<meta http-equiv="refresh" content="1; url=/nojavascript/"> Zero Product Principle ( Read ) | Algebra | CK-12 Foundation
Dismiss
Skip Navigation

Zero Product Principle

%
Best Score
Practice Zero Product Principle
Practice
Best Score
%
Practice Now

Zero Product Principle

What if you knew that the area of a chalkboard was 48 square feet, and you also knew that the area in square feet could be represented by the expression x^2 + 2x ? Suppose you wrote an equation, moved all the terms to one side, and factored. How could you find the value of x ? In this Concept, you'll learn to solve polynomial equations like this one by using the Zero Product Principle.

Watch This

Multimedia Link: For further explanation of the Zero Product Property, watch this: CK-12 Basic Algebra: Zero Product Property

- YouTube video.

Guidance

Polynomials can be written in expanded form or in factored form . Expanded form means that you have sums and differences of different terms:

6x^4+7x^3-26x^2-17x+30

Notice that the degree of the polynomial is four.

The factored form of a polynomial means it is written as a product of its factors.

The factors are also polynomials, usually of lower degree. Here is the same polynomial in factored form.

(x-1)(x+2)(2x-3)(3x+5)

Suppose we want to know where the polynomial 6x^4+7x^3-26x^2-17x+30 equals zero. It is quite difficult to solve this using the methods we already know. However, we can use the Zero Product Property to help.

Zero Product Property: The only way a product is zero is if one or both of the terms are zero.

By setting the factored form of the polynomial equal to zero and using this property, we can easily solve the original polynomial.

Example A

Solve for x : (x-1)(x+2)(2x-3)(3x+5)=0 .

Solution: According to the property, for the original polynomial to equal zero, we have to set each term equal to zero and solve.

(x-1)&=0 \rightarrow x=1\\(x+2)&=0 \rightarrow x=-2\\(2x-3)&=0 \rightarrow x=\frac{3}{2}\\(3x+5)&=0 \rightarrow x=-\frac{5}{3}

The solutions to 6x^4+7x^3-26x^2-17x+30=0 are the following: x=-2,-\frac{5}{3},1,\frac{3}{2} .

Example B

Solve (x-9)(3x+4)=0 .

Solution: Separate the factors using the Zero Product Property: (x-9)(3x+4)=0 .

x-9=0 && \text{or} && 3x+4=0\\x=9 && && 3x=-4\\&& && x=\frac{-4}{3}

Solving Simple Polynomial Equations by Factoring

We already saw how we can use the Zero Product Property to solve polynomials in factored form. Here you will learn how to solve polynomials in expanded form. These are the steps for this process.

Step 1: Rewrite the equation in standard form such that: Polynomial expression = 0 .

Step 2: Factor the polynomial completely.

Step 3: Use the zero-product rule to set each factor equal to zero.

Step 4: Solve each equation from step 3.

Step 5: Check your answers by substituting your solutions into the original equation.

Example C

Solve the following polynomial equation:

10x^3-5x^2=0 .

Solution:

First, factor by removing the greatest common factor. Since both terms have at least 5x^2 as a factor, we will remove that.

10x^3-5x^2&=0\\5x^2(2x-1)&=0

Separate each factor and set equal to zero:

5x^2=0 && \text{or} && 2x-1=0\\x^2=0 && && 2x=1\\x=0&& && x=\frac{1}{2}

Video Review

Guided Practice

Solve the following polynomial equation.

x^2-2x=0

Solution: 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 :

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

Sample explanations for some of the practice exercises below are available by viewing the following video. Note that there is not always a match between the number of the practice exercise in the video and the number of the practice exercise listed in the following exercise set. However, the practice exercise is the same in both. CK-12 Basic Algebra: Polynomial Equations in Factored Form (9:29)

  1. What is the Zero Product Property? How does this simplify solving complex polynomials?

Why can’t the Zero Product Property be used to solve the following polynomials?

  1. (x-2)(x)=2
  2. (x+6)+(3x-1)=0
  3. (x^{-3})(x+7)=0
  4. (x+9)-(6x-1)=4
  5. (x^4)(x^2-1)=0

Solve the following polynomial equations.

  1. x(x+12)=0
  2. (2x+3)(5x-4)=0
  3. (2x+1)(2x-1)=0
  4. 24x^2-4x=0
  5. 60m=45m^2
  6. (x-5)(2x+7)(3x-4)=0
  7. 2x(x+9)(7x-20)=0
  8. 18y-3y^2=0
  9. 9x^2=27x
  10. 4a^2+a=0
  11. b^2-\frac{5}{3b}=0

Image Attributions

Reviews

Please wait...
You need to be signed in to perform this action. Please sign-in and try again.
Please wait...

Original text