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: 27



Work Step by Step

$ A_{1}=\pi r^{2}\qquad$...area of a circle $ A_{1}=\pi(5x)^{2}\qquad$...substitute 5x for r. $A_{1}=25\pi x^{2}\qquad...$simplify. $A_{2}=s^{2}\qquad...$area of a square. $A_{2}=(15x)^{2}\qquad...$substitute 15x for s. $A_{2}=225x^{2}\qquad...$simplify. The area of the metal frame is $A_{2}-A_{1}$, or $225x^{2}-25\pi x^{2}$. $225x^{2}-25\pi x^{2}\qquad...$find the GCF. $225x^{2}=3\cdot 3\cdot(5\cdot 5)\cdot(x\cdot x)$ $-25\pi x^{2}=-1\cdot(5\cdot 5)\cdot\pi\cdot(x\cdot x)$ GCF=$(5\cdot 5)\cdot(x\cdot x)$, or $25x^{2}$. $225x^{2}-25\pi x^{2}\qquad...$factor out the GCF $=25x^{2}(9)-25x^{2}(\pi)\qquad...$apply the Distributive Property. $=25x^{2}(9-\pi)$ The factored form of the area of the metal frame is $25x^{2}(9-\pi)$.
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