e:/cloud/google drive/latex/teaching notes/math-08x …€¦ · 6.1 the greatest common factor;...

Chapter 6

Chapter 6

6.1 The Greatest Common Factor; Factoring byGrouping6.2 Factoring Trinomials of the Form x2 + bx + c

6.3 Factoring Trinomials of the Form ax2 + bx + c

6.4 Factoring Perfect-Square Trinomials and theDifference of Two Squares6.5 Factoring the Sum and Difference of Two Cubes(SKIP)6.6 A Factoring Strategy6.7 Solving Quadratic Equations by Factoring6.8 Application of Quadratic Equations

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6.1 The Greatest Common Factor; Factoring by Grouping Chapter 6

ConceptsGCF - Greatest Common Factor - the largest factor in all terms.

Recall form Ch 5 we learned to distribute2x2(3x− 5) = 6x3 − 10x2

Now as a class, let’s do this the other way 3x(2x)− 3x(1) = 3x(2x− 1)

Factor the following:

Example 2) 9y + 18y3 = 9y(1 + 2y2)

Example 3)5

2x7 −

3

2x5 +

5

2x3 =

1

2x3(5x4 − 3x2 + 5)

Example 4) 5x(x− 1)− (x− 1) = (5x− 1)(x− 1)

Now you factor the following:

d) (x− 5) + (x− 5)4x3 =

e) 5x(x + 1)− 7y(x + 1) =

f) x2(x + 1) + x(x + 1)− 2(x + 1) =

g) (3x2 + 1)− (3x2 + 1)x2 − 3(3x2 + 1)x4 =

h) x2(2x− 1) + 5x(2x− 1)− 7(2x− 1) =

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6.1 The Greatest Common Factor; Factoring by Grouping Chapter 6

Factoring by Grouping - USED for FOUR TERMED POLYNOMIALS!

Example 1)

2x3 − 4x2 + 3x− 6 = (2x3 − 4x2) + (3x− 6)= 2x2(x− 2) + 3(x− 2)= (2x2 + 3)(x− 2)

Example 2)

4y3 + 10y2 + 2y + 5 =2y2(2y + 5) + 1(2y + 5) =(2y2 + 1)(2y + 5)

You try to factor the following:

a) 15z3 − 5z2 + 6z − 2 =

b) 3x3 + 12x2 − 2x− 8 =

c) y3 − 5y2 − 3y + 15 =

Factor the expression. Then make a sketch of a rectangle that illustrates this factoriza-tion.

6 + 3x =

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6.2 Factoring Trinomials of the Form x2 + bx + c Chapter 6

Recall from Ch 5 the FOIL method of multiplying binomials

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

x2 + 5x + 6

+

F

OI

L

(x + n)(x + m) = x2 + mx + nx + mn = x2 + (m + n)x + mn

So the middle term is the SUM of the factors. While the last term is the PRODUCT

of the factors.

Example 1) Factor x2 − 6x− 40 factors of 40 Difference1 40 392 20 184 10 6

(x− 10)(x + 4)

Example 2) Factor x2 − 15x + 50 factors of 50 Sum1 50 512 25 275 10 15

(x− 10)(x− 5)

Example 3) Factor x2 + 4x− 21 factors of 21 Difference1 21 203 7 4

(x + 7)(x− 3)

You try to factor the following:

a) x2 + 7x + 12 =

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b) x2 + 2x− 15 =

c) x2 − 3x− 28 =

d) x2 − 10x + 24 =

e) 55− 16x + x2 =

f) −39− 10x + x2 =

A square has an area of x2 + 8x + 16 Find the length of a side and make a sketch of thesquare.

length of a side =

Create a sketch below

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A box is constructed by cutting out square corners of a rectangular piece of cardboardand folding up the sides. If the cutout corners have sides with length x, then the volume ofthe box is given by the polynomial 4x3 − 40x2 + 96x

Factor the polynomial 4x3 − 40x2 + 96x

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6.3 Factoring Trinomials of the Form ax2 + bx + c Chapter 6

3 ways to factor1) FOIL method2) Special Factoring formulas

a) Difference of Squaresb) Perfect Square Trinomialsc) Sum and Difference of Cubes

3) Master Product method (grouping/ac method)

FOIL - (Trial and Check) method is methodical and will work - if the trinomial isfactorable.

Special Factoring formulas - memorized shortcuts for certain factoring forms.Master Product method - method is methodical as well - definite starting and ending

point.

Helpful hints:1) Always write in descending order.2) Always factor out the GCF3) If the x2 has a negative sign on its coefficient, factor out a −1, then use the factor

method that you prefer.

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FOIL Method

Example 1) Factor 15x2 − 13x− 20

First we know that the signs of the factors must be opposite because of the −20. Sowrite the factors as ( x− )( x + )

I now need to list the Factors of 15 and of 20

Factors of 15 Factors of 201 - 15 1 - 203 - 5 2 - 10

4 - 5

Try combinations of 3 - 5 with all combinations of the factors of 20. If none work thentry combinations of 1 - 15 with all combinations of the factors of 20. If you get a middleterm of +13 then the signs of the constant factors are reversed.

I will start with 3 - 5 and do all possible combinations of factors of 20.

(3x− 1)(5x + 20) =

(3x− 20)(5x + 1) =

(3x− 10)(5x + 2) =

(3x− 5)(5x + 4) =

If the above do not work then use factors 1 - 15 and do all possible combinations offactors of 20.

(x− 20)(15x + 1) =

(x− 10)(15x + 2) =

(x− 5)(15x + 4) =

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Example 2) Factor: 5x2 − 18x− 8 =

First we know that the only factors of 5 are 1 and 5. We also know that the signs of thefactors must be opposite because of the −8. So write the factors as (5x− )(x + ). Atthis point we don’t know if the signs are correct. If they aren’t, all we do is switch them.

Factors of 81 - 82 - 4

Let’s try combinations of 2 - 4, if they don’t work then try combinations of 1 - 8

(5x− 4)(x + 2) =

(5x− 2)(x + 4) =

(5x− 1)(x + 8) =

(5x− 8)(x + 1) =

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Factor the following using the FOIL Method.

a) 24x2 − 17x− 20 =

b) 36x2 − 41x− 5 =

c) 36x2 − 72x + 35 =

d) 49x2 − 4 =

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Master Product method (Also know as ac method or the grouping method.)

Example 1) 10x2 + 19x− 15 =

The a is 10 and the c is −15.

Multiply the coefficients of the first (a) and last (c) term to get (10)(−15) = −150

As this product is negative find all factors of −150 whose difference is +19

If the product is positive then you find the sum!

Calculate the ac product (10)(−15) = −150

First factor Second Factor Difference−1 150 149−2 75 73−3 50 47−4 30 25−6 25 19

So we will use −6 and +25 for the middle term

10x2 + 19x− 1510x2+25x− 6x− 15

Now use the factor by grouping technique!

(10x2 + 25x) + (−6x− 15)5x(2x + 5)− 3(2x + 5)

(5x− 3)(2x + 5)

Example 2) Factor 5n2 + 19n + 12

Calculate the ac product 5 • 12 = 60

First factor Second Factor Sum1 60 612 30 323 20 234 15 19

So we will use 4 and 15 for the middle term

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5n2 + 19n + 125n2+15n + 4n + 12(5n2 + 15n) + (4n + 12)5n(n + 3) + 4(n + 3)(5n + 4)(n + 3)

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Factor the following using the Master Product method.

a) 12x2 − 37x + 21 =

b) 10x2 + 21x + 9 =

c) 28x2 + 45x− 7 =

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6.4 Factoring Perfect-Square Trinomials and the Difference of Two Squares Chapter 6

Special Types of Factoring - Perfect Square Trinomials

Recall: (a+b)2 = (a+b)(a+b) = a2 +2ab+b2 and (a−b)2 = (a−b)(a−b) = a2−2ab+b2

We are going to go from a2 + 2ab + b2 to (a + b)2

Example 1) Factor x2 + 20x + 100

√x2 = x√100 = 10

and +2 • x • 10 = +20x← this is the middle term.

So, x2 + 20x + 100 = (x + 10)2

Example 2) Factor x2 − 6x + 9

√x2 = x√9 = 3

and −2 • x • 3 = −6x← this is the middle term.

So, x2 − 6x + 9 = (x− 3)2

a) x2 + 4x + 4 =

b) z2 − 18z + 36 =

c) 16x2 + 8x + 1 =

d) 25x2 − 60x + 36 =

e) 7x2 + 14x + 7 =

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6.4 Factoring Perfect-Square Trinomials and the Difference of Two Squares Chapter 6

Special Types of Factoring - Difference of Squares

Recall (a− b)(a + b) = a2 − b2

So we will now go the other way.

Example 1) x2 − 4 = (x− 2)(x + 2)

Example 2) x4 − 16 = (x2 − 4)(x2 + 4) = (x− 2)(x + 2)(x2 + 4)

a) x2 − 52 =b) x4 − 81 =

c) z2 − 1 =d) y2 − 16 =

Is x2 − 7 factorable?

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6.5 Factoring the Sum and Difference of Two Cubes (SKIP) Chapter 6

Sum and Difference of Cubes

For any real number a and b

a3 + b3 = (a + b)(a2 − ab + b2) and a3 − b3 = (a− b)(a2 + ab + b2)

(F + L)(FF − FL + LL) and (F − L)(FF + FL + LL)

a) x3 − 73 =

b) 27x3 + 8 =

c) 750x3 − 162 =

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6.6 A Factoring Strategy Chapter 6

Factoring Strategy

1) Factor out the GCF

2) Determine the Number of Terms

3) 4 terms - Factor by grouping

4) 3 terms (ax2 + bx + c)

Is it a perfect square?

Yes - Perfect square trinomial

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

a2 − 2ab + b2 = (a− b)(a− b)

No - Use FOIL or Master Product Method (ac Method)

5) 2 termsDo I have squares or cubes

Squaresi. Sum - Not factorable at this levelii. Difference - a2 − b2 = (a− b)(a + b)

Cubesi. Sum - a3 + b3 = (a + b)(a2 − ab + b2)ii. Difference - a3 − b3 = (a− b)(a2 + ab + b2)

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Example 1) Factor the following using the Factoring Strategy.

4x2 + 28x + 244(x2 + 7x + 6)4(x + 1)(x + 6)

Example 2)

20x2 − 13x + 2( x− 1)( x− 2)(4x− 1)(5x− 2)

Factor the following using the Factoring Strategy.

a) 2x2 + 16x + 24 =

b) 9x2 + 15x− 14 =

c) 8x2 + 14xy − 49y2 =

d) 16x3 + 20x2 + 4x =

e) x3 − x− 3x2 + 3 =

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f) 32x2 + 108x + 81 =

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6.7 Solving Quadratic Equations by Factoring Chapter 6

The zero-product property (zero factor theorem)

For all real numbers a and b, if ab = 0 then a = 0 or b = 0 or both are zero.

We can use this property to solve equations

(x + 2)(x− 5) = 0

by the zero-product property we have

(x + 2) = 0 or (x− 5) = 0x = −2 or x = 5

If we substitute the values into the original equation will I get zero?

Yes.

21x2 − 3x = 0 Factor out the GCF3x(7x− 1) = 03x = 0 or 7x− 1 = 0x = 0 or 7x = 1

x = 0 or x =1

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SOLVING QUADRATIC EQUATIONS

1) If necessary, rewrite the equation in descending order. Factor out a −1 if the coefficientof x2 is negative. (If the GCF is larger then 1 then factor out a negative GCF)

2) Factor the equation - using the method of YOUR choice.3) Apply the zero-product property.4) Solve the resulting linear equations

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Solve the following by any method.

a) 9x2 − 4 = 0

b) 10x2 = 5x

c) 2x2 + 3x = 14

d) x2 − x + 1 = 0

e) x(x− 7) = −12

f) x(3x + 2) = 5

g) 4x2 − 16x− 20 = 0

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h) x3 − x2 + 4x− 4 = 0

i) 3x3 + 2x2 − 27x− 18 = 0

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6.8 Application of Quadratic Equations Chapter 6

SOLVING APPLICATIONS OF QUADRATIC EQUATIONS

Problem types:GeometricPythagorean TheoremConsecutive Integers

Example 1:

The perimeter of the quadrilateral is 114 feet. Find the length of the sides.

x + 10

x + 8

4x− 3

x2 − 8x

(x + 10) + (x + 8) + (4x− 3) + (x2 − 8x) = 114x2 − 2x + 15 = 114x2 − 2x− 99 = 0(x− 11)(x + 9) = 0x = 11 or

✘✘✘✘

x = −9

So the lengths are: 21 ft,19 ft,41 ft, and 33 ft

The perimeter of the quadrilateral is 182 feet. Find the length of the sides.

x + 5

x + 12

4x− 3

x2 − 8x

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Example 2:

Find the value of x

x

2x + 8

2x + 12

a2 + b2 = c2

x2 + (2x + 8)2 = (2x + 12)2

x2 + 4x2 + 32x + 64 = 4x2 + 48x + 1445x2 + 32x + 64 = 4x2 + 48x + 144x2 − 16x− 80 = 0(x− 20)(x + 4) = 0x = 20 or

✘✘✘✘

x = −4

Find the value of x

x

x + 5

x + 10

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Example 3: The sum of a number and its square is 240. Find the number(s).

x + x2 = 240x2 + x− 240 = 0(x + 16)(x− 15) = 0x = −16 or x = 15

The sum of a number and its square is 132. Find the number(s).

Example 4: A smart-phone is thrown upwards from the top of a 96-foot building withan initial velocity of 16 feet per second. The height h of the smart-phone after t seconds isgiven by the quadratic equation h = −16t2 + 16t + 96. When will the smart-phone hit theground?

0 = −16t2 + 16t + 960 = −16(t2 − t− 6)(t− 3)(t + 2) = 0t = 3 or ✘

✘✘✘t = −2

A smart-phone is thrown upwards from the top of a 480-foot building with an initialvelocity of 16 feet per second. The height h of the smart-phone after t seconds is given bythe quadratic equation h = −16t2 + 16t + 480. When will the smart-phone hit the ground?

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Example 5: The length of a rectangle is nine inches more than its width. Its area is 1102square inches. Find the width and length of the rectangle.

1102 square inches w

w + 9

A = LW

1102 = (w + 9)w1102 = w2 + 9w

0 = w2 + 9w − 11020 = (w + 38)(w − 29)w = 29 or ✭

✭✭✭✭

w = −38

Width =

Length =

The length of a rectangle is nine inches more than its width. Its area is 630 square inches.Find the width and length of the rectangle.

Width =

Length =

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Example 6: At the end of 2 years, P dollars invested at an interest rate r, compoundedannually, increases to an amount A dollars given by A = P (1 + r)2

Find the interest rate if $500.00 increased to $551.25 in 2 years.

551.25 = 500(1 + r)2

551.25 = 500(1 + 2r + r2)551.25 = 500 + 1000r + 500r2

500r2 + 1000r − 51.25 = 050000r2 + 100000r − 5125 = 0125(400r2 + 800r − 41) = 0125(20r − 1)(20r + 41) = 0(20r − 1) = 0 or (20r + 41) = 020r = 1 or 20r = −41r = 0.05 = 5% or ✭

✭✭✭

✭✭r = −2.05

Write your answer as a percent. r = % (If needed, round your answer to 1 decimalplace.)

At the end of 2 years, P dollars invested at an interest rate r, compounded annually,increases to an amount A dollars given by A = P (1 + r)2

Find the interest rate if $300.00 increased to $675.00 in 2 years.

Write your answer as a percent. r = % (If needed, round your answer to 1 decimalplace.)

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