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9.4: Polynomial Equations in Factored Form

Difficulty Level: At Grade Created by: CK-12
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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.


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 \begin{align*}(a + b)(x + y) \Rightarrow ax + bx + ay + by\end{align*}(a+b)(x+y)ax+bx+ay+by; they will now learn to “put it back together”: \begin{align*}ax + bx + ay + by \Rightarrow (a + b)(x + y)\end{align*}ax+bx+ay+by(a+b)(x+y).
  • 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 \begin{align*}(x, y, a,\end{align*}(x,y,a, or \begin{align*}b)\end{align*}b) 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.


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

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

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

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

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

(Are students incorrectly setting each factor equal to ?)

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

For example, \begin{align*}6ax^2 - 9ax + 3a \neq 3a(2x^2 - 3x)\end{align*}6ax29ax+3a3a(2x23x); but \begin{align*}6ax^2 - 9ax + 3a = 3a(2x^2 - 3x + 1)\end{align*}6ax29ax+3a=3a(2x23x+1). 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 \begin{align*}6ax^2 - 9ax + 3a = 3a(2x^2 - 3x)\end{align*}6ax29ax+3a=3a(2x23x), students will convince themselves that a \begin{align*}1\end{align*}1 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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