Precalculus (6th Edition) Blitzer

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

Chapter 7 - Cumulative Review Exercises - Page 880: 15


The solution of the equation is $\left\{ \underline{-3,2,\frac{1}{2}} \right\}$

Work Step by Step

We consider the provided equation: $2{{x}^{3}}+{{x}^{2}}-13x+6=0$ Factorize and solve the equation as given below: $\begin{align} & 2{{x}^{3}}-4{{x}^{2}}+5{{x}^{2}}-10x-3x+6=0 \\ & 2{{x}^{2}}\left( x-2 \right)+5x\left( x-2 \right)-3\left( x-2 \right)=0 \\ & \left( x-2 \right)\left( 2{{x}^{2}}+5x-3 \right)=0 \\ & \left( x-2 \right)\left( 2{{x}^{2}}+6x-x-3 \right)=0 \end{align}$ $\begin{align} & \left( x-2 \right)\left( 2x\left( x+3 \right)-1\left( x+3 \right) \right)=0 \\ & \left( x-2 \right)\left( 2x-1 \right)\left( x+3 \right)=0 \end{align}$ Therefore,, $ x=2,\frac{1}{2},-3$. Thus, the solution of the equation is $ x=\left\{ \underline{-3,2,\frac{1}{2}} \right\}$.
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