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 - Practice and Problem-Solving Exercises - Page 495: 28



Work Step by Step

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