Intermediate Algebra (12th Edition)

Published by Pearson
ISBN 10: 0321969359
ISBN 13: 978-0-32196-935-4

Chapter 5 - Review Exercises: 15

Answer

$(2k-h)(5k-3h)$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To factor the given expression, $ 10k^2-11kh+3h^2 ,$ 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. $\bf{\text{Solution Details:}}$ To factor the trinomial expression above, note that the value of $ac$ is $ 10(3)=30 $ and the value of $b$ is $ -11 .$ The possible pairs of integers whose product is $ac$ are \begin{array}{l}\require{cancel} \{1,30\}, \{2,15\}, \{3,10\}, \{5,6\}, \\ \{-1,-30\}, \{-2,-15\}, \{-3,-10\}, \{-5,-6\} .\end{array} Among these pairs, the one that gives a sum of $b$ is $\{ -5,-6 \}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to \begin{array}{l}\require{cancel} 10k^2-5kh-6kh+3h^2 .\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} (10k^2-5kh)-(6kh-3h^2) .\end{array} Factoring the $GCF$ in each group results to \begin{array}{l}\require{cancel} 5k(2k-h)-3h(2k-h) .\end{array} Factoring the $GCF= (2k-h) $ of the entire expression above results to \begin{array}{l}\require{cancel} (2k-h)(5k-3h) .\end{array}
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