College Algebra (6th Edition)

Published by Pearson
ISBN 10: 0-32178-228-3
ISBN 13: 978-0-32178-228-1

Chapter P - Prerequisites: Fundamental Concepts of Algebra - Exercise Set P.5 - Page 75: 92



Work Step by Step

$2x^{3}-98a^{2}x+28x^{2}+98x$ Rearrange the terms to match the proper order. $=2x^{3}+28x^{2}+98x-98a^{2}x$ Factor out the common factor $2x$ from the expression. $=2x(x^{2}+14x+49-49a^{2})$ Group the first three terms. This is the perfect square trinomial. $=2x((x^{2}+14x+49)-49a^{2})$ The formula for it can be used here is $ A^{2}+ 2.A.B+B^{2} =(A+B)^{2}$ Using the formula, factor the perfect square trinomial. $[x^{2}+2.x.7+7^{2}= (x+7) ^{2}$ $x^{2}+14x+49= (x+7) ^{2}]$ $=2x((x+7) ^{2}-49a^{2})$ $=2x((x+7) ^{2}-(7a)^{2})$ Factor the difference of squares. The factors are the sum and difference of the expressions being squared. Using the formula $[(a^{2}-b^{2}) = (a+b)(a-b)]$ $=(2x)(x+7+7a)(x+7-7a)$
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