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

1 Answer
Oct 9, 2015

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]}