Intermediate Algebra (12th Edition)

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

Chapter 5 - Chapter 5 Test: 8

Answer

$(x+1+2z)(x+1-2z)$

Work Step by Step

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