Intermediate Algebra (6th Edition)

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

Chapter 8 - Review - Page 531: 34

Answer

$(-∞,-4)$ U $(-1, 1)$ U $(4,∞)$

Work Step by Step

$(x^2-16)(x^2-1) > 0$ $(x-4)(x+4)(x-1)(x+1) > 0$ $x-4=0$ $x-4+4=0+4$ $x=4$ $x-1=0$ $x-1+1=0+1$ $x=1$ $x+4=0$ $x+4-4=0-4$ $x=-4$ $x+1=0$ $x+1-1=0-1$ $x=-1$ Five regions to test: $(-∞,-4)$, $(-4, -1)$, $(-1, 1)$, $(1,4)$, $(4,∞)$ Let $x=-5$, $x=-2$, $x=0$, $x=2$, $x=5$ $x=-5$ $(x^2-16)(x^2-1) > 0$ $((-5)^2-16)((-5)^2-1) > 0$ $(25-16)(25-1) > 0$ $9*24 >0$ $216 > 0$ (true) $x=-2$ $(x^2-16)(x^2-1) > 0$ $((-2)^2-16)((-2)^2-1) > 0$ $(4-16)(4-1) >0$ $-12*3 > 0$ $-36 > 0$ (false) $x=0$ $(x^2-16)(x^2-1) > 0$ $(0^2-16)(0^2-1) > 0$ $(0-16)(0-1) > 0$ $-16*-1 > 0$ $16 > 0$ (true) $x=2$ $(x^2-16)(x^2-1) > 0$ $((2)^2-16)((2)^2-1) > 0$ $(4-16)(4-1) >0$ $-12*3 > 0$ $-36 > 0$ (false) $x=5$ $(x^2-16)(x^2-1) > 0$ $((5)^2-16)((5)^2-1) > 0$ $(25-16)(25-1) > 0$ $9*24 >0$ $216 > 0$ (true)
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