Algebra 1: Common Core (15th Edition)

Published by Prentice Hall
ISBN 10: 0133281140
ISBN 13: 978-0-13328-114-9

Chapter 8 - Polynomials and Factoring - 8-2 Multiplying and Factoring - Got It? - Page 494: 4



Work Step by Step

$A_{1}=s^{2}\qquad...$area of a square. $A_{1}=(6x)^{2}\qquad...$substitute 6x for s. $A_{1}=36x^{2}\qquad...$simplify. $ A_{2}=\pi r^{2}\qquad$...area of a circle $ A_{2}=\pi(3x)^{2}\qquad$...substitute 3x for r. $A_{2}=9\pi x^{2}\qquad...$simplify. The area of the shaded region is $A_{1}-A_{2}$, or $36x^{2}-9\pi x^{2}$. $36x^{2}-9\pi x^{2}\qquad...$find the GCF. $36x^{2}=2\cdot 2\cdot(3\cdot 3)\cdot(x\cdot x)$ $-9\pi x^{2}=-1\cdot(3\cdot 3)\cdot\pi\cdot(x\cdot x)$ GCF=$(3\cdot 3)\cdot(x\cdot x)$, or $9x^{2}$. $36x^{2}-9\pi x^{2}\qquad...$factor out the GCF $=9x^{2}(4)-9x^{2}(\pi)\qquad...$apply the Distributive Property. $=9x^{2}(4-\pi)$ The factored form of the area of the shaded region is $9x^{2}(4-\pi)$.
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