Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 1 - Section 1.4 - Rational Expressions - 1.4 Exercises: 32

Answer

$\dfrac{x^{2}+2xy+y^{2}}{x^{2}-y^{2}}\cdot\dfrac{2x^{2}-xy-y^{2}}{x^{2}-xy-2y^{2}}=\dfrac{2x+y}{x-2y}$

Work Step by Step

$\dfrac{x^{2}+2xy+y^{2}}{x^{2}-y^{2}}\cdot\dfrac{2x^{2}-xy-y^{2}}{x^{2}-xy-2y^{2}}$ Factor the numerator and denominator of the first fraction: $\dfrac{(x+y)^{2}}{(x-y)(x+y)}\cdot\dfrac{2x^{2}-xy-y^{2}}{x^{2}-xy-2y^{2}}=...$ Simplify the first fraction $...=\dfrac{(x+y)}{(x-y)}\cdot\dfrac{2x^{2}-xy-y^{2}}{x^{2}-xy-2y^{2}}=...$ Evaluate the product $...=\dfrac{2x^{3}-x^{2}y-xy^{2}+2x^{2}y-xy^{2}-y^{3}}{x^{3}-x^{2}y-2xy^{2}-x^{2}y+xy^{2}+2y^{3}}=...$ Simplify like terms: $...=\dfrac{2x^{3}+x^{2}y-2xy^{2}-y^{3}}{x^{3}-2x^{2}y-xy^{2}+2y^{3}}=...$ Factor the numerator and the denominator by grouping terms and simplify: $...=\dfrac{(2x^{3}+x^{2}y)-(2xy^{2}+y^{3})}{(x^{3}-2x^{2}y)-(xy^{2}-2y^{3})}=\dfrac{x^{2}(2x+y)-y^{2}(2x+y)}{x^{2}(x-2y)-y^{2}(x-2y)}=$ $...=\dfrac{(x^{2}-y^{2})(2x+y)}{(x^{2}-y^{2})(x-2y)}=\dfrac{2x+y}{x-2y}$
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