Intermediate Algebra (12th Edition)

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

Chapter 5 - Chapters R-5 - Cumulative Review Exercises: 28



Work Step by Step

$\bf{\text{Solution Outline:}}$ To factor the given expression, $ 4x^2-4x+1-y^2 ,$ group the first $3$ terms since these form a perfect square trinomial. Then factor the trinomial. This makes the entire expression a difference of $2$ squares. Use then the factoring of the difference of $2$ squares. $\bf{\text{Solution Details:}}$ Grouping the first $3$ terms, the expression above is equivalent to \begin{array}{l}\require{cancel} (4x^2-4x+1)-y^2 .\end{array} Using factoring of trinomials, the value of $ac$ in the trinomial expression above is $ 4(1)=4 $ and the value of $b$ is $ -4 .$ The $2$ numbers that have a product of $ac$ and a sum of $b$ are $\left\{ -2,-2 \right\}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to \begin{array}{l}\require{cancel} (4x^2-2x -2x +1 )-y^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} [(4x^2-2x )-(2x -1 )]-y^2 .\end{array} Factoring the $GCF$ in each group results to \begin{array}{l}\require{cancel} [2x(2x-1 )-(2x -1 )]-y^2 .\end{array} Factoring the $GCF= (2x-1 ) $ of the entire expression above results to \begin{array}{l}\require{cancel} [(2x-1 )(2x-1)]-y^2 \\\\= (2x-1 )^2-y^2 .\end{array} The expressions $ (2x-1 )^2 $ and $ y^2 $ are both perfect squares (the square root is exact) and are separated by a minus sign. Hence, $ (2x-1 )^2-y^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} (2x-1 )^2-(y)^2 \\\\= (2x-1+y)(2x-1-y) .\end{array}
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