Intermediate Algebra (12th Edition)

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

Chapter 5 - Section 5.4 - A General Approach to Factoring - 5.4 Exercises - Page 348: 48



Work Step by Step

$\bf{\text{Solution Outline:}}$ To factor the given expression, $ 4r^2-s^2-2st-t^2 ,$ group the last $3$ terms and factor the trinomial. Then use the factoring of the difference of $2$ squares. $\bf{\text{Solution Details:}}$ Grouping the last $3$ terms, the given expression is equivalent to \begin{array}{l}\require{cancel} 4r^2-(s^2+2st+t^2) .\end{array} In the trinomial expression above, $b= 2 ,\text{ and } c= 1 .$ Using the factoring of trinomials in the form $x^2+bx+c,$ the two numbers whose product is $c$ and whose sum is $b$ are $\left\{ 1,1 \right\}.$ Using these two numbers, the factored form of the expression above is \begin{array}{l}\require{cancel} 4r^2-[(s+1t)(s+1t)) \\\\= 4r^2-(s+t)^2 .\end{array} The expressions $ 4r^2 $ and $ (s+t)^2 $ are both perfect squares (the square root is exact) and are separated by a minus sign. Hence, $ 4r^2-(s+t)^2 ,$ is a difference of $2$ squares. Using the factoring of the difference of $2$ squares which is given by $a^2-b^2=(a+b)(a-b),$ the expression above is equivalent to \begin{array}{l}\require{cancel} (2r)^2-(s+t)^2 \\\\= [2r+(s+t)][2r-(s+t)] \\\\= (2r+s+t)(2r-s-t) .\end{array}
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