Question

How do you find the holes for `(2x^3-5x^2-9x+18) / (x^2 - x - 6)`?

Answer 1

There are holes at `x=-2` and `x=3`

Explanation:

Factoring the denominator, we find:

`x^2-x-6 = (x+2)(x-3)`

So the denominator is zero when `x=-2` or `x=3`

At any hole, both the numerator and denominator are zero. Note this condition is required but not sufficient.

Let `f(x) = 2x^3-5x^2-9x+18`

Then `f(-2) = -16-20+18+18 = 0`.

So `x=-2` is a zero of the numerator and `(x+2)` a factor:

`2x^3-5x^2-9x+18`

`=(x+2)(2x^2-9x+9)`

`=(x+2)(x-3)(2x-3)`

So we find:

`(2x^3-5x^2-9x+18)/(x^2-x-6) = (color(red)(cancel(color(black)((x+2))))color(red)(cancel(color(black)((x-3))))(2x-3))/(color(red)(cancel(color(black)((x+2))))color(red)(cancel(color(black)((x-3))))) = 2x-3`

excluding `x=-2` and `x=3`

The singularities at `x=-2` and `x=3` are holes, since the function is defined for any other Real value of `x` and is continuous at these points.