Evaluating limits with constant?

[shocklightnin]

Homework Statement

Is there such a number b such that lim x->-2 (3x^2+bx+b+3)/(x^2+x-2) exists? If so, find b and the limit.

Homework Equations

lim x->-2 (3x^2+bx+b+3)/(x^2+x-2)

The Attempt at a Solution

for the denominator we have zeroes at x = 1 and -2. so we need to get rid of the -2 part right? set up (3x^2+bx+b+3)=0 and solve for b to get b = -3-3x^2/(x+1) which has b = 15 (which is what the answer book has). but if b = 15, the numerator just becomes 3x^2+15x+18 when factored go to (3x+18)(x+1) which doesn't cancel out with anything on the bottom. they also say that the limit is -1 but how??

EDIT::
neeeeeevermind. got the factoring mixed up. limit is -1.

 

[Dick]

You factored 3x^2+15x+18 wrong. Try that again.