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

1 Answer
Jul 28, 2016

Answer:

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.