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: 33

Answer

$(5x-6)(2x+3)$

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, $ 10x^2+3x-18 ,$ the value of $ac$ is $ 10(-18)=-180 $ and the value of $b$ is $ 3 .$ The possible pairs of integers whose product is $ac$ are \begin{array}{l}\require{cancel} \{1,-180\}, \{2,-90\}, \{3,-60\}, \{4,-45\}, \{5,-36\}, \{6,-30\}, \{9,-20\}, \{10,-18\}, \{12,-15\}, \{-1,180\}, \{-2,90\}, \{-3,60\}, \{-4,45\}, \{-5,36\}, \{-6,30\}, \{-9,20\}, \{-10,18\}, \{-12,15\} .\end{array} Among these pairs, the one that gives a sum of $b$ is $\{ -12,15 \}.$ Using these $2$ numbers to decompose the middle term of the given expression results to \begin{array}{l}\require{cancel} 10x^2-12x+15x-18 .\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} (10x^2-12x)+(15x-18) .\end{array} Factoring the $GCF$ in each group results to \begin{array}{l}\require{cancel} 2x(5x-6)+3(5x-6) .\end{array} Factoring the $GCF= (5x-6) $ of the entire expression above results to \begin{array}{l}\require{cancel} (5x-6)(2x+3) .\end{array}
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