Zero Product Property for
Quadratic Equations
Brenda Meery
Kaitlyn Spong
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Printed: April 23, 2015
AUTHORS
Brenda Meery
Kaitlyn Spong
www.ck12.org
C HAPTER
Chapter 1. Zero Product Property for Quadratic Equations
1
Zero Product Property for
Quadratic Equations
Here you’ll learn how to solve a quadratic equation by factoring and using the zero product property.
The area of a particular rectangle was found to be A(w) = w2 − 8w − 58. Determine the dimensions of the rectangle
if the area was known to be 7 units.
Watch This
Khan Academy Factoring Special Products
MEDIA
Click image to the left or use the URL below.
URL: http://www.ck12.org/flx/render/embeddedobject/122
Guidance
Recall that when solving an equation, you are trying to determine the values of the variable that make the equation
true. For the equation 2x2 + 10x + 8 = 0, x = −1 and x = −4 are both solutions. You can check this:
• 2(−1)2 + 10(−1) + 8 = 2(1) − 10 + 8 = 0
• 2(−4)2 + 10(−4) + 8 = 2(16) − 40 + 8 = 32 − 40 + 8 = 0
Here you will focus on solving quadratic equations. One of the methods for quadratic equations utilizes your
factoring skills and a property called the zero product property.
If a · b = 0, what can you say about a or b? What you should realize is that either a or b have to be equal to 0, because
that is the only way that their product will be 0. If both a and b were non-zero, then their product would have to be
non-zero. This is the idea of the zero product property. The zero product property states that if the product of two
quantities is zero, then one or both of the quantities must be zero.
The zero product property has to do with products being equal to zero. When you factor, you turn a quadratic
expression into a product. If you have a quadratic expression equal to zero, you can factor it and then use the zero
product property to solve. So, if you were given the equation 2x2 + 5x − 3 = 0, first you would want to turn the
quadratic expression into a product by factoring it:
2x2 + 5x − 3 = (x + 3)(2x − 1)
You can rewrite the equation you are trying to solve as (x + 3)(2x − 1) = 0.
Now, you have the product of two binomials equal to zero. This means at least one of those binomials must be equal
to zero. So, you have two mini-equations that you can solve to find the values of x that cause each binomial to be
equal to zero.
• x + 3 = 0, which means x = −3 OR
• 2x − 1 = 0, which means x = 12
1
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The two solutions to the equation 2x2 + 5x − 3 = 0 are x = −3 and x = 12 .
Keep in mind that you can only use the zero product property if your equation is set equal to zero! If you have an
equation not set equal to zero, first rewrite it so that it is set equal to zero. Then factor and use the zero product
property.
Example A
Solve for x: x2 + 5x + 6 = 0.
Solution: First, change x2 + 5x + 6 into a product so that you can use the zero product property. Change the
expression into a product by factoring:
x2 + 5x + 6 = (x + 3)(x + 2)
Next, rewrite the equation you are trying to solve:
x2 + 5x + 6 = 0 becomes (x + 3)(x + 2) = 0.
Finally, set up two mini-equations to solve in order to find the values of x that cause each binomial to be equal to
zero.
• x + 3 = 0, which means that x = −3
• x + 2 = 0, which means that x = −2
The solutions are x = −3 or x = −2.
Example B
Solve for x: 6x2 + x − 15 = 0.
In order to solve for x you need to factor the polynomial.
Solution: First, change 6x2 + x − 15 into a product so that you can use the zero product property. Change the
expression into a product by factoring:
6x2 + x − 15 = (3x + 5)(2x − 3)
Next, rewrite the equation you are trying to solve:
6x2 + x − 15 = 0 becomes (3x + 5)(2x − 3) = 0.
Finally, set up two mini-equations to solve in order to find the values of x that cause each binomial to be equal to
zero.
• 3x + 5 = 0, which means that x = − 53
• 2x − 3 = 0, which means that x = 23
The solutions are x = − 53 or x = 32 .
Example C
Solve for x: x2 + 2x − 35 = 0.
Solution: First, change x2 + 2x − 35 into a product so that you can use the zero product property. Change the
expression into a product by factoring:
x2 + 2x − 35 = (x + 7)(x − 5)
2
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Chapter 1. Zero Product Property for Quadratic Equations
Next, rewrite the equation you are trying to solve:
x2 + 2x − 35 = 0 becomes (x + 7)(x − 5) = 0.
Finally, set up two mini-equations to solve in order to find the values of x that cause each binomial to be equal to
zero.
• x + 7 = 0, which means that x = −7
• x − 5 = 0, which means that x = 5
The solutions are x = −7 or x = 5.
Concept Problem Revisited
The area of a particular rectangle was found to be A(w) = w2 − 8w − 58. Determine the dimensions of the rectangle
if the area was known to be 7 units.
In other words, you are being asked to solve the problem:
w2 − 8w − 58 = 7
OR
w2 − 8w − 65 = 0
You can solve this problem by factoring and using the zero product property.
w2 − 8w − 65 = 0 becomes (w + 5)(w − 13) = 0
(w + 5)(w − 13) = 0
.
&
w − 13 = 0
w+5 = 0
w = −5 or
w = 13
Since you are asked for dimensions, a width of –5 units does not make sense. Therefore for the rectangle, the width
would be 13 units.
Guided Practice
1. Solve for the variable in the polynomial: x2 + 4x − 21 = 0
2. Solve for the variable in the polynomial: 20m2 + 11m − 4 = 0
3. Solve for the variable in the polynomial: 2e2 + 7e + 6 = 0
Answers:
1. x2 + 4x − 21 = (x − 3)(x + 7)
(x − 3)(x + 7) = 0
.
(x − 3) = 0
x=3
&
(x + 7) = 0
x = −7
3
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2. 20m2 + 11m − 4 = (4m − 1)(5m + 4)
(4m − 1)(5m + 4) = 0
.
4m − 1 = 0
4m = 1
1
m=
4
&
5m + 4 = 0
5m = −4
−4
m=
5
3. 2e2 + 7e + 6 = (2e + 3)(e + 2)
(2e + 3)(e + 2) = 0
.
2e + 3 = 0
2e = −3
−3
e=
2
Explore More
Solve for the variable in each of the following equations.
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
4
(x + 1)(x − 3) = 0
(a + 3)(a + 5) = 0
(x − 5)(x + 4) = 0
(2t − 4)(t + 3) = 0
(x − 8)(3x − 7) = 0
x2 + x − 12 = 0
b2 + 2b − 24 = 0
t 2 + 3t − 18 = 0
w2 + 3w − 108 = 0
e2 − 2e − 99 = 0
6x2 − x − 2 = 0
2d 2 + 14d − 16 = 0
3s2 + 20s + 12 = 0
18x2 + 12x + 2 = 0
3 j2 − 17 j + 10 = 0
&
e+2 = 0
e = −2