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: 15

Answer

$(2x-3)$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To factor the given expression, $ 4x^2-4x-3 ,$ 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. $\bf{\text{Solution Details:}}$ In the given expression the value of $ac$ is $ 4(-3)=-12 $ and the value of $b$ is $ -4 .$ The possible pairs of integers whose product is $ac$ are \begin{array}{l}\require{cancel} \{1,-12\}, \{2,-6\}, \{3,-4\}, \{-1,12\}, \{-2,6\}, \{-3,4\} .\end{array} Among these pairs, the one that gives a sum of $b$ is $\{ 2,-6 \}.$ Using these $2$ numbers to decompose the middle term of the given expression results to \begin{array}{l}\require{cancel} 4x^2+2x-6x-3 .\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} (4x^2+2x)-(6x+3) .\end{array} Factoring the $GCF$ in each group results to \begin{array}{l}\require{cancel} 2x(2x+1)-3(2x+1) .\end{array} Factoring the $GCF= (2x+1) $ of the entire expression above results to \begin{array}{l}\require{cancel} (2x+1)(2x-3) .\end{array} The missing factor of the given expression is $ (2x-3) .$
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