1. ## Limits

I'm having some troubles reducing this.

I need to find the limit of $x^3 + 2x^2 - 5x - 6 / (x+ 3)$ where x approches -3?

ive forgotten how to factor $x^3 + 2x^2 - 5x - 6$

help would be much appreciated

2. Because the limit approaches $\frac{0}{0}$ when you substitute -4 and $\infty$ when you substitute $\infty$ you can apply Hospital's rule:

$\lim_{x\to a} f(x)=\frac{g(x)}{h(x)}$

if $\lim_{x\to a} g(x) = \lim_{x\to a} h(x) = 0$

then $\lim_{x\to a} f(x)=\frac{g(x)}{h(x)} = \lim_{x\to a} f(x)=\frac{g'(x)}{h'(x)}$

So:
$\lim_{x\rightarrow-3}\frac{x^3 + 2x^2 - 5x - 6}{x+ 3} =
\lim_{x\rightarrow-3}\frac{\frac{d}{dx}(x^3 + 2x^2 - 5x - 6)}{\frac{d}{dx}(x+ 3)} =
\lim_{x\rightarrow-3}\frac{3x^2 + 4x - 5}{1} = 10
$

3. $x^3+2x^2-5x-6 = (x+3)(x-2)(x+1)$

$lim_{x\to-3}\frac{x^3+2x^2-5x-6}{x+3} = \lim_{x\to-3} (x-2)(x+1)$

$= \lim_{x\to-3} x^2 -x -2 = 10$

4. Originally Posted by foreverbrokenpromises
I'm having some troubles reducing this.

I need to find the limit of $x^3 + 2x^2 - 5x - 6 / (x+ 3)$ where x approches -3?

ive forgotten how to factor $x^3 + 2x^2 - 5x - 6$

help would be much appreciated
The only reason you want to factor that is because both it and the denominator become 0 when x= -3. That means that x-(-3)= x+ 3 is a factor. Now just do a "long division" to find the other factor: x goes into $x^3$ $x^2$ times. Multiplying $x^2$ by x+ 3 we get $\begin{array}{cccc}x^3& +2x^2 & -5x& -6 \\ x^2 & +3x^2 & 0 & 0\\0 & -x^2& -5x& -6\end{array}$

Now, x goes into $-x^2$ -x times. Multiplying -x by x+ 3 we get $\begin{array}{ccc}-x^2 & -5x& -6 \\-x^2 & -3x& -6 \\ 0 & -2x & -6\end{array}$.

Finally, x goes into -2x -2 times and, multiplying -2 by x+ 3 we get $\begin{array}{cc}-2x & -6 \\ -2x & -6 \\ 0 & 0\end{array}$

$x^3+ 2x^2- 5x- 6= (x+ 3)(x^2- x- 2)$.