# factoring to solve

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```					Solving Equations by
Factoring

Module 8, Lesson 6
Online Algebra
VHS@PWCS

So far we have solved quadratic equations by:
1. Taking the square root of both sides.
 We can do this when the equation is of the form ax2
+ c = 0.
 Note that the only variable in this type is squared.

 We can use this to solve any quadratic equations.

b  b 2  4ac
x
2a
Solve the following equations

x2 = 25                x2 - 63 = 81

Take the square root of     Get the x2 by itself.
both sides.            x2 – 63 + 63 = 81 + 63
x=5                     x2 = 144
x = -5
Take the square root of
Remember that               both sides.
quadratics can have 2           x = 12
solutions!
x = -12
Solve the following equation.
2y2 + 3 = -7y                      7  7 2  4(2)(3)
x
2(2)
1.   Get everything on the same side.
7  49  24
2y2 + 7y + 3 = -7y + 7y              x
4
2y2 + 7y + 3 = 0
7  25
2.   Identify a, b and c.                         x
4
a = 2, b = 7 and c = 3
7  5
3.   Substitute into the quadratic formula.        x
4
b  b 2  4ac                          x  3
x                                              1
2a                                 x
2
Zero Product Property

The Zero Product Property states that:
For all numbers a and b, if ab = 0, then a = 0,
b= 0 or both a and b equal 0.

This makes sense since 0 times any number will
always give you zero.

We are going to use this property to solve
Solve the following equation.
(y + 2)(3y + 5) = 0
 According to the zero product property either (y + 2) or
(3y + 5) must equal zero.
 We will assume that each equals zero and set them
equal to zero.
y+2=0                       3y + 5 = 0
 Then we will solve.
y + 2 – 2 = 0 + -2            3y + 5 + -5 = 0 + -5
y = -2                        3y = -5
y = -5/3
Solve the following equation.

y(y – 12) = 0

1. Set each factor equal to zero.
y=0                     y – 12 = 0
2. Solve.
y=0              y – 12 + 12 = 0 + 12
y = 12.

Remember quadratics can have 2 solutions.
Solving equations by factoring
Sometimes we need to factor an equation before setting it
equal to zero.
Solve m2 + 36m = 0
1. Factor. Here we can pull out the GCF of m.
m(m + 36) = 0
2. Set each factor equal to zero.
m=0                          m + 36 = 0
3. Solve.
m=0                   m + 36 + -36 = 0 + -36
m = -36
4. We have 2 solutions m = 0 and m = -36.
Solve: a2 + 4a = 21

1. Just like when using the             a2 + 4a = 21
quadratic formula, we need     a2 + 4a + -21 = 21 + -21
one side to be zero.               a2 + 4a – 21 = 0
2. Factor
3. Set both factors equal to
(a + 7)(a – 3) = 0
zero.
4. Solve
a+7=0            a–3=0
a=3
a = -7
Try these! Click for the
answers.        c – 17c + 60 = 0 2

(c -12)(c – 5) = 0
1. c2 – 17c + 60 = 0             c – 12 = 0           c–5=0
c = 12              c=5
2. a2 + 64 = -16a
a2 + 64 = -16a
a2 + 64 + 16 a = -16a + 16a
Since these factors are the same
a2 + 16a + 64 = 0
we only need to set one equal to
zero. This quadratic will have only        (a + 8)(a + 8) = 0
one solution. a = -8
a+8=0
a = -8

We now have 3 different ways to solve quadratic
equations.
 Taking the square root of both sides.
   This only works when the equation is of the form
ax2 + c = 0.