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Transcript
Zero Product Principle
Andrew Gloag
Eve Rawley
Anne Gloag
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Complete terms can be found at http://www.ck12.org/terms.
Printed: December 11, 2014
AUTHORS
Andrew Gloag
Eve Rawley
Anne Gloag
www.ck12.org
C HAPTER
Chapter 1. Zero Product Principle
1
Zero Product Principle
Here you’ll learn how to apply the zero-product property and how to factor polynomials to solve for their unknown
variables.
What if you had a polynomial equation like 3x2 + 4x − 4 = 0? How could you factor the polynomial to solve the
equation? After completing this Concept, you’ll be able to solve polynomial equations by factoring and by using the
zero-product property.
Watch This
MEDIA
Click image to the left or use the URL below.
URL: http://www.ck12.org/flx/render/embeddedobject/133003
CK-12 Foundation: 0907S Factoring to Solve Polynomials
Guidance
The most useful thing about factoring is that we can use it to help solve polynomial equations.
Example A
Consider an equation like 2x2 + 5x − 42 = 0. How do you solve for x?
Solution:
There’s no good way to isolate x in this equation, so we can’t solve it using any of the techniques we’ve already
learned. But the left-hand side of the equation can be factored, making the equation (x + 6)(2x − 7) = 0.
How is this helpful? The answer lies in a useful property of multiplication: if two numbers multiply to zero, then at
least one of those numbers must be zero. This is called the Zero-Product Property.
What does this mean for our polynomial equation? Since the product equals 0, then at least one of the factors on
the left-hand side must equal zero. So we can find the two solutions by setting each factor equal to zero and solving
each equation separately.
Setting the factors equal to zero gives us:
(x + 6) = 0
OR
(2x − 7) = 0
Solving both of those equations gives us:
1
www.ck12.org
2x − 7 = 0
x+6 = 0
x = −6
2x = 7
OR
x=
7
2
Notice that the solution is x = −6 OR x = 27 . The OR means that either of these values of x would make the product
of the two factors equal to zero. Let’s plug the solutions back into the equation and check that this is correct.
7
2
(x + 6)(2x − 7) =
7
7
+6
2· −7 =
2
2
19
(7 − 7) =
2
19
(0) = 0
2
Check : x = −6;
Check : x =
(x + 6)(2x − 7) =
(−6 + 6)(2(−6) − 7) =
(0)(−19) = 0
Both solutions check out.
Factoring a polynomial is very useful because the Zero-Product Property allows us to break up the problem into
simpler separate steps. When we can’t factor a polynomial, the problem becomes harder and we must use other
methods that you will learn later.
As a last note in this section, keep in mind that the Zero-Product Property only works when a product equals zero.
For example, if you multiplied two numbers and the answer was nine, that wouldn’t mean that one or both of the
numbers must be nine. In order to use the property, the factored polynomial must be equal to zero.
Example B
Solve each equation:
a) (x − 9)(3x + 4) = 0
b) x(5x − 4) = 0
c) 4x(x + 6)(4x − 9) = 0
Solution
Since all the polynomials are in factored form, we can just set each factor equal to zero and solve the simpler
equations separately
a) (x − 9)(3x + 4) = 0 can be split up into two linear equations:
x−9 = 0
x=9
2
3x + 4 = 0
or
3x = −4
4
x=−
3
www.ck12.org
Chapter 1. Zero Product Principle
b) x(5x − 4) = 0 can be split up into two linear equations:
5x − 4 = 0
x=0
5x = 4
4
x=
5
or
c) 4x(x + 6)(4x − 9) = 0 can be split up into three linear equations:
4x = 0
0
x=
4
4x − 9 = 0
x+6 = 0
or
x = −6
or
4x = 9
x=
x=0
9
4
Solve Simple Polynomial Equations by Factoring
Now that we know the basics of factoring, we can solve some simple polynomial equations. We already saw how we
can use the Zero-Product Property to solve polynomials in factored form—now we can use that knowledge to solve
polynomials by factoring them first. Here are the steps:
a) If necessary, rewrite the equation in standard form so that the right-hand side equals zero.
b) Factor the polynomial completely.
c) Use the zero-product rule to set each factor equal to zero.
d) Solve each equation from step 3.
e) Check your answers by substituting your solutions into the original equation
Example C
Solve the following polynomial equations.
a) x2 − 2x = 0
b) 2x2 = 5x
c) 9x2 y − 6xy = 0
Solution
a) x2 − 2x = 0
Rewrite: this is not necessary since the equation is in the correct form.
Factor: The common factor is x, so this factors as x(x − 2) = 0.
Set each factor equal to zero:
x=0
or
x−2 = 0
Solve:
3
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x=0
x=2
or
Check: Substitute each solution back into the original equation.
x = 0 ⇒ (0)2 − 2(0) = 0
works out
x = 2 ⇒ (2)2 − 2(2) = 4 − 4 = 0
works out
Answer: x = 0, x = 2
b) 2x2 = 5x
Rewrite: 2x2 = 5x ⇒ 2x2 − 5x = 0
Factor: The common factor is x, so this factors as x(2x − 5) = 0.
Set each factor equal to zero:
x=0
2x − 5 = 0
or
Solve:
x=0
2x = 5
5
x=
2
or
Check: Substitute each solution back into the original equation.
x = 0 ⇒ 2(0)2 = 5(0) ⇒ 0 = 0
2
5
5
5
25 25
25 25
x= ⇒2
= 5· ⇒ 2·
=
⇒
=
2
2
2
4
2
2
2
Answer: x = 0, x =
works out
works out
5
2
c) 9x2 y − 6xy = 0
Rewrite: not necessary
Factor: The common factor is 3xy, so this factors as 3xy(3x − 2) = 0.
Set each factor equal to zero:
3 = 0 is never true, so this part does not give a solution. The factors we have left give us:
x=0
Solve:
4
or
y=0
or
3x − 2 = 0
www.ck12.org
Chapter 1. Zero Product Principle
x=0
y=0
2
x=
3
or
or
3x = 2
Check: Substitute each solution back into the original equation.
x = 0 ⇒ 9(0)y − 6(0)y = 0 − 0 = 0
2
y = 0 ⇒ 9x (0) − 6x(0) = 0 − 0 = 0
2
2
2
4
2
y − 6 · y = 9 · y − 4y = 4y − 4y = 0
x = ⇒ 9·
3
3
3
9
Answer: x = 0, y = 0, x =
works out
works out
works out
2
3
Watch this video for help with the Examples above.
MEDIA
Click image to the left or use the URL below.
URL: http://www.ck12.org/flx/render/embeddedobject/133004
CK-12 Foundation: Factoring to Solve Polynomials
Vocabulary
• Polynomials can be written in expanded form or in factored form. Expanded form means that you have
sums and differences of different terms:
• The factored form of a polynomial means it is written as a product of its factors.
• Zero Product Property: The only way a product is zero is if one or more of the terms are equal to zero:
a · b = 0 ⇒ a = 0 or b = 0.
Guided Practice
Solve the following polynomial equation.
9x2 − 3x = 0
Solution: 9x2 − 3x = 0
Rewrite: This is not necessary since the equation is in the correct form.
5
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Factor: The common factor is 3x, so this factors as: 3x(3x − 1) = 0.
Set each factor equal to zero.
3x = 0
or
x−2 = 0
Solve:
x=0
x=2
or
Check: Substitute each solution back into the original equation.
x=0
(0)2 − 2(0) = 0
x=2
(2)2 − 2(2) = 0
Answer x = 0, x = 2
Explore More
Solve the following polynomial equations.
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
6
x(x + 12) = 0
(2x + 1)(2x − 1) = 0
(x − 5)(2x + 7)(3x − 4) = 0
2x(x + 9)(7x − 20) = 0
x(3 + y) = 0
x(x − 2y) = 0
18y − 3y2 = 0
9x2 = 27x
4a2 + a = 0
b2 − 53 b = 0
4x2 = 36
x3 − 5x2 = 0