Intermediate Algebra (12th Edition)

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

Chapter 5 - Review Exercises: 19

Answer

$(y^2-2)(y^2+4)$

Work Step by Step

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

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.