Precalculus (6th Edition)

Published by Pearson
ISBN 10: 013421742X
ISBN 13: 978-0-13421-742-0

Chapter R - Review of Basic Concepts - R.4 Factoring Polynomials - R.4 Exercises - Page 45: 120



Work Step by Step

With $\frac{121}{25}y^4=(\frac{11}{5}y^2)^2$ and $49x^2=(7x)^2$, the given binomial is equivalent to: $=(\frac{11}{5}y^2)^2-(7x)^2$ Factor using the formula $a^2-b^2=(a+b)(a-b)$ with $a=\frac{11}{5}y^2$ and $b=7x$ to obtain: $=\color{blue}{(\frac{11}{5}y^2+7x)(\frac{11}{5}y^2-7x)}$
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