# 9.7: Zero Product Principle

**Basic**Created by: CK-12

**Practice**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 \begin{align*}x^2 + 2x\end{align*}

### 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:

\begin{align*}6x^4+7x^3-26x^2-17x+30\end{align*}

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.

\begin{align*}(x-1)(x+2)(2x-3)(3x+5)\end{align*}

Suppose we want to know where the polynomial \begin{align*}6x^4+7x^3-26x^2-17x+30\end{align*}

**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 \begin{align*}x\end{align*} x: \begin{align*}(x-1)(x+2)(2x-3)(3x+5)=0\end{align*}(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.

\begin{align*}(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}\end{align*}

The solutions to \begin{align*}6x^4+7x^3-26x^2-17x+30=0\end{align*}

#### Example B

*Solve* \begin{align*}(x-9)(3x+4)=0\end{align*}

**Solution:** Separate the factors using the Zero Product Property: \begin{align*}(x-9)(3x+4)=0\end{align*}

\begin{align*}x-9=0 && \text{or} && 3x+4=0\\
x=9 && && 3x=-4\\
&& && x=\frac{-4}{3}\end{align*}

**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 \begin{align*}= 0\end{align*}

**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:*

\begin{align*}10x^3-5x^2=0\end{align*}

**Solution:**

First, factor by removing the greatest common factor. Since both terms have at least \begin{align*}5x^2\end{align*}

\begin{align*}10x^3-5x^2&=0\\
5x^2(2x-1)&=0
\end{align*}

Separate each factor and set equal to zero:

\begin{align*}5x^2=0 && \text{or} && 2x-1=0\\ x^2=0 && && 2x=1\\ x=0&& && x=\frac{1}{2}\end{align*}

### Guided Practice

*Solve the following polynomial equation.*

\begin{align*}x^2-2x=0\end{align*}

**Solution:** \begin{align*}x^2-2x=0\end{align*}

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

**Factor**: The common factor is \begin{align*}x\end{align*}, so this factors as: \begin{align*}x(x-2)=0\end{align*}.

**Set each factor equal to zero.**

\begin{align*}x=0 && \text{or} && x-2=0\end{align*}

**Solve**:

\begin{align*}x=0 && \text{or} && x=2\end{align*}

**Check**: Substitute each solution back into the original equation.

\begin{align*}x=0 && (0)^2-2(0)=0\\ x=2 && (2)^2-2(2)=0\end{align*}

**Answer** \begin{align*}x=0, \ x=2\end{align*}

### 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)

- 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?

- \begin{align*}(x-2)(x)=2\end{align*}
- \begin{align*}(x+6)+(3x-1)=0\end{align*}
- \begin{align*}(x^{-3})(x+7)=0\end{align*}
- \begin{align*}(x+9)-(6x-1)=4\end{align*}
- \begin{align*}(x^4)(x^2-1)=0\end{align*}

Solve the following polynomial equations.

- \begin{align*}x(x+12)=0\end{align*}
- \begin{align*}(2x+3)(5x-4)=0\end{align*}
- \begin{align*}(2x+1)(2x-1)=0\end{align*}
- \begin{align*}24x^2-4x=0\end{align*}
- \begin{align*}60m=45m^2\end{align*}
- \begin{align*}(x-5)(2x+7)(3x-4)=0\end{align*}
- \begin{align*}2x(x+9)(7x-20)=0\end{align*}
- \begin{align*}18y-3y^2=0\end{align*}
- \begin{align*}9x^2=27x\end{align*}
- \begin{align*}4a^2+a=0\end{align*}
- \begin{align*}b^2-\frac{5}{3b}=0\end{align*}

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factored form

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

expanded form

Polynomials can be written in**expanded form**or in

**factored form**. The expanded form contains sums and differences of different terms.

Zero Product Property

The only way a product is zero is if one or more of the terms are equal to zero:Zero Product Rule

The zero product rule states that if the product of two expressions is equal to zero, then at least one of the original expressions much be equal to zero.### Image Attributions

Here you'll find the solutions to polynomial equations by using the Zero Product Principle.