Intermediate Algebra (6th Edition)

Published by Pearson
ISBN 10: 0321785045
ISBN 13: 978-0-32178-504-6

Chapter 6 - Section 6.4 - Dividing Polynomials: Long Division and Synthetic Division - Exercise Set - Page 370: 56

Answer

$P\left( \dfrac{2}{3} \right)=\dfrac{968}{243}$

Work Step by Step

Substituting $x$ with $ \dfrac{2}{3} $ in $ P(x)=x^5-2x^3+4x^2-5x+6 $, then \begin{array}{l} P\left( \dfrac{2}{3} \right)=\left( \dfrac{2}{3} \right)^5-2\left( \dfrac{2}{3} \right)^3+4\left( \dfrac{2}{3} \right)^2-5\left( \dfrac{2}{3} \right)+6 \\\\ P\left( \dfrac{2}{3} \right)=\dfrac{32}{243}-2\left( \dfrac{8}{27} \right)+4\left( \dfrac{4}{9} \right)-5\left( \dfrac{2}{3} \right)+6 \\\\ P\left( \dfrac{2}{3} \right)=\dfrac{32}{243}-\dfrac{16}{27}+\dfrac{16}{9}-\dfrac{10}{3}+6 \\\\ P\left( \dfrac{2}{3} \right)=\dfrac{968}{243} .\end{array}
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