Algebra: A Combined Approach (4th Edition)

Published by Pearson
ISBN 10: 0321726391
ISBN 13: 978-0-32172-639-1

Chapter 11 - Review - Page 829: 33

Answer

$(-infinity, -4) U (-1, 1) U (4, infinity)$

Work Step by Step

$(x^2-16)(x^2-1)>0$ $x^2-16=0$ $x^2=16$ $\sqrt {x^2} =\sqrt {16}$ $x =±4$ $x=4, -4$ $x^2-1=0$ $x^2=1$ $\sqrt {x^2} =\sqrt {1}$ $x =±1$ $x=1, -1$ 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) $x=-2$ $(x^2-16)(x^2-1)>0$ $((-2)^2-16)((-2)^2-1)>0$ $(4-16)(4-1)>0$ $-13*3 > 0$ $-39 > 0$ (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) $x=2$ $(x^2-16)(x^2-1)>0$ $((2)^2-16)((2)^2-1)>0$ $(4-16)(4-1)>0$ $-13*3 > 0$ $-39 > 0$ (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)
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.