Intermediate Algebra (12th Edition)

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

Chapter 5 - Review Exercises: 20

Answer

$(2k^2+1)(k^2-3)$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To factor the given expression, $ 2k^4-5k^2-3 ,$ 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. $\bf{\text{Solution Details:}}$ To factor the trinomial expression above, note that the value of $ac$ is $ 2(-3)=-6 $ and the value of $b$ is $ -5 .$ The possible pairs of integers whose product is $ac$ are \begin{array}{l}\require{cancel} \{1,-6\}, \{2,-3\}, \\ \{-1,6\}, \{-2,3\} .\end{array} Among these pairs, the one that gives a sum of $b$ is $\{ 1,-6 \}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to \begin{array}{l}\require{cancel} 2k^4+k^2-6k^2-3 .\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} (2k^4+k^2)-(6k^2+3) .\end{array} Factoring the $GCF$ in each group results to \begin{array}{l}\require{cancel} k^2(2k^2+1)-3(2k^2+1) .\end{array} Factoring the $GCF= (y^2-2) $ of the entire expression above results to \begin{array}{l}\require{cancel} (2k^2+1)(k^2-3) .\end{array}
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