## Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole

# Chapter 1 - Review - Exercises: 42

#### Answer

$x+2x\sqrt{x}-\sqrt{x}$ or also $x-x^{1/2}+2x^{3/2}$

#### Work Step by Step

$\sqrt{x}(\sqrt{x}+1)(2\sqrt{x}-1)$ Multiply the first two factors: $\sqrt{x}(\sqrt{x}+1)(2\sqrt{x}-1)=[(\sqrt{x})^{2}+\sqrt{x}](2\sqrt{x}-1)=...$ $...=(x+\sqrt{x})(2\sqrt{x}-1)=...$ Evaluate the remaining product and simplify: $...=2x\sqrt{x}-x+2(\sqrt{x})^{2}-\sqrt{x}=...$ $...=2x\sqrt{x}-x+2x-\sqrt{x}=...$ $...=x+2x\sqrt{x}-\sqrt{x}$ There is another way to express the answer. It is using rational exponents. Change the roots to powers with rational exponent and simplify: $x+2(x)(x^{1/2})-x^{1/2}=...$ $...=x+2x^{3/2}-x^{1/2}=...$ Rearrange: $...=x-x^{1/2}+2x^{3/2}$

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