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

##### 1 Answer

There are holes at

#### Explanation:

Factoring the denominator, we find:

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

So the denominator is zero when

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

Let

Then

So

#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

The singularities at