Solving Equations by Factoring
Objective Learn to solve equations that are
in factored form.
This lesson begins with a review of factoring, and then uses
factored trinomials to solve equations. This technique
(factoring) is very useful in problem solving.
Factoring Expressions
The following are good examples.
Example 1
Factor x 2 - 4x - 21.
Solution
Since the coefficient of x 2 is 1, factor this
expression as ( x + a )( x + b ) for some integer values of a and
b. Now, ( x + a ) and ( x + b ) are multiplied to get x 2
+ ( a + b )x + ab , which is supposed to equal x 2 -
4x - 21. Comparing x 2 + ( a + b )x + ab and x 2
- 4x - 21, we see that the sum a + b = -4 and the product ab =
-21. So, the first step is to find the integral factors of - 21,
for possible choices of a and b .
| - 21 = -21 · 1 |
- 21 = 21 · (-1) |
| - 21 = -3 · 7 |
- 21 = 3 · (-7) |
Look at the above factor pairs of -21 to see if any of them
have a sum equal to - 4. If a = 3 and b = -7, then a + b = -4.
The factorization is given below.
| x 2 - 4x - 21 |
= ( x + a )( x + b ) |
|
| |
= ( x + 3 )[ x + (-7)] |
a = 3, b = -7 |
| |
= ( x + 3 )( x - 7 ) |
|
You should multiply ( x + 3) and ( x - 7) to check that the
factorization is correct.
Example 2
Factor 2x 3 - 3x 2 - 2x .
Solution
There is a common factor of x in each of the terms. So begin
by factoring out x.
2x 3 - 3x 2 - 2x = x( 2x 2 -
3x - 2 )
The next step is to factor 2x 2 - 3x - 2. That is,
write this expression as ( ax + b )( cx + d ) for some values of
a, b, c, and d. When we multiply this out we get acx 2
+ ( ad + bc )x + bd , which is supposed to equal 2x 2
- 3x - 2. So, ac = 2. Since 2 only has factors 2 and 1, let a = 2
and c = 1. We then have to find b and d. Since the coefficient of
x , ad + bc , is now 2d + b · 1 or 2d + b , 2d + b = -3 and bd =
-2. The factors of - 2 are  2 and 1, so b and d must be 2 or 1. On the other hand,
2d + b = -3. By checking the possibilities, we see that we must
have d = -2 and b = 1. Substitute these values for a, b, c, and
d.
| 2x 3 - 3x 2
- 2x |
= ( ax + b )( cx + d ) |
|
| |
= ( 2x + 1 )( x - 2 ) |
a = 2, b = 1, c = 1, d = -2 |
So,
| 2x 3 - 3x 2
- 2x |
= x( 2x 2 - 3x - 2 ) |
| |
= x( 2x + 1 )( x - 2 ) |
Again, you should multiply out x( 2x + 1 )( x - 2 ) to check
the factoring. |
