# VII SOLVING QUADRATIC EQUATIONS BY FACTORING ```VII SOLVING QUADRATIC EQUATIONS BY FACTORING
The Zero Product Property is used to solve quadratic equations that can be
factored easily. This property states that if the product of two or more numbers is equal
to zero, then one or more of the factors must be equal to zero.
Example 1. 6n = 0. Since the product of 6 and n equals zero, the zero product
property states that one of the factors must be zero. Since 6 is one of the factors, then n
must be equal to zero.
Example 2. x( x  1)  0 . Since the product of the two factors equals zero, the
zero product property states that one of the factors must be zero. In this example, either x
could be equal to zero, or x + 1 could be equal to zero. If x + 1 = 0, this means that the
value of x that would make this true is x = -1. So, there are two solutions to the equation,
x = 0 or x = -1.
To solve a quadratic equation :
1) Transform the equation into standard form : ax 2  bx  c  0
2) Factor the equation completely.
3) Set each factor equal to zero.
4) Solve the resulting linear equations.
5) Check the solution.
Example 3. Solve : 3x 2  2 x  8  0
3x 2  2 x  8  0
(3 x  4)  x  2   0
3 x  4  0 or
3x  4
x
or
x20
x2
4
3
Example 4. Solve : a 2  24  5a
1) Write equation in standard form : a 2  5a  24  24  5a  24  5a
a 2  5a  24  0
2) Then, factor the equation : (a  8)(a  3)  0
3) Set each factor equal to zero : a  8  0 or
4) Solve each equation : a = -8
or
a=3
a 3 0
5) Check solutions in original equations :
a 2  24  5a
(8) 2  24  5(8)
64  24  40
or
64  64
(3) 2  24  5(3)
9  24 15
99
Example 5. Solve : t 3  t 2  4t  4
1) Write equation in standard form : t 3  t 2  4t  4  4t  4  4t  4
t 3  t 2  4t  4  0
(t 3  t 2 )  (  4t  4)  0
t 2 (t  1)  4 (t  1)  0
2) Then, factor the equation :


(t  1) t 2  4  0
(t  1)(t  2)  t  2   0
3 - 4) Set each factor equal to zero and solve linear equations :
t 1 0
or
t 20
or
t 20
t   1 or
t   2 or
t 2
5) Check solutions in original equation :
(1)3  (1) 2  4(1)  4
1  1   4  4
0 0
(2)3  (2) 2  4(2)  4
8  4   8  4
4  4
(2)3  (2) 2  4(2)  4
8 4 8 4
12  12
The solutions may be checked mentally.
```