Trouble Factoring X^4-15X^3+32X+372X-1440)/X^2-X-30

[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?

 

[marcusl]

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

 

[mathwonk]

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

 

[murshid_islam]

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

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.

 

[Gib Z]

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

 

[MattsAli1108]

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

 

[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