Precalculus (6th Edition) Blitzer

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

Chapter P - Section P.5 - Factoring Polynomials - Exercise Set - Page 70: 141

Answer

The required expression is $\underline{\frac{-10}{{{\left( x+5 \right)}^{\frac{1}{2}}}{{\left( x-5 \right)}^{\frac{3}{2}}}}}$.

Work Step by Step

Consider the given polynomial: ${{\left( x-5 \right)}^{\frac{-1}{2}}}{{\left( x+5 \right)}^{\frac{-1}{2}}}-{{\left( x+5 \right)}^{\frac{1}{2}}}{{\left( x-5 \right)}^{\frac{-3}{2}}}$ ………..….. (1) Apply the technique, ‘‘factoring the polynomial by greatest common factor.’’ $\begin{align} & {{\left( x-5 \right)}^{\frac{-1}{2}}}{{\left( x+5 \right)}^{\frac{-1}{2}}}-{{\left( x+5 \right)}^{\frac{1}{2}}}{{\left( x-5 \right)}^{\frac{-3}{2}}}={{\left( x+5 \right)}^{\frac{-1}{2}}}{{\left( x-5 \right)}^{\frac{-3}{2}}}\left( {{\left( x-5 \right)}^{\frac{-1}{2}-\left( \frac{-3}{2} \right)}}-{{\left( x+5 \right)}^{\frac{1}{2}-\left( \frac{-1}{2} \right)}} \right) \\ & ={{\left( x+5 \right)}^{\frac{-1}{2}}}{{\left( x-5 \right)}^{\frac{-3}{2}}}\left( \left( x-5 \right)-\left( x+5 \right) \right) \\ & ={{\left( x+5 \right)}^{\frac{-1}{2}}}{{\left( x-5 \right)}^{\frac{-3}{2}}}\left( x-5-x-5 \right) \end{align}$ Further solve, $\begin{align} & {{\left( x-5 \right)}^{\frac{-1}{2}}}{{\left( x+5 \right)}^{\frac{-1}{2}}}-{{\left( x+5 \right)}^{\frac{1}{2}}}{{\left( x-5 \right)}^{\frac{-3}{2}}}=-10{{\left( x+5 \right)}^{\frac{-1}{2}}}{{\left( x-5 \right)}^{\frac{-3}{2}}} \\ & =\frac{-10}{{{\left( x+5 \right)}^{\frac{1}{2}}}{{\left( x-5 \right)}^{\frac{3}{2}}}} \end{align}$
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