Precalculus (6th Edition) Blitzer

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

Chapter 2 - Section 2.4 - Dividing Polynomials; Remainder and Factor Theorems - Exercise Set - Page 365: 71


The given statement is true.

Work Step by Step

Let us try to perform synthetic division over division of $f\left( x \right)$ by $g\left( x \right)$. $\frac{1}{2}\overline{\left){\begin{align} & 10\,\,\,\,\,-6\,\,\,\,\,\,\,4\,\,\,\,\,\,-1 \\ & \frac{\downarrow \,\,\,\,\,\,\,\,\,5\,\,\,\,\,-\frac{1}{2}\,\,\,\,\,\,\,\frac{7}{4}\,}{\begin{align} & 10\,\,\,\,\,-1\,\,\,\,\,\,\,\,\frac{7}{2}\,\,\,\,\,\,\,\,\frac{3}{4}\, \\ & \frac{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\downarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\downarrow \,\,}{10{{x}^{2}}-x+\frac{7}{2}\,\,\,\,\,\,\,R}\, \\ \end{align}} \\ \end{align}}\right.}$ Here, the quotient obtained is $h\left( x \right)=10{{x}^{2}}-x+\frac{7}{2}$ and the remainder is $\frac{3}{4}$. By checking out the solution: $\begin{align} & f\left( x \right)=g\left( x \right).h\left( x \right)+R \\ & =\left( x-\frac{1}{2} \right)\left( 10{{x}^{2}}-x+\frac{7}{2} \right)+\frac{3}{4} \\ & =10{{x}^{3}}-{{x}^{2}}+\frac{7}{2}x-5{{x}^{2}}+\frac{1}{2}x-\frac{7}{4}+\frac{3}{4} \\ & =10{{x}^{3}}-6{{x}^{2}}+4x-1. \end{align}$ Therefore, synthetic division can be used to find the quotient of $f\left( x \right)=10{{x}^{3}}-6{{x}^{2}}+4x-1$ when divided by $g\left( x \right)=x-\frac{1}{2}$.
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