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 360: 52


The solution set is $\left\{-2, 0, 5 \right\}$.

Work Step by Step

Simplify the equation to obtain: \begin{align*} w^3-3w^2-3w+9-7w-9&=0\\ w^3-3w^2-10w&=0 \end{align*} Factor out $w$: $$w(w^2-3w-10)=0$$ Factor the trinomial to obtain: $$w(w-5)(w+2)=0$$ Use the Zero-Product Property by equating each factor to zero, then solve each equation to obtain: \begin{align*} w&=0 &\text{or}& &w-5=0& &\text{or}& &w+2=0\\ w&=0 &\text{or}& &w=5& &\text{or}& &w=-2\\ \end{align*} Thus, the solution set is $\left\{-2, 0, 5 \right\}$.
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