1. Solving Quadratic
Equations
by Factoring

2. Zero Product Property
For any real numbers a and b,
if ab=0, then either
a=0, b=0, or both.

3. In this first example, the equation is
already factored and is set equal to zero.
To solve, simply set the individual
factors equal to zero.
x + 3( ) 2x −1( ) = 0
x + 3 = 0 or 2x − 1 = 0
x = −3 or 2x = 1
x = −3 or x =
1
2
The solutions are -3 and 1/2.

4. In this example, you must first factor
the equation. Notice the familiar
pattern. After factoring, set the
individual factors equal to zero.
9x2
− 4 = 0
3x + 2( ) 3x − 2( )= 0
3x + 2 = 0 or 3x − 2 = 0
3x = −2 or 3x = 2
2 2
or
3 3
x x= − =
Factor using “difference of two squares.”

5. In the next example, you must set the
equation equal to zero before factoring.
Then set the individual factors equal to
zero and solve.
x2
− 6x − 27 = 0
x − 9( ) x + 3( )= 0
x − 9 = 0 or x + 3 = 0
9 or 3x x= = −
x2
− 27 = 6x

6. Re-write this example in the proper form.
Notice that the leading coefficient is not
one. Use an appropriate factoring
technique. Then solve as you have
done before.
2x2
+ 5x − 3 = 0
2x −1( ) x + 3( ) = 0
2x −1 = 0 or x + 3 = 0
1
2
or 3x x= = −
2x2
−3 = −5x

7. This one uses a different technique
than the previous ones. Really, this is
something you should consider at the
beginning of every factoring problem.
See if you can solve it.
2x x + 4( )= 0
2x = 0 or x + 4 = 0
0 or 4x x= = −
2x2
+8x = 0
Did you take out GCF?

8. Now, try several problems. Write these
on your own paper, showing all steps
carefully.
1. 3y − 5( ) 2y + 7( )= 0
2. x
2
+ x = 12
3. d
2
− 5d = 0
4. 4c2
= 25
5. 18u2
−1 = 3u
After completion,
click here.

9. Here are the answers.
For help, click on the
numbers.
 1.
 2.
 3.
 4.
 5.
If all are correct,
you’re finished!
y = 5 3 or y = −7 2
x = −4 or x = 3
d = 0 or d = 5
c = −5 2 or c = 5 2
u = −1 6 or u = 1 3

10. 3y − 5( ) 2y + 7( )= 0
3y − 5 = 0 or 2y + 7 = 0
3y = 5 or 2y = −7
y = 5 3 or y = −7 2
Back to questions

11. x
2
+ x =12
x2
+ x −12 = 0
x + 4( ) x − 3( ) = 0
x + 4 = 0 or x − 3 = 0
x = −4 or x = 3 Back to questions

12. d2
− 5d = 0
d d − 5( ) = 0
d = 0 or d − 5 = 0
d = 0 or d = 5
Back to questions

13. 4c2
= 25
4c2
− 25 = 0
2c + 5( ) 2c − 5( ) = 0
2c + 5 = 0 or 2c − 5 = 0
2c = −5 or 2c = 5
c = −5 2 or c = 5 2
Back to questions

14. 18u
2
−3u = 1
18u2
−3u −1 = 0
6u +1( ) 3u −1( ) = 0
6u +1 = 0 or 3u −1 = 0
6u = −1or 3u = 1
u = −1 6 or u = 1 3
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