# How do you solve (x+8)^2(x+5)(x+7)^2>=0?

Jul 2, 2016

$x \in \left\{- 8 , - 7\right\} \cup \left[- 5 , \infty\right)$

#### Explanation:

Both ${\left(x + 8\right)}^{2}$ and ${\left(x + 7\right)}^{2}$ are squares, so always non-negative for Real values of $x$.

As a result, the left hand side is negative when $x + 5 < 0$ unless $x + 7 = 0$ or $x + 8 = 0$, when it is zero.

Hence the inequality is satisfied when: $x \in \left\{- 8 , - 7\right\} \cup \left[- 5 , \infty\right)$

graph{(x+8)^2(x+5)(x+7)^2 [-13.96, 6.04, -7.68, 2.32]}