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 - Page 337: 16



Work Step by Step

$\bf{\text{Solution Outline:}}$ To factor the given expression, $ 6z^2-11z+4 ,$ 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 $ 6(4)=24 $ and the value of $b$ is $ -11 .$ The possible pairs of integers whose product is $ac$ are \begin{array}{l}\require{cancel} \{1,24\}, \{2,12\}, \{3,8\}, \{4,6\}, \{-1,-24\}, \{-2,-12\}, \{-3,-8\}, \{-4,-6\} .\end{array} Among these pairs, the one that gives a sum of $b$ is $\{ -3,-8 \}.$ Using these $2$ numbers to decompose the middle term of the given expression results to \begin{array}{l}\require{cancel} 6z^2-3z-8z+4 .\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} (6z^2-3z)-(8z-4) .\end{array} Factoring the $GCF$ in each group results to \begin{array}{l}\require{cancel} 3z(2z-1)-4(2z-1) .\end{array} Factoring the $GCF= (2z-1) $ of the entire expression above results to \begin{array}{l}\require{cancel} (2z-1)(3z-4) .\end{array} The missing factor of the given expression is $ (2z-1) .$
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