Intermediate Algebra (12th Edition)

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

Chapter 5 - Section 5.2 - Factoring Trinomials - 5.2 Exercises: 35

Answer

$(4k+3)(5k+8)$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To factor the quadratic expression $ax^2+bx+c,$ find two numbers whose product is $ac$ and whose sum is $b$. Use these $2$ numbers to decompose the middle term of the quadratic expression and then use factoring by grouping. $\bf{\text{Solution Details:}}$ In the given expression, $ 20k^2+47k+24 ,$ the value of $ac$ is $ 20(24)=480 $ and the value of $b$ is $ 47 .$ The $2$ numbers that have a product $ac$ and a sum of $b$ are $\{ 15,32 \}.$ Using these $2$ numbers to decompose the middle term of the given expression results to \begin{array}{l}\require{cancel} 20k^2+15k+32k+24 .\end{array} Grouping the first and third terms and the second and fourth terms, the given expression is equivalent to \begin{array}{l}\require{cancel} (20k^2+15k)+(32k+24) .\end{array} Factoring the $GCF$ in each group results to \begin{array}{l}\require{cancel} 5k(4k+3)+8(4k+3) .\end{array} Factoring the $GCF= (4k+3) $ of the entire expression above results to \begin{array}{l}\require{cancel} (4k+3)(5k+8) .\end{array}
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