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 69: 93

Answer

$x^{\frac{1}{2}}(x-1)$

Work Step by Step

$x^{3/2}=x^{1+\frac{1}{2}}=x^{1}\cdot x^{\frac{1}{2}}$ So, $ x^{\frac{3}{2}}-x^{\frac{1}{2}}=x^{1}\cdot x^{\frac{1}{2}}-x^{\frac{1}{2}}\qquad$... factor out $x^{\frac{1}{2}}$ =$x^{\frac{1}{2}}(x-1)$
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