Question

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

Answer 1

This reduces to finding out when `x^3-x^2-4x+4 < 0`, which happens when `x in (-oo, -2) uu (1, 2)`

Explanation:

`5/(x^3-x^2-4x+4) < 0` when `x^3-x^2-4x+4 < 0`,

so let's find where that happens...

Let `f(x) = x^3-x^2-4x+4`

We can factor `f(x)` by grouping...

`f(x) = x^3-x^2-4x+4`

`= (x^3-x^2)-(4x-4)`

`= x^2(x-1) - 4(x-1)`

`= (x^2-4)(x-1)`

`= (x-2)(x+2)(x-1)`

The graph of `f(x)` 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(x) = 0` are repeated.

Since the coefficient of `x^3` is positive, this means that `f(x) < 0` when `x in (-oo, -2) uu (1, 2)`

`x^3-x^2-4x+4` ...

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

`5/(x^3-x^2-4x+4)` ...

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