Intermediate Algebra: Connecting Concepts through Application

Published by Brooks Cole
ISBN 10: 0-53449-636-9
ISBN 13: 978-0-53449-636-4

Chapter 3 - Exponents, Polynomials and Functions - 3.5 Special Factoring Techniques - 3.5 Exercises: 63

Answer

$(2r^{\frac{1}{2}}-5)(3r^{\frac{1}{2}}+4)$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To factor the given expression, $ 6r+7r^{\frac{1}{2}}-20 ,$ find two numbers whose product is $ac$ and whose sum is $b$ in the quadratic expression $ax^2+bx+c.$ Use these $2$ numbers to decompose the middle term of the given quadratic expression and then use factoring by grouping. $\bf{\text{Solution Details:}}$ Using factoring of trinomials, the value of $ac$ in the trinomial expression above is $ 6(-20)=-120 $ and the value of $b$ is $ 7 .$ The $2$ numbers that have a product of $ac$ and a sum of $b$ are $\left\{ -15,8 \right\}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to \begin{array}{l}\require{cancel} 6r-15r^{\frac{1}{2}}+8r^{\frac{1}{2}}-20 .\end{array} Grouping the first and second terms and the third and fourth terms, the given expression is equivalent to \begin{array}{l}\require{cancel} (6r-15r^{\frac{1}{2}})+(8r^{\frac{1}{2}}-20) .\end{array} Factoring the $GCF$ in each group results to \begin{array}{l}\require{cancel} 3r^{\frac{1}{2}}(2r^{\frac{1}{2}}-5)+4(2r^{\frac{1}{2}}-5) .\end{array} Factoring the $GCF= (2r^{\frac{1}{2}}-5) $ of the entire expression above results to \begin{array}{l}\require{cancel} (2r^{\frac{1}{2}}-5)(3r^{\frac{1}{2}}+4) .\end{array}
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