Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter P - Mid-Chapter Check Point - Page 71: 36

Answer

$ \displaystyle \frac{(x-3)(x+3)}{ (x^{2}+1)^{1/2}}$

Work Step by Step

$(x^{2}+1)^{\frac{1}{2}}=(x^{2}+1)^{-\frac{1}{2}+1}=(x^{2}+1)^{-\frac{1}{2}}\cdot(x^{2}+1)$ factor out $(x^{2}+1)^{-1/2}$ $= (x^{2}+1)^{-1/2}\left[ (x^{2}+1)-10 \right]$ $=(x^{2}+1)^{-1/2}\left[ x^{2}-9 \right]$ the brackets hold a difference of squares, $x^{2}-3^{2}$, use $a^{2}-b^{2}=(a+b)(a-b)$ = $(x^{2}+1)^{-1/2}(x-3)(x+3)\qquad $... apply $a^{-n}=\displaystyle \frac{1}{a^{n}}$ $= \displaystyle \frac{(x-3)(x+3)}{ (x^{2}+1)^{1/2}}$
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