Precalculus (6th Edition) Blitzer

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

Chapter 2 - Mid-Chapter Check Point - Page 381: 22


The solutions of the equation are $x=1$ , $x=\frac{1}{3}$ , and $x=\frac{1}{2}$.

Work Step by Step

Consider the polynomial equation $f\left( x \right)=6{{x}^{3}}-11{{x}^{2}}+6x-1$. Use the rational root theorem to observe: ${{a}_{0}}=1\text{ , }{{\text{a}}_{n}}=6$ Factors of ${{a}_{0}}$ are: 1 and the factors of ${{a}_{n}}$ are: 1, 2, 3, 6. The possible roots are given as: $\pm \frac{1}{1,2,3,6}$. $\frac{1}{1}$ is a root of the expression, so factor out is $x-1$. Compute $\frac{6{{x}^{3}}\ -\ 11{{x}^{2}}\ +\ 6x\ -\ 1}{x\ -\ 1}$ to get the rest of the equation: $6{{x}^{2}}\ -\ 5x\ +\ 1$. $\begin{align} & \left( x\ -\ 1 \right)\ \left( 6{{x}^{2}}\ -\ 5x\ +\ 1 \right)=0 \\ & \left( x-1 \right)\left( 2x-1 \right)\left( 3x-1 \right)=0 \\ & x=1,\frac{1}{2},\frac{1}{3} \end{align}$ Hence, the roots of the polynomial equation $f\left( x \right)=6{{x}^{3}}-11{{x}^{2}}+6x-1$ are $x=1,\frac{1}{2},\frac{1}{3}$.
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