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Zero Product Principle

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What if you had a polynomial equation like 3x^2 + 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.

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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 2x^2 + 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 && \text{OR} && (2x - 7) = 0

Solving both of those equations gives us:

& x + 6 = 0 && && 2x - 7 =0\\& \underline{\underline{x = -6}} && \text{OR} && 2x = 7\\& && && \underline{\underline{x = \frac{7}{2}}}

Notice that the solution is x = -6 OR x = \frac{7}{2} . 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.

& Check: x = - 6; && Check: x = \frac{7}{2}\\& (x + 6)(2x - 7)= && (x + 6)(2x - 7)=\\& (-6 +6)(2(-6) -7)= && \left ( \frac{7}{2} + 6 \right ) \left (2 \cdot \frac{7}{2} - 7 \right ) =\\& (0)(-19) = 0 && \left (\frac{19}{2} \right ) (7 - 7) = \\& && \left (\frac{19}{2} \right ) (0) = 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 && && 3x + 4 = 0\\& \underline{\underline{x = 9}} && \text{or} && 3x = -4\\& && && \underline{\underline{x = - \frac{4}{3}}}

b) x(5x - 4) = 0 can be split up into two linear equations:

& && && 5x - 4 = 0\\& \underline{\underline{x = 0}} && \text{or} && 5x = 4\\& && && \underline{\underline{x = \frac{4}{5}}}

c) 4x(x + 6)(4x - 9) =0 can be split up into three linear equations:

& 4x = 0 && && x + 6 = 0 && && 4x - 9 =0\\& x= \frac{0}{4} && \text{or} && x = -6 && \text{or} && 4x = 9\\& \underline{\underline{x = 0}} && && && && \underline{\underline{x = \frac{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) x^2 - 2x =0

b) 2x^2 = 5x

c) 9x^2 y - 6xy = 0

Solution

a) x^2 - 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 && \text{or} && x - 2 = 0

Solve:

\underline{x = 0} && \text{or} && \underline{x = 2}

Check: Substitute each solution back into the original equation.

x & = 0 \Rightarrow (0)^2 - 2(0) = 0 && \text{works out}\\x & = 2 \Rightarrow (2)^2 - 2(2) = 4 - 4 = 0 && \text{works out}

Answer: x = 0, x = 2

b) 2x^2 = 5x

Rewrite: 2x^2 = 5x \Rightarrow 2x^2 - 5x = 0

Factor: The common factor is x , so this factors as x(2x - 5)=0 .

Set each factor equal to zero:

x = 0 && \text{or} && 2x - 5 = 0

Solve:

& \underline{x = 0} && \text{or} &&  2x = 5\\& &&&& \underline{x = \frac{5}{2}}

Check: Substitute each solution back into the original equation.

x & = 0 \Rightarrow 2(0)^2 = 5(0) \Rightarrow 0 = 0 && \text{works out}\\x & = \frac{5}{2} \Rightarrow 2 \left ( \frac{5}{2} \right )^2 = 5 \cdot \frac{5}{2} \Rightarrow 2 \cdot \frac{25}{4} = \frac{25}{2} \Rightarrow \frac{25}{2} = \frac{25}{2} && \text{works out}

Answer: x = 0, x =\frac{5}{2}

c) 9x^2y - 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 && \text{or} && y = 0 && \text{or} && 3x - 2 = 0

Solve:

& \underline{x = 0} && \text{or} && \underline{y = 0} && \text{or} && 3x = 2\\& &&&& \underline{x = \frac{2}{3}}

Check: Substitute each solution back into the original equation.

& x = 0 \Rightarrow 9(0)y - 6(0)y = 0 - 0 = 0 && \text{works out}\\& y = 0 \Rightarrow 9x^2 (0) - 6x(0) = 0 - 0 = 0 && \text{works out}\\& x = \frac{2}{3} \Rightarrow 9 \cdot \left ( \frac{2}{3} \right)^2 y - 6 \cdot \frac{2}{3} y = 9 \cdot \frac{4}{9} y - 4y = 4y - 4y = 0 && \text{works out}

Answer: x = 0, y = 0, x = \frac{2}{3}

Watch this video for help with the Examples above.

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\cdot b=0 \Rightarrow a=0 \text{ or } b=0.

Guided Practice

Solve the following polynomial equation.

9x^2-3x=0

Solution: 9x^2-3x=0

Rewrite : This is not necessary since the equation is in the correct form.

Factor : The common factor is 3x , so this factors as: 3x(3x-1)=0 .

Set each factor equal to zero.

3x=0 && \text{or} && x-2=0

Solve :

x=0 && \text{or} && x=2

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

Practice

Solve the following polynomial equations.

  1. x(x + 12) = 0
  2. (2x + 1)(2x - 1) = 0
  3. (x - 5)(2x + 7)(3x - 4) = 0
  4. 2x(x + 9)(7x - 20) = 0
  5. x(3 + y) = 0
  6. x(x - 2y) = 0
  7. 18y - 3y^2 = 0
  8. 9x^2 = 27x
  9. 4a^2 + a = 0
  10. b^2 - \frac{5}{3}b = 0
  11. 4x^2 = 36
  12. x^3 - 5x^2 = 0

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