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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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6.2 Factoring Trinomials of the Form x2 + bx + c Chapter 6
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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6.2 Factoring Trinomials of the Form x2 + bx + c Chapter 6
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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6.3 Factoring Trinomials of the Form ax2 + bx + c Chapter 6
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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6.3 Factoring Trinomials of the Form ax2 + bx + c Chapter 6
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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6.3 Factoring Trinomials of the Form ax2 + bx + c Chapter 6
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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6.3 Factoring Trinomials of the Form ax2 + bx + c Chapter 6
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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6.3 Factoring Trinomials of the Form ax2 + bx + c Chapter 6
5n2 + 19n + 125n2+15n + 4n + 12(5n2 + 15n) + (4n + 12)5n(n + 3) + 4(n + 3)(5n + 4)(n + 3)
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6.3 Factoring Trinomials of the Form ax2 + bx + c Chapter 6
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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6.6 A Factoring Strategy Chapter 6
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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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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6.7 Solving Quadratic Equations by Factoring Chapter 6
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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6.7 Solving Quadratic Equations by Factoring Chapter 6
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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6.8 Application of Quadratic Equations Chapter 6
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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6.8 Application of Quadratic Equations Chapter 6
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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6.8 Application of Quadratic Equations Chapter 6
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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6.8 Application of Quadratic Equations Chapter 6
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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