# Trouble with Limits

1. Jan 16, 2007

### MattsAli1108

(X^4-15X^3+32X+372X-1440)/X^2-X-30 as X approaches 6

I know that somehow I am supposed to be able to factor this, but I'm having trouble doing so, and am stuck with a TI-82 that isn't very much help, either. Could someone please show me how to do this by hand?

2. Jan 16, 2007

### marcusl

EDIT: Sorry, I was blind to the answer. mathwonk has it right in the next post...

Last edited: Jan 16, 2007
3. Jan 16, 2007

### mathwonk

the root factor theorem says that if x=6 makes a polynomial zero, then x-6 is a factor, and vice versa.

4. Jan 17, 2007

### murshid_islam

did you write the question correctly?
is the third term in the numerator $32x^2$?
then it becomes
$$\frac{x^4-15x^3+32x^2+372x-1440}{x^2-x-30}=\frac{(x+5)(x-8)(x-6)^2}{(x+5)(x-6)}=(x-8)(x-6)$$

and this approaches 0 as x approaches 6.

5. Jan 17, 2007

### Gib Z

I would have a feeling you assumption is correct, murshid. Nice work

6. Jan 18, 2007

### MattsAli1108

Thank you so much for your help!! All of you! It is very much apprecitated!

7. Jan 18, 2007

### mr_homm

Since the trouble was originally that you got 0/0 by plugging 6 in directly, you could also have used L'Hopital's rule. Of course, it is conceptually more elementary to factor the polynomial, but it may actually be less work to do 1 derivative than to do the polynomial long division.

Just offering an alternative.

--Stuart Anderson