Integrated Review 6: THE PRINCIPLE OF ZERO PRODUCTS
The principle of zero products gives us a method for solving polynomial equations.
The Principle of Zero Products
For any real numbers a and b:
If ab  0, then a  0 or b  0 (or both).
If a  0 or b  0, then ab  0.
Example 1 Solve: x 2  x  6  0.
Factor.
Set each factor equal to 0.
Solve separately.
Check.
 x  3 x  2  0
x  3  0 or
x  3 or
2
x  x6  0
32  3  6
9 36
x20
x  2
x2  x  6  0
2
 2    2   6 0
0
4 26
0 0 True
0 0 True
The numbers 3 and 2 are both solutions.
Note that we must have 0 on one side of the equation in order to use the principle of zero
products.
Example 2 Solve: 7 y  3 y 2  2.
Get 0 on one side and
3y2  7 y  2  0
write in descending order.
Factor.  3 y  1 y  2  0
Use the principle of
3y 1  0
or y  2  0
zero products.
Solve separately.
3 y  1 or
y  2
y
Check.
1
or
3
y  2
7 y  3 y 2  2
2
 1  1
7     3  
2
 3  3
 1 1
7     3 
 3  9
7 1
 
3 3
6

3
2 2 True
1
The solutions are  and 2 .
3
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7 y  3 y 2  2
7  2   3  2 
2
2
7  2  3  4
14  12
2
2 True
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Example 3 Solve: 5b 2  10b.
Get 0 on one side. 5b 2  10b  0
Factor. 5b  b  2  0
5b  0 or b  2  0
Use the principle of zero products.
b  0 or
Solve separately.
b2
The solutions are 0 and 2.
Example 4 Solve: 6 x  x 2  9.
Get 0 on one side and the leading
0  x2  6x  9
coefficient on the other side positive.
Factor. 0   x  3 x  3
Use the principle of zero products.
Solve separately.
x  3  0 or x  3  0
x  3 or
x 3
There is only one solution, 3.
Example 5 Solve: 3x3  9 x 2  30 x.
3x3  9 x 2  30 x  0
2
Factor out a common factor. 3 x  x  3 x  10   0
Get 0 on one side.
Factor the trinomial. 3x  x  2 x  5  0
Use the principle of zero products. 3x  0 or
Solve separately.
x20
x  0 or
or x  5  0
x  2 or
The solutions are 0, 2, and 5.
Check Your Understanding
Determine whether each statement is true or false.
1. If  s  6 s  8  0, then both s  6 and s  8 must equal 0.
2. If  x  12 x  9  0, then x 12  0 or x  9  0.
3. If  y  7  y  3  21, then y  7  21 or y  3  21.
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x5
Exercises
Solve using the principle of zero products. In Exercises 1 and 2, fill-in the blanks in key
steps.
x 2  7 x  10
1.
x2  7 x 
x 
0
  x  2  0
x 5 
x5
or
x2
or
x
The solutions are
and
.
15 z 2  3 z
2.
15 z 2  3 z  0
 5 z  1  0
3z  0
z
or 5 z  1  0
or
The solutions are
z
1
5
and
.
3.
y 2  2 y  63
4. 18 x 2  9 x
6.
x 2  20 x  100  0
7.
32  4 x  x 2  0
10.
2 x3  2 x 2  12 x
9. 10  r  21r 2  0
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5.
9 x  x 2  20  0
8. 11x  4 x 2  6
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Notes:
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