Intermediate Algebra (6th Edition)

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

Chapter 5 - Section 5.3 - Polynomials and Polynomial Functions - Exercise Set: 24

Answer

$\frac{7}{4}$

Work Step by Step

We can find the value of $P(\frac{1}{2})$ by plugging $\frac{1}{2}$ into the formula for $P(x)$. $P(x)=x^{2}+x+1$ $P(\frac{1}{2})=(\frac{1}{2})^{2}+\frac{1}{2}+1=(\frac{1}{2}\times\frac{1}{2})+\frac{1}{2}+1=\frac{1}{4}+\frac{1}{2}+1=\frac{1}{4}+\frac{2}{4}+\frac{4}{4}=\frac{7}{4}$
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