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 74: 86



Work Step by Step

$x^{2}-10x+25-36y^{2}$ To factor the polynomial group the first three terms. $=(x^{2}-10x+25)-36y^{2}$ $x^{2}-10x+25$ is the perfect square trinomial, so 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 $x^{2}-2.x.5+5^{2}= (x-5) ^{2}$ $x^{2}-10x+25= (x-5) ^{2}$ $x^{2}-10x+25-36y^{2}$ $ =(x-5)^{2}-36y^{2}$ $ =(x-5)^{2}-(6y)^{2}$ Factor the difference of squares. The factors are the sum and difference of the expressions being squared. Using the formula $[(a^{2}-b^{2}) = (a+b)(a-b)]$ $=(x-5+6y)(x-5-6y)$
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