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# 9.4: Polynomial Equations in Factored Form

Difficulty Level: At Grade Created by: CK-12

## Learning Objectives

At the end of this lesson, students will be able to:

• Use the zero-product property.
• Find greatest common monomial factor.
• Solve simple polynomial equations by factoring.

## Vocabulary

Terms introduced in this lesson:

factoring, factoring a polynomial
expanded form
factored form
zero product property
factoring completely
common factor
greatest common monomial factor
polynomial equation

## Teaching Strategies and Tips

Use the introduction to motivate factoring.

• The reverse of distribution is called factoring.
• Whereas before students were learning the direction (a+b)(x+y)ax+bx+ay+by\begin{align*}(a + b)(x + y) \Rightarrow ax + bx + ay + by\end{align*}; they will now learn to “put it back together”: ax+bx+ay+by(a+b)(x+y)\begin{align*}ax + bx + ay + by \Rightarrow (a + b)(x + y)\end{align*}.
• Students realize that polynomials can be expressed in expanded or factored form

Teachers may decide to have their students pull common factors out one at a time, instead of factoring the GCF in one step.

## Error Troubleshooting

In Review Questions 9 and 12-16, remind students to set the monomial factor (x,y,a,\begin{align*}(x, y, a,\end{align*} or b)\begin{align*}b)\end{align*} equal to zero.

• Caution students against dividing by variables. In doing so, they will lose as a solution. See also Example 6.

General Tip: Check that students are using the zero-product property correctly.

Examples:

a. Solve for x\begin{align*}x\end{align*}.

(x+3)(x4)=8\begin{align*}(x + 3)(x - 4) = 8\end{align*}

(Are students incorrectly setting each factor equal to 8\begin{align*}8\end{align*}?)

b. Solve for x\begin{align*}x\end{align*}.

(x+3)(x4)2=0\begin{align*}(x + 3)(x - 4) - 2 = 0\end{align*}

(Are students incorrectly setting each factor equal to ?)

General Tip: Remind students when factoring the GCF out of itself to leave a 1\begin{align*}1\end{align*}.

For example, 6ax29ax+3a3a(2x23x)\begin{align*}6ax^2 - 9ax + 3a \neq 3a(2x^2 - 3x)\end{align*}; but 6ax29ax+3a=3a(2x23x+1)\begin{align*}6ax^2 - 9ax + 3a = 3a(2x^2 - 3x + 1)\end{align*}. See Example 5b and Review Questions 3 and 15.

General Tip: Have students check their work by expanding the factored polynomial.

• By checking a problem worked out as 6ax29ax+3a=3a(2x23x)\begin{align*}6ax^2 - 9ax + 3a = 3a(2x^2 - 3x)\end{align*}, students will convince themselves that a 1\begin{align*}1\end{align*} is missing.

General Tip: Suggest that students look carefully over the remaining terms after having factored out the GCF so as to not leave any other common factors.

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