Intermediate Algebra: Connecting Concepts through Application

Published by Brooks Cole
ISBN 10: 0-53449-636-9
ISBN 13: 978-0-53449-636-4

Chapter 4 - Quadratic Functions - 4.5 Solving Equations by Factoring - 4.5 Exercises - Page 362: 94


The solution set is $\left\{-3, 0, 3\right\}$.

Work Step by Step

Factor out $5c$ to obtain: \begin{align*} 5c(c^2-9)&=0\\ 5c(c-3)(c+3)&=0 \end{align*} Use the Zero-Product Property by equating each factor to zero. Then, solve each equation to obtain: \begin{align*} 5c&=0 &\text{or}& &c-3=0& &\text{or}& &c+3=0\\ c&=0 &\text{or}& &c=3& &\text{or}& &c=-3\\ \end{align*} Checking: \begin{align*} 5(0^3)-45(0)&=0\\ 5(0)-0&=0\\ 0&\stackrel{\checkmark}=0 \end{align*} \begin{align*} 5(3^3)-45(3)&=0\\ 5(27)-135&=0\\ 135-135&=0\\ 0&\stackrel{\checkmark}=0 \end{align*} \begin{align*} 5(-3)^3-45(-3)&=0\\ 5(-27)+135&=0\\ -135+135&=0\\ 0&\stackrel{\checkmark}=0 \end{align*} Therefiore, the solution set is $\left\{-3, 0, 3\right\}$.
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