Answer
(-infinity, -5] U $[-3$, infinity)
Work Step by Step
$x^2+8x+15 \geq0$
$(x+3)(x+5)\geq0$
$x+3=0$
$x=-3$
$x+5=0$
$x=-5$
(-infinity, -5]
$[-5, -3]$
$[-3$, infinity)
Let $x=-10$, $x=-4$, and $x=0$
$x=-10$
$(x+3)(x+5)\geq0$
$(-10+3)(-10+5)\geq0$
$(-7)(-5)\geq0$
$35 \geq 0$ (true)
$x=-4$
$(x+3)(x+5)\geq0$
$(-4+3)(-4+5)\geq0$
$-1*1\geq 0$
$-1\geq 0$ (false)
$x=0$
$(x+3)(x+5)\geq0$
$(0+3)(0+5)\geq0$
$3*5\geq 0$
$15\geq 0$ (true)
