```5.12: Solving Equations by Factoring
Zero Product Property: For all real numbers a and b:
ab = 0 if and only if a = 0 or b = 0. (Multiplicative Property of Zero)
(A product of factors is zero if and only if one or more of the factors is zero). (Converse)
-The zero-product property is true for any number of factors. You can use this property to solve certain equations.
Polynomial Equations: both sides of an equation are polynomials. Polynomial equations are usually named by the term
of highest degree.
If a ≠0:
+  = 0 is a linear equation.
2 +  +  = 0 is a quadratic equation.
3 +  2 +  +  = 0 is a cubic equation.
-Many polynomial equations can be solved by factoring and then using the zero-product property. Often, the
first step is to transform the equation into standard form in which one side is zero. The other side should be a
simplified polynomial arranged in order of decreasing degree of the variable.
-If a factor occurs twice in the factored form of an equation, it is a double root or multiple root. It is only listed one time
in the solution set.
Ex: #3, p.232
( + ) =
1. Set each factor equal to zero and solve.
15( + 15) = 0
( + 15) = 0
15 = 0
=0
= −15
The solution set is {0, −15}.
Ex: #15, p.232
=  +
1. Transform the equation into standard form.
2. Factor the polynomial.
2 − 4 − 32 = 0
( 2 − 8) + (4 − 32) = 0
( − 8) + 4( − 8) = 0
( − 8) ( + 4) = 0
−8 = 0
+4 =0
=8
= −4
The solution set is {8, −4}.
3. Set each factor equal to zero and solve.
Ex: #39, p.232
+  =
1. Transform the equation into standard form.
2. Factor completely.
3. Set each factor equal to zero and solve.
9 3 − 30 2 + 9 = 0
3 ( 3 2 − 10 + 3) = 0
3 ( 3 2 − 9 −  + 3) = 0
3 {(3 2 − 9 + (− + 3)} = 0
3 {3( − 3) − 1( − 3)} = 0
3 (3 − 1) ( − 3) = 0
3 = 0
3 − 1 = 0
−3 = 0
=0
=
1
3
The solution set is {0,
=3
1
,
3
3}.
5.12: Solving Equations by Factoring
( − ) ( + ) =
Ex: #45, p.233
1. Since you have factors that are equal to something besides
zero, you must turn this into a polynomial (by foiling).
2 + 3 − 2 − 6 = 6
2. Transform the equation into standard form.
2 +  − 12 = 0
{( 2 + 4) + (−3 − 12)} = 0
3. Factor the polynomial.
{( + 4) − 3( + 4)} = 0
( + 4) ( − 3) = 0
4. Set each factor equal to zero and solve.
+4=0
−3=0
= −4
=3
The solution set is {−4, 3}.
```