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: 45

Answer

$3(2x+1)(4x+5)$

Work Step by Step

$\bf{\text{Solution Outline:}}$ First, factor the GCF of the terms. Then, 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. $\bf{\text{Solution Details:}}$ Using the $GCF= 3 ,$ the given expression, $ 24x^2+42x+15 ,$ is equivalent to \begin{array}{l}\require{cancel} 3(8x^2+14x+5) .\end{array} In the expression above, the value of $ac$ is $ 8(5)=40 $ and the value of $b$ is $ 14 .$ The possible pairs of integers whose product is $ac$ are \begin{array}{l}\require{cancel} \{1,40\}, \{2,20\}, \{4,10\}, \{5,8\}, \{-1,-40\}, \{-2,-20\}, \{-4,-10\}, \{-5,-8\} .\end{array} Among these pairs, the one that gives a sum of $b$ is $\{ 4,10 \}.$ Using these $2$ numbers to decompose the middle term of the expression above results to \begin{array}{l}\require{cancel} 3(8x^2+4x+10x+5) .\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} 3[(8x^2+4x)+(10x+5)] .\end{array} Factoring the $GCF$ in each group results to \begin{array}{l}\require{cancel} 3[4x(2x+1)+5(2x+1)] .\end{array} Factoring the $GCF= (2x+1) $ of the entire expression above results to \begin{array}{l}\require{cancel} 3[(2x+1)(4x+5)] \\\\= 3(2x+1)(4x+5) .\end{array}
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