &1 or 1 & 4)

(3x+2)(x+2) =3x2 +6x +2x+4 or 3x2 +8x +4

(3x+4)(x+1) = 3x2 +3x +4x +4 or 3x2 7x +4

(3x +1)(x+4) = 3x2 +12x +x+4 or 3x2 +13x +4

Final Answer: (3x+1)(x+4)

The first thing I notice is that there is a GCF that I

can factor out. 4 is the GCF. We then have:

4(9x2 + 3x +1) Now I can factor inside of the

parenthesis:

4(9x2 + 3x +1)


ac=9 (9 &1 or 3 & 3 ) and bd= 1 (1 &1)

Since my sum is 3x, I am not going to start with 9

&1, I will start with 3 & 3.

(3x +1)(3x+1) = 9x2 + 3x +3x +1 or 9x2 + 6x +1

This cannot be factored, so our final answer is:

Final Answer: 4(9x2 + 3x +1)



5. 2x2 – 9x + 4 10. 8x2 – 14x + 6

Part 2: Mixed Factoring! Factor each trinomial completely using ANY method (or combination of

methods.)

1. x2 + 9x + 18 9. 24x2 + 80x + 24


ac=2 (2 &1 )and bd= 4 – Since we have -9x, the

factors of 4 must both be negative. (-2 &-2)(-4&-1)

(-1&-4)

(2x – 2)(x-2) = 2x2 -4x -2x +4 or 2x2 -6x +4

(2x -1)(x-4) = 2x2 -8x –x + 4 or 2x2 -9x +4

I can stop since I have found the factors.

Final Answer: (2x-1)(x-4)

Notice that 2 is the GCF for all 3 terms. We’ll factor

a 2 out first.

2(4x2 – 7x + 3)


ac=4 (4 &1 )(2 & 2)and bd= 3. Since we have -7x,

the factors of 3 must both be negative

(-3 &-1)

(2x -3)(2x-1) = 4x2 -2x – 6x + 3 or 4x2 -8x +3

(4x-3)(x-1) = 4x2 – 4x – 3x + 3 or 4x2 – 7x + 3.

Final Answer: 2(4x-3)(x-1)

Since the lead coefficient is 1, I need to think of

two positive numbers that have a product of 18

and a sum of 9. (6&3)


Since the lead coefficient is greater than 1, I will

first see if there is a GCF for all three terms. 8 is

the GCF, so first we’ll factor out an 8.

8(3x2 + 10x + 3)

The lead coefficient inside of the parenthesis is

still greater than 1, so:


ac=3 (3 &1 ) and bd= 3.(3&1)(1&3)

Factor: 3x2 + 10x + 3

(3x +3) (x+1) = 3x2 + 3x+3x + 3 or 3x2 + 6x + 3

(3x+1)(x+3) = 3x2 + 9x + x + 3 or 3x2 + 10x + 3

Final Answer: 8(3x+1)(x+3)



2. 3x3 – 33x2 + 84x 10. 6x3 -24x2 -126x

3. 2x2 + 9x – 5 11. 5x3 -35x2 -40x


first see if there is a GCF for all three terms. 3x is

the GCF, so first we’ll factor out a 3x.

3x(x2 - 11x + 28)

The lead coefficient inside of the parenthesis is 1,

so: We need two negative numbers whose

product is 28 and sum is -11. (-7 & -4)

Factor: x2 - 11x + 28

(x-7)(x-4)

Final Answer: 3x(x-7)(x-4)



the GCF, so first we’ll factor out a 6x.

6x(x2 - 4x - 21)

The lead coefficient inside of the parenthesis is 1,

so: We need a positive and negative number

whose product is -21 and whose sum is -4. (-7&3

- Since 4 is negative, the largest number (7) must

be negative.

Factor: x2 - 4x -21

(x -7)(x+3)

Final Answer: 6x(x-7)(x+3)


first see if there is a GCF for all three terms. There

is no GCF.


ac=2 (2 &1 ) and bd= -5.(-5&1)(-1&5)

(2x – 5)(x+1) = 2x2 +2x – 5x -5 or 2x2 -3x – 5

(2x – 1)(x +5) = 2x2 + 10x – x -5 or 2x2 +9x -5

Final Answer: (2x-1)(x+5)


first see if there is a GCF for all three terms. The

GCF is 5x, so I will first factor out 5x.

5x(x2 – 7x – 8)

Now the lead coefficient inside of the parenthesis

is 1, so I will factor: x2 – 7x – 8. I need a positive

and negative number whose product is -8 and

whose sum is -7. (-8 & 1) (8 must be negative

since 7 is negative.

(x – 8)(x+1)

Final Answer: 5x (x-8)(x+1)



4. 10x2 + 5x – 15 12. 18x2 -21x -9

5. x2 –x – 56 13. 10x2 -32x +6


first see if there is a GCF for all three terms. The

GCF is 5.

5(2x2 + x – 3)


greater than 1, so:


ac=2 (2 &1 ) and bd= -3.(-3&1)(-1&3)

(2x –3)(x+1) = 2x2 +2x – 3x -3 or 2x2 -x – 3

We need a sum of positive x, so let’s switch the

signs:

(2x+3)(x-1) = 2x2 – 2x + 3x – 3 or 2x2+x - 3

Final Answer: 5(2x+3)(x-1)



the GCF. We’ll first factor out a 3.

3(6x2 – 7x – 3)

Since the lead coefficient inside of the parenthesis

is greater than 1, we need to:

Use the model (ax + b)(cx+d), we need:

ac=6 (2 &3)(6&1) and bd= -3.(-3&1)(-1&3)

Factor: 6x2 – 7x – 3

(3x -3)(2x +1) = 6x2 +3x – 6x – 3 or 6x2 – 3x – 3

(3x+1)(2x-3) = 6x2 – 9x + 2x -3 or 6x2 -7x -3

Final Answer: 3(3x+1)(2x-3)

Since the lead coefficient is 1, we need to find a

positive and negative number whose product is -

56 and whose sum is -1. (-8&7) (Since x is

negative, our largest number (8) must be negative

because -8 + 7 = -1.

(x-8)(x+7)

Final Answer: (x-8)(x+7)



the GCF.

2(5x2 – 16x + 3)


still greater than one, so:


ac=5 (5 &1 ) and bd= 3.(-3&-1)(-1&-3)(Both

factors of 3 must be negative since 16 is negative)

(5x – 1)(x-3) = 5x2 – 15x – x +3 or 5x2 -16x +3

Final Answer: 2(5x-1)(x-3)



6. 6x2 – 10x – 4 14. 12x3 +36x2 + 27x

7. 3x2 – 33x + 54 15. 4x2 – 1



the GCF.

2(3x2 – 5x – 2)


greater than 1, so:


ac=3 (3 &1 ) and bd= -2.(-2&1)(-1&2)

(3x + 1)(x – 2) = 3x2 – 6x + x – 2 or 3x2 – 5x – 2

Final Answer: 2(3x+1)(x-2)



the GCF.

3x(4x2 + 12x + 9)


greater than 1, so:


ac=4 (2 &2)(4&1) and bd= 9.(9&1)(3&3)

I know that 2 x 3 = 6 and 6+6 = 12, so let’s try 2

&2 for ac, and 3 & 3 for bd.

(2x+3)(2x+3) = 4x2 + 6x +6x +9 or 4x2 + 12x + 9

Final Answer: 3x(2x+3)(2x+3) or 3x(2x+3)2



the GCF.

3(x2 – 11x + 18)

Now we will factor inside of the parenthesis. I

need two negative numbers whose product is 18

and sum is -11. (-9&-2)

(x -9)(x-2)

Final Answer: 3 (x-9)(x-2)

Since the middle term is missing, I know that this

is a difference of two squares. What number can

you square to get 4 (2) and what number can you

square to get 1 (1)?

(2x+1)(2x-1)

Multiply to check: 4x2 – 2x + 2x – 1 or 4x2 -1

Final Answer: (2x +1)(2x-1)



8. 2x3 + 20x2 + 42x 16. 27x2 -12

1. 6x2 + 3x – 3 3. 24x2 +52x – 20



the GCF.

2x(x2 + 10x + 21)

Now factor inside of the parenthesis. The lead

coefficient is 1, so we need two positive numbers

whose product is 21 and whose sum is 10. (7&3)

(x+7)(x+3)

Final Answer: 2x (x+7)(x+3)


first see if there is a GCF for both terms. 3 is the

GCF.

3(9x2 – 4)

Inside of the parenthesis we have no middle term,

so we have a difference of two squares. What

can we square to get 9 (3) and what can we

square to get 4 (2)?

(3x +2)(3x-2)

Final Answer: 3(3x+2)(3x-2)



the GCF.

3(2x2 + x - 1)


coefficient is greater than 1, so:


ac=2 (2 &1) and bd= -1.(1&-1)(-1&1)

Factor: 2x2 + x – 1

(2x -1)(x+1)= 2x2 +2x – x -1 or 2x2 + x -1

Final Answer: 3 (2x-1)(x+1)



the GCF.

4(6x2 + 13x – 5)

(Now factor inside of the parenthesis. The lead



ac=6 (6 &1)(3&2) and bd= -5.(5&-1)(-1&5)

Factor: 6x2 + 13x - 5

(3x -1)(2x + 5) = 6x2 + 15x -2x – 5 or 6x2 + 13x -5

Final Answer: 4(3x-1)(2x+5)




2. 8x2 +14x +6 4. 18x2 -12x + 2



the GCF.

2(4x2 +7 x +3)




ac=4 (4 &1)(2&2) and bd= 3.(1&3)(3&1)

Factor: 4x2 +7 x+ 3

(4x +3)(x+1) = 4x2 +4x +3x+3 or 4x2 +7x +3

Final Answer: 2(4x+3)(x+1)



the GCF.

2(9x2 -6x +1)




ac=9 (9 &1)(3&3) and bd= 1.(-1&-1)

Factor: 9x2 – 6x +1

(3x -1)(3x-1) = 9x2 -3x -3x +1 or 9x2 – 6x +1

Final Answer: 2 (3x-1)(3x-1) or 2(3x-1)2



Unit 9: Factoring Chapter Test

Part 1: Factor Completely. Remember to factor out the greatest common factor first if it is other

than 1.

1. x2 + 12x + 32 7. 3y2 + 21y + 36

2. y2 – 2y – 35 8. 2x3 – 18x

3. n2-8n + 16 9. 12x4y2 + 18x3y3 – 24x2y4

4. x2 – 13x + 36 10. 6x2 + 5x - 4

5. 8x4 + 2x2 – 4x 11. 4p2 – 7p - 2

6. 2x3 + 6x2 – 20x 12. 4x2 + 10x – 6

Part 2: Answer each problem completely.

1. What polynomial when factored gives: (3x+ 2y)( 2x – 5y)

2. One factor of the polynomial x2 + 27x + 180 is (x+15). What is the other factor?

3. Find two values for b. Explain how you determined your answer.

x2 + bx – 12

4. Find a value for c that will make this a perfect square trinomial.

x2 + 18x + c



Unit 9: Factoring Chapter Test – Answer Key

Part 1: Factor Completely. Remember to factor out the greatest common factor first if it is other

than 1. (1 point each)

1. x2 + 12x + 32 7. 3y2 + 21y + 36

2. y2 – 2y – 35 8. 2x3 – 18x

3. n2-8n + 16 9. 12x4y2 + 18x3y3 – 24x2y4

We need to find two positive integers whose

product is 32 and whose sum is 12.

8 & 4 8●4 = 32 8+4 = 12


Since the lead coefficient is greater than1, we

need to check to see if there is a GCF that can

be factored out. The GCF is 3.

3(y2 + 7y + 12)

Now we need to factor the polynomial inside of

the parenthesis. We need two positive

numbers whose product is 12 and whose sum

is 7. 4&3

Final Answer: 3(x+4)(x+3)

We need to find two integers whose product is

-35 and whose sum is -2. Since the product is

negative, one integer is positive and one

negative. The larger integer is negative since

the sum is negative.

-7 & 5 -7●5 = -35 -7+5 = -2

Final Answer: (y-7)(y+5)

Since the lead coefficient is greater than1, we

need to check to see if there is a GCF that can

be factored out. The GCF is 2x.

2x(x2 - 9)

Now we need to factor the polynomial inside of

the parenthesis. Since there is no middle term

and 9 is a perfect square, we know this factors

as a difference of two squares:

Final Answer: 2x(x+3)(x-3)


16 and whose sum is -8. Since the product is

positive and the sum is negative, both integers

must be negative.

-4 & -4 -4●(-4) = 16 -4+(-4) = -8

Final Answer: (n-4)(n-4) or (n-4)2

First we need to see if there is a GCF. The

GCF is 6x2y2. Factor out 6x2y2.

6x2y2 (2x2+ 3xy – 4y2)

There is no way to factor inside of the

parenthesis. Therefore:

Final Answer: 6x2y2 (2x2+ 3xy – 4y2)



4. x2 – 13x + 36 10. 6x2 + 5x – 4

5. 8x4 + 2x2 – 4x 11. 4p2 – 7p – 2

6. 2x3 + 6x2 – 20x 12. 4x2 + 10x – 6


36 and whose sum is -13. Since the product is

positive and the sum is negative, both integers

must be negative.

-9 & -4 -9●(-4) = 36 -9+(-4) = -13

Final Answer: (x-9)(x-4)

Since the lead coefficient is greater than 1, look to

see if there is a GCF. There is no GCF for this

polynomial, so we must use the guess and check

method for factoring.

Factors of 6: 6,1 & 3,2

Factors of -4: (2,-2) & (-4,1) & (-1,4)

(2x -1) (3x+4) = 6x2 + 8x – 3x – 4 or 6x2 +5x -4

Final Answer: (2x -1) (3x+4)

First we need to check to see if there is a GCF

that can be factored out. 2x is the GCF.

2x(4x3 + x – 2)

The inside of the parenthesis cannot be

factored any further, therefore:

Final Answer: 2x(4x3 + x – 2)


see if there is a GCF. There is no GCF for this

polynomial, so we must use the guess and check

method for factoring.

Factors of 4: 2,2 & 4,1

Factors of -2: (1,-2) & (-2,1)

(4p +1) (p-2) = 4p2 – 8p + p – 2 or 4p2 -7x -2

Final Answer: (4p +1) (p-2)

First we need to check to see if there is a GCF

that can be factored out. 2x is the GCF.

2x(x2 + 3x - 10)

Now we must factor the inside of the

parenthesis. We need a positive and negative

integer whose product is -10 and whose sum is

3. The larger integer will be positive since the

sum is positive. 5 & -2

Final Answer: 2x(x+5)(x-2)


see if there is a GCF. 2 is the GCF.

2(2x2 + 5x – 3)

Now factor the inside of the parenthesis using the

guess and check method.

Factors of 2: 2,1

Factors of -3: (1,-3) & (-3,1)

2(2x-1) (x+3) =2(2x2 +6x - x – 3) or 2(2x2 +5x -3)

Final Answer: 2(2x-1)(x+3)



Part 2: Answer each problem completely. (2 points each)

1. What polynomial when factored gives: (3x+ 2y)( 2x – 5y)

2. One factor of the polynomial x2 + 27x + 180 is (x+15). What is the other factor?

3. Find two values for b. Explain how you determined your answer.

x2 + bx – 12

4. Find a value for c that will make this a perfect square trinomial.

x2 + 18x + c

We can multiply to figure out the polynomial: Use the foil method.

(3x+2y)(2x-5y)

3x(2x) + 3x(-5y) + 2y(2x) + 2y(-5y) Use FOIL

6x2 – 15xy + 4xy – 10y2 Multiply the terms

6x2 – 11xy – 10y2 Simplify the middle terms.

Final Answer: 6x2 – 11xy – 10y2

We know that our two chosen integers must have a product of 180 and a sum of 27. Since we know

that one factor is (x+15), we know that one of our integers is 15. The other integer must be 12 since

180/15 = 12. Also, 15+ 12 = 27, so this confirms our choice.

Therefore the other factor is: (x+ 12)

Final Answer: (x+12)

We know that we need two integers, one positive and one negative, whose product is -12. The sum

of these two integers is the variable b in the polynomial.

Integers whose product is -12: (-4)(3); (-3)(4); (-6)(2); (-2)(6); (-12)(1); (12)(-1)

b could be equal to: -1, 1, -4, 4,-11, 11 (If you add the integers together, you will end up with their

sum, which is the variable b.)

Letter C must be a perfect square. The square root of this number multiplied by 2 is 18. The only

number that can represent letter c is 81. The factors would be (x+9)2

This test is worth 20 points



Cumulative Test on Polynomials and Factoring

Part 1: Complete each problem by circling the correct answer.

For problems 1-8, simplify the expression. State the answer in standard form.

1. (3x2+8x – 11) + (2x2 – 6x +4)

A. 5x2 + 14x + 15 C. 5x2 + 2x – 7

B. 5x2 + 2x + 7 D. 5x2 + 2x – 15

2. (4x4 + 2x2 – 3x + 1) + (3x3 – 2x2 + 8x – 6)

A. 4x4 + 3x3 + 5x – 5 C. 4x4 + 3x2 + 4x2 – 3x +8x – 5

B. 7x7 + 5x – 5 D. 4x4 + 3x3 + 2x2 + 5x – 5

3. (6x2 – 3x + 9) – (4x2 + 8x – 6)

A. 2x2 + 5x + 3 C. 2x2 + 11x + 15

B. 2x2 – 11x + 15 D. 10x2 + 5x + 3

4. 3(2x2 – 4x +2) – 2(x2 + 4)

A. 4x2 – 12x + 14 C. 8x2 – 12x – 2

B. 8x2 – 12x + 14 D. 4x2 – 12x – 2

5. (5r-3)2

A. 25r2 – 9 C. 5r2 + 9

B. 25r2 – 30r + 9 D. 25r2 + 9

6. (x+6)(x-6)

A. x2 + 12x – 36 C. x2 + 36

B. x2 – 36 D. x2 -6x – 36



7. (3x-2)(2x+1)

A. 6x2 + 7x – 2 C. 6x2 – x + 2

B. 6x2 + x – 2 D. 6x2 – x – 2

8. (x+4)2(3x-2)

A. 3x3 + 22x2 +32x – 32 C. 3x2 + 10x – 8

B. 3x3 – 2x2 + 48x – 32 D. 3x3 + 26x2 + 48x – 32

For problems 9-10, factor the polynomial.

9. x2 -3x -70

A. (x+10)(x-7) C. (x+7)(x-10)

B. (x+5)(x-14) D. (x+14)(x-5)

10. 6x2 – 2x – 8

A. (3x-4)(2x+2) B. (3x+2)(2x-4)

C. (2x-2)(3x+4) D. (2x+4)(3x-2)

Part 2: Short Answer

1. A square table measures 3x +2y units. Write an expression in simplified form that represents the

area of the table.

2. A rectangular sheet of paper has a width of x units. The area of the paper is x2+8x. Write an

expression that represents the length of the rectangular sheet of paper.

3. Factor the polynomial completely.

2x2 + 11x + 12




9x3 +6x2 -48x


y2 – 121

6. A rectangle has the following dimensions: width: (2x + y) length: (5x – 2y)

Express the area of the rectangle as a polynomial in standard form.

7. A triangle has the following dimensions: Side 1: x2 +2x -1 Side 2: x2 + 2x -1

Side 3: 2x2 +x – 3

Write an expression that represents the perimeter of the triangle.

8. A rectangle has a length of 8x -4 and a width of 4x +3. The perimeter of the rectangle is 58 units.

Find the width of the rectangle.



Part 3: Extended Response.

1. You have a rectangular piece of wood in which

you are making a carnival game. You plan

to cut out 2 circles that children can use to try to

throw a bean bag through at the carnival.

Complete each bullet below given

the following dimensions:

Length of rectangular piece of wood: (5x+2)

Width of rectangular piece of wood: (3x+1)

Radius of each circle: (x+1)

• Write an algebraic expression for the area of the

rectangular piece of wood before the circles are cut.

• Write an algebraic expression for the area of 1 circle.

• Write an algebraic expression for the area of the wood after both circles are cut out. Explain

how you determined your answer.



Cumulative Test on Polynomials and Factoring - Answer Key

Part 1: Complete each problem by circling the correct answer. (1 point each)

For problems 1-8, simplify the expression. State the answer in standard form.

1. (3x2+8x – 11) + (2x2 – 6x +4)

A. 5x2 + 14x + 15 C. 5x2 + 2x – 7

B. 5x2 + 2x + 7 D. 5x2 + 2x – 15

2. (4x4 + 2x2 – 3x + 1) + (3x3 – 2x2 + 8x – 6)

A. 4x4 + 3x3 + 5x – 5 C. 4x4 + 3x2 + 4x2 – 3x +8x – 5

B. 7x7 + 5x – 5 D. 4x4 + 3x3 + 2x2 + 5x – 5

3. (6x2 – 3x + 9) – (4x2 + 8x – 6)

A. 2x2 + 5x + 3 C. 2x2 + 11x + 15

B. 2x2 – 11x + 15 D. 10x2 + 5x + 3

Write the expression vertically. 3x2 + 8x – 11

+ 2x2 – 6x + 4

5x2 +2x - 7

Write the expression vertically. 4x4 + 0x3 +2x2 – 3x + 1

+ 3x3 -2x2 + 8x – 6

4x4 + 3x3 +0x2 +5x - 5

Step 1: Rewrite as an addition problem. (6x2 – 3x + 9) + (-4x2 – 8x + 6)

Write the new expression vertically.

6x2 – 3x + 9

+ -4x2 – 8x + 6

2x2 – 11x + 15



4. 3(2x2 – 4x +2) – 2(x2 + 4)

A. 4x2 – 12x + 14 C. 8x2 – 12x – 2

B. 8x2 – 12x + 14 D. 4x2 – 12x – 2

5. (5r-3)2

A. 25r2 – 9 C. 5r2 + 9

B. 25r2 – 30r + 9 D. 25r2 + 9

6. (x+6)(x-6)

A. x2 + 12x – 36 C. x2 + 36

B. x2 – 36 D. x2 -6x – 36

Step 1: Distribute the 3 throughout the first set of parenthesis and the -2 throughout the second set of

parenthesis:

(6x2 – 12x + 6) + (-2x2 – 8)

Step 2: Write the new expression vertically.

6x2 – 12x + 6

+ -2x2 +0x -8

4x2 – 12x -2

Step 1: Use the Difference of a Square rule.

(a-b)2 = a2 – 2ab + b2

(5r-3)2 = (5r)2 – 2(5r)(3) + 32

(5r-3)2 = 25r2 – 30r +9

Step 1: Use the Difference of Two Square Rule:

(a+b)(a-b) = a2 – b2

(x+6)(x-6) = x2 – 62

(x+6)(x-6) = x2 - 36



7. (3x-2)(2x+1)

A. 6x2 + 7x – 2 C. 6x2 – x + 2

B. 6x2 + x – 2 D. 6x2 – x – 2

8. (x+4)2(3x-2)

A. 3x3 + 22x2 +32x – 32 C. 3x2 + 10x – 8

B. 3x3 – 2x2 + 48x – 32 D. 3x3 + 26x2 + 48x – 32

Step 1: Using FOIL multiply the two binomials.

(3x-2)(2x+1) = 3x(2x) + 3x(1) + (-2)(2x)+ (-2)(1)

6x2 + 3x – 4x – 2

Step 2: Combine like terms. 6x2 – x – 2

Final Answer: 6x2 – x - 2

Step 1: Using the Square of A sum rule, multiply (x+4)2

(x+4)2 = x2 + 8x + 16 (a2 + 2ab + b2)

Step 2: Multiply (3x-2) (x2 + 8x + 16) Use the extended distributive property.

3x(x2) + 3x(8x) + 3x(16) + (-2)(x2) + (-2)(8x) + (-2)(16)

3x3 + 24x2 + 48x – 2x2 - 16x – 32

Step 3: Rewrite with like terms together.

3x3 + 24x2 – 2x2 + 48x – 16x – 32

Step 4: Combine like terms: 3x3 + 22x2 + 32x - 32



For problems 9-10, factor the polynomial.

9. x2 -3x -70

A. (x+10)(x-7) C. (x+7)(x-10)

B. (x+5)(x-14) D. (x+14)(x-5)

10. 6x2 – 2x – 8

A. 2(3x-4)(x+1) B. 2(x-2) (3x+2)

C. 2(x-1)(3x+4) D. 2(x+2)(3x-2)

Since the lead coefficient is 1, we need to think of two numbers that have a product of -70 and a sum

of -3. One integer must be positive and one must be negative. The larger integer will be negative

since the sum is -3.

**Since this is multiple choice – we should take a look at the answers given. The answers are using

factors of 10 and 7 and 5 and 14. Of those two choices, let’s see which one works.

***I know that the larger integer must be negative, since the sum is -3; therefore, I can eliminate A

because the larger integer, 10 is positive. For the same reason, I can eliminate D. Let’s try letter C.

We’ll multiply: (x+7)(x-10) = x2 – 10x + 7x – 70

x2 – 3x – 70 This is the answer we need; therefore, letter C is the correct choice.

Since this is multiply choice, we should use the answers to our advantage. It may be best to guess

one of the answer choices below and check by multiplying. We would have to use

the guess and check strategy anyway to factor this polynomial since the lead

coefficient is greater than 1.

A: 2(3x-4)(x+1)

(6x – 8)(x+1) - Distribute the 2 throughout the first parenthesis

6x2 +6x – 8x -8 = 6x2 – 2x – 8 Our first try works!!!



Part 2: Short Answer (2 points each)

1. A square table measures 3x +2y units. Write an expression in simplified form that represents the

area of the table top.

2. A rectangular sheet of paper has a width of x units. The area of the paper is x2+8x. Write an

expression that represents the length of the rectangular sheet of paper.

A square table has four sides with the same measurement (3x+2y).

Area of a square is: A = l● w. or A = s2

A = (3x+2y)2

We can use the Square of a Sum rule in order to multiply and simplify this expression.

(a+b)2 = a2 + 2ab +b2

(3x+2y)2 = (3x)2 + 2(3x)(2y) + (2y)2

A = 9x2 + 12xy + 4y2

width= x length = ? Area = x2 + 8x

The formula for area of a rectangle is: A = lw

Substitute our given information into this formula:

x2 + 8x = Lx Now we need to solve for L by dividing by x.

x2 + 8x /x = l/x

x2 + 8x = L

x

An x can be factored out of the numerator: x(x+ 8) = L

X

Now we can simplify the fraction: x(x+ 8) = L

X

x + 8 = L We end up with an expression of x + 8 = L

Let’s check:

A = lw A = (x+8)(x)

A = x2 + 8x This expression matches the original problem; therefore, our answer is correct

****You may also, have solved

this problem mentally by

thinking:

x(?) = x2 + 8x What can I

multiply by x in order to get

x2+ 8x?

x(x +8) = x2 + 8x

Length = x+8




2x2 + 11x + 12


9x3 +6x2 -48x


y2 – 121

Factors of 2: (2●1)

Factors of 12: (12 ●1) (6●2) (3●4)

Guess and check: We need the inside terms to add up to 11.

(2x + 6)(x+2) = 2x2 + 4x + 6x + 12 = 2x2+ 10x + 12

(2x + 3)(x+4) = 2x2 + 8x + 3x + 12 = 2x2 + 11x + 12

Final Answer: (2x+3)(x+4)

First factor out the GCF which is 3x.

3x(3x2 + 2x – 16)

Now factor: 3x2 + 2x – 16 Factors of 3: 3,1 Factors of -16: -4(4), -8(2) or -2(8)

We must use the guess and check method:

3x (3x+8) (x-2)

Let’s check: (3x+8) (x-2) = 3x2 – 6x + 8x – 16 = 3x2 + 2x – 16 (Yeah ! This worked)

Final answer: 3x(3x+8)(x-2)

Since y2 and 121 are both perfect squares and there is no middle term, I recognize this product as a

Difference of Two Squares. I know that 112 = 121. Therefore, my factors are:

(y+11)(y-11)

Final Answer: (y+11)(y-11)



6. A rectangle has the following dimensions: width: (2x + y) length: (5x – 2y)

Express the area of the rectangle as a polynomial in standard form.

7. A triangle has the following dimensions: Side 1: x2 +2x -1 Side 2: x2 + 2x -1

Side 3: 2x2 +x – 3

Write an expression that represents the perimeter of the triangle.

8. A rectangle has a length of 8x -4 and a width of 4x +3. The perimeter of the rectangle is 58 units.

Find the width of the rectangle.

The formula for area of a rectangle is: A = lw

A = (2x+y)(5x-2y) Use FOIL to multiply.

A = 2x(5x) +2x(-2y) + y(5x) + y(-2y)

A = 10x2 – 4xy + 5xy – 2y2

A = 10x2 + xy – 2y2

Final Answer: 10x2 + xy – 2y2

To find the perimeter of the triangle, we must add all sides together. I will set it up vertically.

x2 + 2x – 1

x2 + 2x – 1

+ 2x2 + x – 3

4x2 + 5x – 5

The perimeter of the triangle is: 4x2 + 5x - 5

The formula for perimeter of a rectangle is: P = 2L + 2w

P = 2(8x-4) + 2(4x+3)

P = 16x – 8 + 8x + 6 Distribute.

P = 16x + 8x – 8 + 6 Write like terms together.

P = 24x – 2 Expression that represents the perimeter.

58 = 24x – 2 Now substitute 58 for P since we know the perimeter is 58 units.

58 +2 = 24x -2 + 2 Add 2 to both sides

60 = 24x

60/24 = 24x/24 Divide by 24 on both sides.

2.5 = x This is the value for x

4(2.5) + 3 = 13 Substitute 2.5 for x into the expression for width (4x+3) The width is 13 units.



Part 3: Extended Response. (4 points)

1. You have a rectangular piece of wood in which

you are making a carnival game. You plan

to cut out 2 circles that children can use to try to

throw a bean bag through at the carnival.

Complete each bullet below given

the following dimensions:

Length of rectangular piece of wood: (5x+2)

Width of rectangular piece of wood: (3x+1)

Radius of each circle: (x+1)

• Write an algebraic expression for the area of the

rectangular piece of wood before the circles are cut.

• Write an algebraic expression for the area of 1 circle.

Area of a rectangle: A = lw

A = (5x+2)(3x+1)

A = 15x2 + 5x + 6x + 2 or A = 15x2 + 11x + 2 Use FOIL to multiply

The Area of the Wood before circles are cut is: 15x2 + 11x + 2

The formula for area of a circle is: A = πr2

A = π (x+1)2

A = π (x2 + 2x+ 1) Use square of a sum rule to multiply the expression.

The area of one circle is: π (x2 + 2x+ 1) or 3.14(x2 + 2x + 1)



• Write an algebraic expression for the area of the wood after both circles are cut out. Explain

how you determined your answer.

Step 1: We need to find the expression that represents the total area of both circles that will be cut

out. Since the area of 1 circle is: 3.14(x2 + 2x + 1) , we need to multiply this by 2.

3.14(2)(x2 + 2x +1) or 6.28(x2 + 2x + 1)

Step 2: Let’s distribute the 6.28

6.28x2 + 12.56x + 6.28

This represents the total area of both circles that will be cut out.

Step 3: Subtract the area of the circles from the total area of the wood.

15x2 + 11x + 2 – (6.28x2 + 12.56x + 6.28)

Step 4: Rewrite as an addition problem.

15x2 + 11x + 2 + (-6.28x2 - 12.56x - 6.28)

Step 4: Add. Set up vertically.

15x2 + 11x + 2

+ -6.28x2 – 12.56x – 6.28

8.72x2 – 1.56x - 4.28

The expression that represents the area of the board after two circles are cut is:

8.72x2 – 1.56x - 4.28. I first multiplied the expression that represents the area of the circle

by 2 since there are two circles. Then I subtracted the expression that represents the area of

two circles, from the area of the rectangle.

This test is worth 30 points

lesson 1: introduction to factoring - algebra class e-course 9 factoring.pdf · the gcf for 9 and...

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