Introductory Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-805-X
ISBN 13: 978-0-13417-805-9

Chapter 6 - Section 6.2 - Factoring Trinomials Whose Leading Coefficient is 1 - Exercise Set: 60

Answer

$x^2(x-10)(x-12)$

Work Step by Step

Factoring the $GCF= x^2 ,$ the given expression, $ x^4-22x^3+120x^2 ,$ is equivalent to \begin{array}{l}\require{cancel} x^2(x^2-22x+120) .\end{array} The trinomial expression above has $c= 120 $ and $b= -22 .$ The possible factors of $c$ are $ \{ 1,120 \} ,\{ 2,60 \} ,\{ 3,40 \} ,\{ 4,30 \} ,\{ 5,24 \} ,\{ 6,20 \} ,\{ 8,15 \} ,\{ 10,12 \} ,\{ -1,-120 \} ,\{ -2,-60 \} ,\{ -3,-40 \} ,\{ -4,-30 \} ,\{ -5,-24 \} ,\{ -6,-20 \} ,\{ -8,-15 \} ,\{ -10,-12 \} $. Among these factors, the pair whose sum is equal to $b$ is $\{ -10,-12 \}.$ Hence, the factored form of the given expression is \begin{array}{l}\require{cancel} x^2(x-10)(x-12) .\end{array}
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