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## 4-4 Solving Equations Using Factoring - Two Days

Warm-up

1. Find the area of a square whose sides are 12 m long.
2. Find the product of 2x(x + 1).
3. Find the product (2x + 1) (x + 1).
4. If 2ab = 0 and a = -1, what is the value of b?
5. What is the greatest common factor of 6, 15, and 21?

1. Area is 144 m2
2. 2x2 + 2x
3. 2x2 + 3x + 1
4. b = 0
5. GCF is 3

## 4-4 Solving quadratic equations by factoring

A trinomial is an expression that can be written as the sum of 3 unlike terms Factoring is undoing FOILing. The trinomial will factor into two binomials.
Examples 1-4 show how to factor when a = 1. Example 5 shows how to factor when a ≠1.
Factoring when a = 1 • If c is positive, then the factors are either both positive or both negative.
o If b is positive, then the factors are positive
o If b is negative, then the factors are negative.
o In either case, you're looking for factors that add to b.

• If c is negative, then the factors you're looking for are of alternating signs; that is, one
is negative and one is positive.
o If b is positive, then the larger factor is positive.
o If b is negative, then the larger factor is negative.
In either case, you're looking for factors that are b units apart.

Example 1

Factor x2 - 9x + 18 (ax2 + bx + c)

 Solution (x ____)(x ____) List the factors of 18 (the c) -9 ·- 2 Factors must add up to –9. So, one factor + and one factor is - 2 · 9 -6 · -3 3 · 6 -18 · -1 1 · 18

Guess and check
( x-9)( x-2) =x2-11x+18
no
( x+9)( x-2) =x2+11x+18
no
(x- 3)( x-6) =x2-3x-6x+18
yes

Example 2

Factor x2 +3x + 2

 Solution (x ____)(x ____) Factors of 2 Add up to 3 2·1 Yes 1·2 Yes

Guess and check

(x + 2)(x +1) = x2 + 2x + x +1= x2 + 3x +1 yes!

Example 3

Factor x2 -3x - 40

 Solution (x ____)(x ____) Factors of –40 Add up to -3 -8 · 5 Yes -5 · -8 No -2· 20 No -20 · 2 No

(x-8)(x+5)
Check
(x-8)(x+5)=x2 -8x +5x-40= x2 -3x - 40

Example 4

Factor x2-x - 6

 Solution (x ____)(x ____) Factors of –6 Add up to -1 -2 · 3 No -3 · 2 Yes

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

Sometimes a (the x2coefficient) is not 1.

Example 5

Factor 2x2 − 7x −30
Solution
x2 − 7x −30
Use guess and check! ## 4-4 Solving equations by factoring – Day 2

Factoring when a ≠1

Example 1
Factor 3x2 +11x +10
Solution

 Factors of Factors of c Product 3 • 1 10 • 1 (3x + 10)(x + 1) 1 • 3 10 • 1 (x + 10)(3x + 1) 3 • 1 5 • 2 (3x + 5)(x + 2) → 3x2 + 11x + 10 ←This is correct! 1 • 3 5 • 2 (x + 5)(3x + 2)

(3x + 5)(x + 2)

Example 2

Factor 4x2 −10x −14
Solution
Factor out the greatest common factor.
2(2x2  – 5x – 7)

 Factors of Factors of c Product 2 • 1 -7 • 1 2(2x – 7)(x + 1) →  4x2– 10x – 14 ←This is correct! 1 • 2 -7 • 1 2(x – 14)(4x – 1) 2 • 1 7 • -1 2(2x – 14)(2x + 1) 1 • 2 7 • -1

(4x – 14)(x + 1)

Special Factoring Patterns – Yellow box on page 208

 To factor a difference of two squares: Examples a2 −b2 = (a + b)(a − b) 9x2−100 = (3x +10)(3x −10) To factor a perfect square trinomial: a2 + 2ab + b2 = (a + b)2 16x2 + 20x +169 = (4x +13)2 a2 − 2ab + b2 = (a − b)2 4x2 − 20x + 25 = (2x − 5)2

Guidelines for factoring completely – Yellow box on page 210

1. Factor out the greatest common factor first.
2. Look for a difference of two squares.
3. Look for a perfect square trinomial.
4. If a trinomial is not a perfect square, use trial and error to look for a
pair of factors.

Example 3

Factor 25x2 - 81
Solution
Test whether the expression is a difference of two squares. Ask these questions:
• Is the expression a difference? Yes
• Is the first term a square? Yes
•  Is the second term a square? Yes
To factor a difference of two squares:
a2 −b2 = (a + b)(a − b)
(5x + 9)(5x – 9)

Example 4

Factor 16x2 + 56x + 49

Solution
Test whether the trinomial is a perfect square trinomial. Ask these questions:
• Is the first term a square? Yes
•  Is the last term a square? Yes
•   Is the middle term twice the product of ? Yes, 4 • 7 = 28, which is ½ of 56.
To factor a perfect square trinomial:
a2 + 2ab + b2 = (a + b)2
a2 − 2ab + b2 = (a − b)2
(4x + 7)2

Solving equations by factoring
Some quadratic equations can be solved by factoring.
• First the equation must be written in standard form, a x2 + bx + c
•  Then, if the trinomial is factorable, the equation can be solved using the zero-product
property.

Zero-product property (ZPP)
If ab = 0, then a = 0 or b = 0 or both are zero.
Example: If y (x + 5) = 0, then y = 0 or x + 5 = 0, or both.

Example 6

Solve 6x = x2 + 9
Solution
x2 - 6x + 9 = 0 ← Rewrite equation in standard form.
The equation is a perfect square trinomial.
To factor a perfect square trinomial:
a2 + 2ab + b2 = (a + b)2
a2 − 2ab + b2 = (a − b)2
So, the equation factors into
(x + 3)(x + 3) = 0
Set each factor equal to zero (ZPP) and solve for x.
x + 3 = 0 or x + 3 = 0
x = -3 or x = -3

Example 7

Solve 8x2 - 18x = -4
Solution
8x2 - 18x + 4 = 0
Rewrite equation in standard form.
2(4x2 - 9x + 2) = 0
Factor out the greatest common factor, 2.
4x2 - 9x + 2 = 0
Divide both sides by 2
(4x – 1)(x – 2) = 0
Factor the trinomial
4x – 1 = 0 or x – 2 = 0
Set each factor equal to 0 and solve for x. (ZPP)
x = ¼ or x = 2

There where three factors that multiplied to 0. Why didn’t we set all three of them equal to
zero when solving?