Intermediate Algebra (12th Edition)

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

Chapter 5 - Chapter 5 Test: 12

Answer

$(3k^2-7)(2k^2+5)$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To factor the given expression, $ 6k^4-k^2-35 ,$ 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. $\bf{\text{Solution Details:}}$ Using factoring of trinomials, the value of $ac$ in the trinomial expression above is $ 6(-35)=-210 $ and the value of $b$ is $ -1 .$ The $2$ numbers that have a product of $ac$ and a sum of $b$ are $\left\{ -14,15 \right\}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to \begin{array}{l}\require{cancel} 6k^4-14k^2+15k^2-35 .\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} (6k^4-14k^2)+(15k^2-35 ) .\end{array} Factoring the $GCF$ in each group results to \begin{array}{l}\require{cancel} 2k^2(3k^2-7)+5 (3k^2-7) .\end{array} Factoring the $GCF= (3k^2-7) $ of the entire expression above results to \begin{array}{l}\require{cancel} (3k^2-7)(2k^2+5) .\end{array}
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