# How do you solve 5/(x^3-x^2-4X+4) < 0?

Oct 9, 2015

This reduces to finding out when ${x}^{3} - {x}^{2} - 4 x + 4 < 0$, which happens when $x \in \left(- \infty , - 2\right) \cup \left(1 , 2\right)$

#### Explanation:

$\frac{5}{{x}^{3} - {x}^{2} - 4 x + 4} < 0$ when ${x}^{3} - {x}^{2} - 4 x + 4 < 0$,

so let's find where that happens...

Let $f \left(x\right) = {x}^{3} - {x}^{2} - 4 x + 4$

We can factor $f \left(x\right)$ by grouping...

$f \left(x\right) = {x}^{3} - {x}^{2} - 4 x + 4$

$= \left({x}^{3} - {x}^{2}\right) - \left(4 x - 4\right)$

$= {x}^{2} \left(x - 1\right) - 4 \left(x - 1\right)$

$= \left({x}^{2} - 4\right) \left(x - 1\right)$

$= \left(x - 2\right) \left(x + 2\right) \left(x - 1\right)$

The graph of $f \left(x\right)$ will cross the $x$ axis when $x = - 2$, $x = 1$ and $x = 2$. I say cross rather than intersect, since none of these roots of $f \left(x\right) = 0$ are repeated.

Since the coefficient of ${x}^{3}$ is positive, this means that $f \left(x\right) < 0$ when $x \in \left(- \infty , - 2\right) \cup \left(1 , 2\right)$

${x}^{3} - {x}^{2} - 4 x + 4$ ...

graph{x^3-x^2-4x+4 [-9.92, 10.08, -3.12, 6.88]}

$\frac{5}{{x}^{3} - {x}^{2} - 4 x + 4}$ ...

graph{5/(x^3-x^2-4x+4) [-18.6, 17.45, -10.69, 7.33]}