# How do you solve (x^2-9)*(x^2-16)<=0?

Jun 22, 2015

$f \left(x\right) = \left({x}^{2} - 9\right) \left({x}^{2} - 16\right)$ is a continuous function with zeros at $x = \pm 4$ and $x = \pm 3$ of multiplicity $1$.

Hence solution: $x \in \left[- 4 , - 3\right] \cup \left[3 , 4\right]$

#### Explanation:

Let $f \left(x\right) = \left({x}^{2} - 9\right) \left({x}^{2} - 16\right)$

$= \left(x - 3\right) \left(x + 3\right) \left(x - 4\right) \left(x + 4\right)$

$f \left(x\right)$ is a continuous function with zeros at $x = \pm 4$ and $x = \pm 3$.

Each of the roots of $f \left(x\right) = 0$ is of multiplicity $1$, hence $f \left(x\right)$ changes sign at those points.

For large positive and negative values of $x$, $f \left(x\right)$ is large and positive.

Hence $f \left(x\right) \le 0$ when $x \in \left[- 4 , - 3\right] \cup \left[3 , 4\right]$

Here's a graph of $f \frac{x}{20}$...

graph{(x^2-9)(x^2-16)/20 [-9.68, 10.32, -1.52, 8.48]}