Intermediate Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-894-7
ISBN 13: 978-0-13417-894-3

Chapter 5 - Section 5.6 - A General Factoring Strategy - Exercise Set - Page 379: 85

Answer

a) $8x^2-2\pi x^2$ b) $(8-2\pi)x^2$

Work Step by Step

a) We write the area of the shaded blue region as the difference between the area of the rectangle and the areas of the two circles: $$\begin{align*} (4x)(2x)-2(\pi x^2)=8x^2-2\pi x^2. \end{align*}$$ b) We factor the expression: $$8x^2-2\pi x^2=(8-2\pi)x^2.$$
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