# 9.4: Polynomial Equations in Factored Form

**At Grade**Created by: CK-12

We have been multiplying polynomials by using the Distributive Property, where all the terms in one polynomial must be multiplied by all terms in the other polynomial. In this lesson, you will start learning how to do this process using a different method called **factoring**.

**Factoring:** Take the factors that are common to all the terms in a polynomial. Then multiply the common factors by a parenthetical expression containing all the terms that are left over when you divide out the common factors.

Let’s look at the areas of the rectangles again: \begin{align*}Area = length \times width\end{align*}

**Method 1:** Find the areas of all the small rectangles and add them.

Blue rectangle \begin{align*}= ab\end{align*}

Orange rectangle \begin{align*}= ac\end{align*}

Red rectangle \begin{align*}=ad\end{align*}

Green rectangle \begin{align*}=ae\end{align*}

Purple rectangle \begin{align*}=2a\end{align*}

Total area \begin{align*}=ab+ac+ad+ae+2a\end{align*}

**Method 2:** Find the area of the big rectangle all at once.

\begin{align*}\text{Length} & = a\\
\text{Width} & = b+c+d+e+2\\
\text{Area} & = a(b+c+d+e=2)\end{align*}

The answers are the same no matter which method you use:

\begin{align*}ab+ac+ad+ae+2a=a(b+c+d+e+2)\end{align*}

## Using the Zero Product Property

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.

**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.

\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 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*} are \begin{align*}x=-2,-\frac{5}{3},1,\frac{3}{2}\end{align*}.

**Example 2:** *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*}

## Finding the Greatest Common Monomial Factor

Once we get a polynomial in factored form, it is easier to solve the polynomial equation. But first, we need to learn how to factor. Factoring can take several steps because we want to factor completely so we cannot factor any more.

A **common factor** can be a number, a variable, or a combination of numbers and variables that appear in **every term** of the polynomial.

When a common factor is factored from a polynomial, you divide each term by the common factor. What is left over remains in parentheses.

**Example 3:** *Factor:*

- \begin{align*}15x-25\end{align*}
- \begin{align*}3a+9b+6\end{align*}

**Solution:**

1. We see that the factor of 5 divides evenly from all terms.

\begin{align*}15x-25=5(3x-5)\end{align*}

2. We see that the factor of 3 divides evenly from all terms.

\begin{align*}3a+9b+6=3(a+3b+2)\end{align*}

Now we will use examples where different powers can be factored and there is more than one common factor.

**Example 4:** *Find the greatest common factor.*

(a) \begin{align*}a^3-3a^2+4a\end{align*}

(b) \begin{align*}5x^3y-15x^2y^2+25xy^3\end{align*}

**Solution:**

(a) Notice that the factor \begin{align*}a\end{align*} appears in all terms of \begin{align*}a^3-3a^2+4a\end{align*} but each term has a different power of \begin{align*}a\end{align*}. The common factor is the lowest power that appears in the expression. In this case the factor is \begin{align*}a\end{align*}.

Let’s rewrite \begin{align*}a^3-3a^2+4a=a(a^2)+a(-3a)+a(4)\end{align*}

Factor \begin{align*}a\end{align*} to get \begin{align*}a(a^2-3a+4)\end{align*}

(b) The common factors are \begin{align*}5xy\end{align*}.

When we factor \begin{align*}5xy\end{align*}, we obtain \begin{align*}5xy(x^2-3xy+5y^2)\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 5:** *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 Set

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?

Factor the common factor from the following polynomials.

- \begin{align*}36a^2+9a^3-6a^7\end{align*}
- \begin{align*}4x^3 y+12x+16y\end{align*}
- \begin{align*}3x^3-21x\end{align*}
- \begin{align*}5x^6+15x^4\end{align*}
- \begin{align*}4x^3+10x^2-2x\end{align*}
- \begin{align*}-10x^6+12x^5-4x^4\end{align*}
- \begin{align*}12xy+24xy^2+36xy^3\end{align*}
- \begin{align*}5a^3-7a\end{align*}
- \begin{align*}45y^{12}+30y^{10}\end{align*}
- \begin{align*}16xy^2z+4x^3y\end{align*}

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)(\frac{1}{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^2-3b=0\end{align*}

**Mixed Review**

- Rewrite in standard form: \begin{align*}-4x+11x^4-6x^7+1-3x^2\end{align*}. State the polynomial's degree and leading coefficient.
- Simplify \begin{align*}(9a^2-8a+11a^3)-(3a^2+14a^5-12a)+(9-3a^5-13a)\end{align*}.
- Multiply \begin{align*}\frac{1}{3} a^3\end{align*} by \begin{align*}(36a^4+6)\end{align*}.
- Melissa made a trail mix by combining \begin{align*}x\end{align*} ounces of a 40% cashew mixture with \begin{align*}y\end{align*} ounces of a 30% cashew mixture. The result is 12 ounces of cashews.
- Write the equation to represent this situation.
- Graph using its intercepts.
- Give three possible combinations to make this sentence true.

- Explain how to use mental math to simplify 8(12.99).

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