Introductory Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-805-X
ISBN 13: 978-0-13417-805-9

Chapter 6 - Section 6.1 - The Greatest Common Factor and Factoring by Grouping - Exercise Set - Page 428: 96


$16x^{2}-2\cdot\pi\cdot x^{2}=2x^{2}(8-\pi)$

Work Step by Step

The area of a square = $($side length$)^{2}$ The area of a circle = $\pi\cdot($radius$)^{2}$ Shaded area = (area of square) - 2$\times$(area of circle) $=(4x)^{2}-2\cdot\pi\cdot x^{2}$ $=16x^{2}-\pi\cdot 2x^{2}$ GCF of the terms is $2x^{2}$. Use the distributive property to factor: $=2x^{2}\cdot 8-2x^{2}\cdot\pi$ $=2x^{2}(8-\pi)$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.