How do you find that: $x^3+4x^2+x-6=(x+2)(x^2+2x-3)$? How do you find that: $x^3+4x^2+x-6=(x+2)(x^2+2x-3)$?
I get that you can take out the $x$ like so: $x^3+4x^2+x-6=x(x^2+4x+1)-6$ but how do you get the $2$ from here?

Observe that $x^3 + 4x^2 + x - 6 = x^3 + 4x^2 + 4x - 3x - 6$.  Therefore,
\begin{align*}
x^3 + 4x^2 + x - 6 & = x^3 + 4x^2 + 4x - 3x - 6\\
                   & = x(x^2 + 4x + 4) - 3(x + 2)\\
                   & = x(x + 2)^2 - 3(x + 2)\\
                   & = (x + 2)[x(x + 2) - 3)]\\
                   & = (x + 2)(x^2 + 2x - 3)
\end{align*}
That said, I recommend using the Rational Roots Theorem.  In this case, a rational root must be a factor of $-6/1 = -6$, so the possible rational roots are $\pm 1, \pm 2, \pm 3, \pm 6$.  If you test the roots, you will find that $1, -2, -3$ are roots.  Once you find that $k$ is a root, you can divide by $x - k$ to find the other factor.  Thus, if you discover that $-2$ is a root, you can divide $x^3 + 4x^2 + x - 6$ by $x - (-2) = x + 2$ to discover that $x^2 + 2x - 3$ is the other factor.  As you can check, the roots of $x^2 + 2x - 3$ are $1$ and $-3$.
