Intermediate Algebra (6th Edition)

Published by Pearson
ISBN 10: 0321785045
ISBN 13: 978-0-32178-504-6

Chapter 8 - Cumulative Review - Page 534: 22

Answer

$a^2+8a-2 /(a-4)(a+4)(a-2)$

Work Step by Step

$(a+1)/(a^2-6a+8) – 3/ (16-a^2)$ $(a+1)/(a-2)(a-4) -3/(-1)(-1)(16-a^2)$ $(a+1)/(a-2)(a-4) +3/ (a^2-16)$ $(a+1)/(a-2)(a-4) +3/ (a-4)(a+4)$ $(a+1)*(a+4)/(a-2)(a-4)*(a+4) +3/(a-4)(a+4)$ $(a+1)(a+4)/(a-2)(a-4)(a+4) +3*(a-2)/(a-4)(a+4)*(a-2)$ $(a+1)(a+4)/(a-2)(a-4)(a+4) +3(a-2)/(a-4)(a+4)(a-2)$ $(a+1)(a+4) +3(a-2) /(a-4)(a+4)(a-2)$ $a*a+a*1+a*4+1*4 + 3*a+3*-2 /(a-4)(a+4)(a-2)$ $a^2+a+4a+4+3a-6/(a-4)(a+4)(a-2)$ $a^2+5a+4+3a-6/(a-4)(a+4)(a-2)$ $a^2+8a-2 /(a-4)(a+4)(a-2)$
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