Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 2 - Cumulative Review Exercises - Page 435: 10

Answer

The equation has the solution $\left\{ -1,2,-3 \right\}$.

Work Step by Step

Consider the cubic equation ${{x}^{3}}+2{{x}^{2}}-5x-6=0$. Apply the hit and trial method and obtain that $\left( x+1 \right)$ is one of the factors of ${{x}^{3}}+2{{x}^{2}}-5x-6=0$ Simplify: ${{x}^{3}}+2{{x}^{2}}-5x-6=\left( x+1 \right)\left( {{x}^{2}}+x-6 \right)$ Factorize $\left( {{x}^{2}}+x-6 \right)$: $\begin{align} & \left( {{x}^{2}}+x-6 \right)={{x}^{2}}+3x-2x-6 \\ & =x\left( x+3 \right)-2\left( x+3 \right) \\ & =\left( x-2 \right)\left( x+3 \right) \end{align}$ Thus, ${{x}^{3}}+2{{x}^{2}}-5x-6=\left( x+1 \right)\left( x-2 \right)\left( x+3 \right)$ Equate the value of the expression to 0. $\begin{align} & \left( x+1 \right)\left( x-2 \right)\left( x+3 \right)=0 \\ & x=-1,2,-3 \end{align}$
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