Answer
$(-2, -1)$ U $(2,4)$
Work Step by Step
$(x-2)(x+2)/(x+1)(x-4)) \le 0$
$x-2=0$
$x-2+2=0+6$
$x=2$
$x+2=0$
$x+2-2=0-2$
$x=-2$
$x-4=0$
$x-4+4=0+4$
$x=4$
$x+1=0$
$x+1-1=0-1$
$x=-1$
The denominator is zero when $x=4$ or $x=-1$, and the numerator is zero when $x=-2$ or $x=2$.
We have five sections: $(−∞,-2)$, $(-2, -1)$, $(-1,2)$, $(2,4)$, and $(4,∞)$. We need to test one value for x in each section to determine if the section would be a solution set. Since we have the $\le$ sign, we include the end points and use brackets instead of parentheses.
Let $x=-3$, $x=-1.5$, $x=0$, $x=3$, and $x=10$
$x=-3$
$(x-2)(x+2)/(x+1)(x-4)) \le 0$
$(-3-2)(-3+2)/(-3+1)(-3-4)) \le 0$
$-5*-1/-2*-7 \ge 0$
$5/14 \le 0$ (false)
$x=-1.5$
$(x-2)(x+2)/(x+1)(x-4)) \le 0$
$(-1.5-2)(-1.5+2)/(-1.5+1)(-1.5-4)) \le 0$
$-3.5*.5/-.5*-5.5 \le0$
$-3.5/5.5 \le 0$
$-7/11 \le 0$ (true)
$x=0$
$(x-2)(x+2)/(x+1)(x-4)) \le 0$
$(0-2)(0+2)/(0+1)(0-4)) \le 0$
$-2*2/1*-4 \le0$
$-4/-4 \le 0$
$1 \le 0$ (false)
$x=3$
$(x-2)(x+2)/(x+1)(x-4)) \le 0$
$(3-2)(3+2)/(3+1)(3-4)) \le 0$
$1*5/4*-1 \le0$
$5/-4 \le0$
$-5/4 \le 0$ (true)
$x=10$
$(x-2)(x+2)/(x+1)(x-4)) \le 0$
$(10-2)(10+2)/(10+1)(10-4)) \le 0$
$8*12/11*6 \le 0$
$96/66 \le 0$
$48/33 \le 0$
$16/11 \le 0$ (false)
