Intermediate Algebra (12th Edition)

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

Chapter 5 - Review Exercises: 34



Work Step by Step

$\bf{\text{Solution Outline:}}$ To factor the given expression, $ x^2+6x+9-25y^2 ,$ group the first $3$ terms since these form a perfect square trinomial. Factor this trinomial. The resulting expression becomes a difference of $2$ squares. Factor this using the factoring of the difference of $2$ squares. $\bf{\text{Solution Details:}}$ Grouping the first $3$ terms, the given expression is equivalent to \begin{array}{l}\require{cancel} (x^2+6x+9)-25y^2 .\end{array} To factor the trinomial expression above, note that the value of $ac$ is $ 1(9)=9 $ and the value of $b$ is $ 6 .$ The possible pairs of integers whose product is $ac$ are \begin{array}{l}\require{cancel} \{1,9\}, \{3,3\}, \\ \{-1,-9\}, \{-3,-3\} .\end{array} Among these pairs, the one that gives a sum of $b$ is $\{ 3,3 \}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to \begin{array}{l}\require{cancel} (x^2+3x+3x+9)-25y^2 .\end{array} Grouping the first and second terms and the third and fourth terms of the trinomial, the given expression is equivalent to \begin{array}{l}\require{cancel} [(x^2+3x)+(3x+9)]-25y^2 .\end{array} Factoring the $GCF$ in each group results to \begin{array}{l}\require{cancel} [x(x+3)+3(x+3)]-25y^2 .\end{array} Factoring the $GCF= (x+3) $ of the entire expression above results to \begin{array}{l}\require{cancel} [(x+3)(x+3)]-25y^2 \\\\= (x+3)^2-25y^2 .\end{array} The expressions $ (x+3)^2 $ and $ 25y^2 $ are both perfect squares (the square root is exact) and are separated by a minus sign. Hence, $ (x+3)^2-25y^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+3)^2-(5y)^2 \\\\= [(x+3)+5y][(x+3)-5y] \\\\= (x+3+5y)(x+3-5y) .\end{array}
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