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: 54

Answer

$P\left( \dfrac{1}{2} \right)=-\dfrac{17}{4}$

Work Step by Step

Substituting $x$ with $ \dfrac{1}{2} $ in $ P(x)=4x^4-2x^3+x^2-x-4 $, then \begin{array}{l} P\left( \dfrac{1}{2} \right)=4\left( \dfrac{1}{2} \right)^4-2\left( \dfrac{1}{2} \right)^3+\left( \dfrac{1}{2} \right)^2-\left( \dfrac{1}{2} \right)-4 \\\\ P\left( \dfrac{1}{2} \right)=4\left( \dfrac{1}{16} \right)-2\left( \dfrac{1}{8} \right)+\left( \dfrac{1}{4} \right)-\left( \dfrac{1}{2} \right)-4 \\\\ P\left( \dfrac{1}{2} \right)=\dfrac{1}{4}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{2}-4 \\\\ P\left( \dfrac{1}{2} \right)=-\dfrac{17}{4} .\end{array}
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