Intermediate Algebra (12th Edition)

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

Chapter 5 - Section 5.3 - Special Factoring - 5.3 Exercises: 23

Answer

$(2z+w)^2$

Work Step by Step

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