how to factorise this

• Aug 3rd 2008, 07:09 PM
Craka
how to factorise this
Trying to jog my memory here, how would I go about factorising this, as I'm trying to find its limit as x approaches 3 from the left.

$\frac{{x^2 - x + 12}}{{x + 3}}$
• Aug 3rd 2008, 07:20 PM
skeeter
fyi, the numerator will not factor, but why factor at all? there is no discontinuity at x = 3 ... just plug in 3 for x and calculate the limit straight up.
• Aug 3rd 2008, 07:49 PM
Craka
sorry should be as x approaches -3 (not 3). How would I go about finding the limit then in this case?
• Aug 3rd 2008, 08:02 PM
mr fantastic
Quote:

Originally Posted by Craka
sorry should be as x approaches -3 (not 3). How would I go about finding the limit then in this case?

It's already been pointed out that the numerator does not factorise. So you can't cancel a common factor of x + 3.

The limit approaches -oo as x --> -3 from the left.

If you don't like this answer, I suggest you go back to the original question and check whether you made a mistake in copying it. If you haven't made a mistake in the equation posted, then I'm afraid you're stuck with the answer I've given.
• Aug 3rd 2008, 08:27 PM
Craka
Sorry to clarify, the question ask to find the value for the for the following expression.

$
\mathop {\lim }\limits_{x \to ( - 3)} \frac{{x^2 - x + 12}}{{x + 3}}
$

• Aug 3rd 2008, 08:47 PM
mr fantastic
Quote:

Originally Posted by Craka
Sorry to clarify, the question ask to find the value for the for the following expression.

$
\mathop {\lim }\limits_{x \to ( - 3)} \frac{{x^2 - x + 12}}{{x + 3}}
$

Did you check the question like I asked?

To get that answer the numerator must be $x^2 - x {\color{red}-} 12$ NOT $x^2 - x {\color{red}+} 12$ as you've posted several times now.

$x^2 - x - 12 = (x - 4)(x + 3)$.
• Aug 3rd 2008, 09:05 PM
Craka
Yes I have checked the question several times. That is the question being asked. From what you are saying it appears there is a mistake in the question being asked on the paper.
• Aug 3rd 2008, 10:50 PM
mr fantastic
Quote:

Originally Posted by Craka
Yes I have checked the question several times. That is the question being asked. From what you are saying it appears there is a mistake in the question being asked on the paper.

Looks like it.
• Aug 5th 2008, 09:06 AM
tutor
Apply the L'HOSPITAL RULE
differentiate with respect rule
u will get
=lim x-->-3 (2x-1)
=2(-3)-1
=-6-1
=-7
• Aug 5th 2008, 02:49 PM
mr fantastic
Quote:

Originally Posted by Craka
Trying to jog my memory here, how would I go about factorising this, as I'm trying to find its limit as x approaches 3 from the left.

$\frac{{x^2 - x + 12}}{{x + 3}}$

Quote:

Originally Posted by Craka
sorry should be as x approaches -3 (not 3). How would I go about finding the limit then in this case?

Quote:

Originally Posted by tutor
Apply the L'HOSPITAL RULE
differentiate with respect rule
u will get
=lim x-->-3 (2x-1)
=2(-3)-1
=-6-1
=-7

NO!!

The limit does NOT have an indeterminant form 0/0. Using l'Hopital's Rule here is totally wrong.

This is the problem with l'Hopital's Rule - it's a lazy approach that is easily abused.

(And in a pre-algebra forum I doubt that l'Hopital's Rule has ever even been heard of).