I'm having some troubles reducing this.
I need to find the limit of $\displaystyle x^3 + 2x^2 - 5x - 6 / (x+ 3) $ where x approches -3?
ive forgotten how to factor $\displaystyle x^3 + 2x^2 - 5x - 6$
help would be much appreciated
Because the limit approaches $\displaystyle \frac{0}{0}$ when you substitute -4 and $\displaystyle \infty$ when you substitute $\displaystyle \infty$ you can apply Hospital's rule:
$\displaystyle \lim_{x\to a} f(x)=\frac{g(x)}{h(x)} $
if $\displaystyle \lim_{x\to a} g(x) = \lim_{x\to a} h(x) = 0 $
then $\displaystyle \lim_{x\to a} f(x)=\frac{g(x)}{h(x)} = \lim_{x\to a} f(x)=\frac{g'(x)}{h'(x)} $
So:
$\displaystyle \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
$
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 $\displaystyle x^3$ $\displaystyle x^2$ times. Multiplying $\displaystyle x^2$ by x+ 3 we get $\displaystyle \begin{array}{cccc}x^3& +2x^2 & -5x& -6 \\ x^2 & +3x^2 & 0 & 0\\0 & -x^2& -5x& -6\end{array}$
Now, x goes into $\displaystyle -x^2$ -x times. Multiplying -x by x+ 3 we get $\displaystyle \begin{array}{ccc}-x^2 & -5x& -6 \\-x^2 & -3x& -6 \\ 0 & -2x & -6\end{array}$.
Finally, x goes into -2x -2 times and, multiplying -2 by x+ 3 we get $\displaystyle \begin{array}{cc}-2x & -6 \\ -2x & -6 \\ 0 & 0\end{array}$
$\displaystyle x^3+ 2x^2- 5x- 6= (x+ 3)(x^2- x- 2)$.