# Determining k so that there is a hole.

• Apr 28th 2013, 06:08 PM
FatimaA
Determining k so that there is a hole.
I've been staring at this problem for a while. I would appreciate a nudge in the right direction.

Determine k so that f(x)=x^3+kx-4/x-1 has a hole at x=1.

Here's what I've been able to figure out. In order for there to be a hole at x=1 there has to be a (x-1) in the numerator, right? But I'm not sure how to go about determining k to get (x-1) on top.
• Apr 28th 2013, 06:23 PM
Prove It
Re: Determining k so that there is a hole.
Some brackets would be nice. Assuming you meant f(x) = (x^3 + kx - 4)/(x - 1), you are correct that you need a factor of (x - 1) in the numerator.

Now remember that by the remainder and factor theorems, that for (x - a) to be a factor of a polynomial P(x), then P(a) = 0. Since you need x - 1 to be a factor, substitute x = 1 into the numerator and equate it to 0. You should be able to find k then.
• Apr 28th 2013, 06:41 PM
FatimaA
Re: Determining k so that there is a hole.
0 = (1)^3 + k(1) - 4

0 = 1 + k - 4

0 = -3 + k

3 = k

So f(x) = (x^3+3x-4)/(x-1)
• Apr 28th 2013, 07:15 PM
Prove It
Re: Determining k so that there is a hole.
Correct :)