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)