1. Solving
Equations
by Factoring

2. Solving Equations
Previously, we have learned how to factor and have explored various factoring techniques. First, we studied how to find and use the GCF. Next, we looked at the distributive property, and then we looked at factoring trinomials. Some examples of factoring are shown below.
GCF & Distributive Property
Factor into the product of two polynomials.
Difference of two squares.
Perfect Square Trinomials

3. Solving Equations
Now that we know how to factor, we can apply this knowledge to the solution of equations.
How would you solve the following equation?
x2 – 36 = 0
Step 1: Factor the polynomial.
(x - 6)(x + 6) = 0

4. Solving Equations
Step 2: Apply the zero product property which states that
For all numbers a and b, if ab = 0, then
a = 0, b = 0, or both a and b equal 0.
(x - 6)(x + 6) = 0
Therefore (x – 6) = 0 or (x + 6) = 0.
x + 6 = 0
x – 6 = 0 or
x = 6
x = -6
This equation has two solutions or zeros: x = 6 or x = -6.

5. Summary of Steps Get a value of zero on one side of the equation.
Factor the polynomial if possible.
Apply the zero product property by setting each factor equal to zero.
Solve for the variable.

6. Example Solve the following equation: x2 – 10x + 5 = 29.
Get zero on one side of the equation
x2 – 10x –24 = 0
Factor the trinomial.
(x – 12) (x + 2) = 0
The factors of –24 are:
-1,24, 1,-24 -2,12 2,-12
-3,8 3,-8 -4,6 4,-6
x – 12 = 0
x + 2 = 0 or
Set each factor equal to 0.
Solve for the variable.
x = 12
x = -2
x = or x = -2

7. You Try It Solve the following equations. x2 – 25 = 0 x2 + 7x – 8 = 0
c2 – 8c = 0
5b3 + 34b2 = 7b

8. Problem 1 x - 5 = or x + 5 = + 5 + 5 - 5 - 5 x = 5 x = -5 x2 – 25 = 0
Factor the polynomial.
Set each factor equal to zero.
Solve for x. x - 5 = or x + 5 = + 5 + 5 - 5 - 5 x = 5 x = -5
Solution: x = 5 or x = -5

9. Problem 2 - = x + 8 = x 1 - 8 - 8 + 1 + 1 x = -8 x = 1 x2 + 7x – 8 = 0
Factor the polynomial.
Set each factor equal to zero. - = x + 8 = or x 1
Solve for x. - 8 - 8 + 1 + 1 x = -8 x = 1
Solution: x = 1 or x = -8

10. Problem 3
x2 – 12x + 36 = 0
Factor the polynomial.
Set each factor equal to zero. x - 6 =
Solve for x. + 6 + 6 x = 6
Solution: x = 6 There is only one solution for this equation since both of the factors are the same.

11. Problem 4 c - 8 = + 8 + 8 c = 8 c2 – 8c = 0 Factor the polynomial.
Set each factor equal to zero. or c - 8 =
Solve for c. + 8 + 8 c = 8
Solution: c = 0 or c = 8 .

12. Problem 5
5b3 + 34b2 = 7b
Get a value of zero on one side of the equation.
5b3 + 34b = 7b
-7b -7b
5b3 + 34b2 –7b = 0
Factor the polynomial.
5b3 + 34b2 – 7b = b(5b2 + 34b – 7)
=b(5b – 1)(b + 7)
Set each factor equal to zero.
b = 0 or 5b – 1 = or b + 7 = 0
Solve for b.
b = 0 or
5b – 1 = 0 or
b + 7 = 0
Solution:
b = 0 or b = 1/5 or b = -7.
5b = 1
b = -7
b = 1/5