## Intermediate Algebra (12th Edition)

Published by Pearson

# Chapter 7 - Section 7.5 - Multiplying and Dividing Radical Expressions - 7.5 Exercises: 105

#### Answer

$\dfrac{6+2\sqrt{6p}}{3}$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$ To simplify the given expression, $\dfrac{6p+\sqrt{24p^3}}{3p} ,$ simplify the radicand that contains a factor that is a perfect power of the index Then, find the $GCF$ of all the terms and express all terms as factors using the $GCF.$ Finally, cancel the $GCF$ in all the terms. $\bf{\text{Solution Details:}}$ Writing the radicand as an expression containing a factor that is a perfect power of the index and extracting the root of that factor result to \begin{array}{l}\require{cancel} \dfrac{6p+\sqrt{4p^2\cdot6p}}{3p} \\\\= \dfrac{6p+\sqrt{(2p)^2\cdot6p}}{3p} \\\\= \dfrac{6p+2p\sqrt{6p}}{3p} .\end{array} The $GCF$ of the constants of the terms $\{ 6,2,3 \}$ is $1$ since it is the highest number that can divide all the given constants. The $GCF$ of the common variable/s is the variable/s with the lowest exponent. Hence, the $GCF$ of the common variable/s $\{ p,p,p \}$ is $p .$ Hence, the entire expression has $GCF= 1p \text{ or } p .$ Writing the given expression as factors using the $GCF$ results to \begin{array}{l}\require{cancel} \dfrac{p\cdot6+p\cdot2\sqrt{6p}}{p\cdot3} .\end{array} Cancelling the $GCF$ in every term results to \begin{array}{l}\require{cancel} \dfrac{\cancel{p}\cdot6+\cancel{p}\cdot2\sqrt{6p}}{\cancel{p}\cdot3} \\\\= \dfrac{6+2\sqrt{6p}}{3} .\end{array}

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