Suppose that you know that the area of a chalkboard is 48 square feet, and you also know that the area in square feet can be represented by the expression \begin{align*}x^2 + 2x\end{align*}. How can you find the value of \begin{align*}x\end{align*} if you wrote an equation, moved all the terms to one side, and factored?

### Zero Product Principle

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*} 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.

The **Zero Product Property **states that 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.

#### Let's solve for \begin{align*}x\end{align*} in the following equations:

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

As shown earlier, the factored version of this equation is:

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

According to the property, for the original polynomial to equal zero, we have to set each factor 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*} are the following: \begin{align*}x=-2,-\frac{5}{3},1,\frac{3}{2}\end{align*}.

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

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.

#### Let's solve the following polynomial equation:

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

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

\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*}

### Examples

#### Example 1

Earlier, you were told that the area of a chalkboard is 48 square feet and that the area in square feet could be represented by the expression \begin{align*}x^2 + 2x\end{align*}. What is the value of \begin{align*}x\end{align*}?

The starting equation is:

\begin{align*}x^2 + 2x = 48\end{align*}

To find the value of \begin{align*}x\end{align*}, we need to follow the steps for solving an equation in expanded form.

Rewrite:

\begin{align*}x^2 + 2x &= 48\\ x^2 + 2x -48&=0\end{align*}

Factor:

Set each factor equal to zero and solve:

\begin{align*}(x+8)&=0 \rightarrow x=-8\\ (x-6)&=0 \rightarrow x=6\end{align*}

Check:

\begin{align*}x=-8 && (-8)^2+2(-8)=64-16=48\\ x=6 && (6)^2 +2(6)=36+12 = 48\end{align*}

The check shows us that the answers are correct. \begin{align*}x\end{align*} is either -8 or 6.

#### Example 2

Solve the following polynomial equation.

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

\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*}

The answer is \begin{align*}x=0, \ x=2\end{align*}.

### Review

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

In 2-6, explain why the Zero Product Property can't be used to solve the polynomial.

- \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*}

In 7-16, solve the 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*}

### Review (Answers)

To see the Review answers, open this PDF file and look for section 9.7.