Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

Published by Pearson
ISBN 10: 0-32184-874-8
ISBN 13: 978-0-32184-874-1

Appendix C - Synthetic Division and the Remainder Theorem - C Exercise Set - Page 992: 7


The value of $\left( {{x}^{3}}-4{{x}^{2}}-2x+5 \right)\div \left( x-1 \right)$is ${{x}^{2}}-3x-5$.

Work Step by Step

Consider the expression: $\left( {{x}^{3}}-4{{x}^{2}}-2x+5 \right)\div \left( x-1 \right)$ The constant term of the divisor$-1$ is written to the left. The coefficients of the dividend 1,$-4,-2$, 5 are written to the right. Evaluate the value of $\left( {{x}^{3}}-4{{x}^{2}}-2x+5 \right)\div \left( x-1 \right)$ using synthetic division as follows. $\begin{array}{c|rrrr}{1}&1&-4&-2&{5}\\&&1&-3&{1} \cdot {-5}={-5}\\\hline&1&-3&{-5}&\left({5}\right)+{-5}={0}\end{array} $ The obtained numbers are the coefficients of a polynomial and the last digit is the remainder. So, the polynomial has coefficients $1,-3,-5$is ${{x}^{2}}-3x-5$ and the remainder is 0. Therefore, the value of$\left( {{x}^{3}}-4{{x}^{2}}-2x+5 \right)\div \left( x-1 \right)$ is ${{x}^{2}}-3x-5$.
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