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?
 A: Use the remainder theorem for polynomials. 
$$
f(a) = 0 \implies f(x) = (x-a)p(x)
$$
Where $p(x)$ is reduced a degree. To find the $2$ you would trail solutions to find a root using a change of sign. 
$\textbf{update}$
Since we have found one root we can assume that the we have an equation of the form 
$$
(x+2)\left(ax^2+bx+c\right)
$$
Thus we have to compare the coefficents leading to
$$
a = 1\\
2a+ b = 4\\
2b + c = 1\\
2c = -6
$$
A: *

*Struggle to find a real root. By trying small positive and negative integers, you find that $P(-2)=0$. So $x+2$ is a factor.

*Apply the change of variable $y=x+2$:
$$P(x)=P(y-2)=(y-2)^3+4(y-2)^2+(y-2)+6=y^3-2y^2-3y.$$
Now you easily factor as
$$P(y-2)=y(y^2-2y-3),$$
and revert to the original
$$P(x)=(x+2)\left((x+2)^2-2(x+2)-3\right)=(x+2)(x^2+2x-3).$$

Similarly, $x=1$ is a root of $x^2+2x-3$; set $z=x-1$, and
$$x^2+2x-3=(z+1)^2+2(z+1)-3=z^2+4z=z(z+4).$$
Hence, $$x^2+2x-3=(x-1)((x-1)+4)=(x-1)(x+3).$$
Tedious, but "without" long division.
A: $x=-2$ is a solution of the equation $$x^3+4x^2+x-6$$ thus you can divide $$x^3+4x^2+x-6$$ by $$x+2$$
A: 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 $-3$.
A: Knowing that $(x+2)$ is a factor, you know that the solution will be of the form
$$x^3+4x^2+x-6=(x+2)(ax^2+bx+c).$$
Execute the product 
$$(x+2)(ax^2+bx+c)=ax^3+(2a+b)x^2+(2b+c)x+(2c).$$
Identify the coefficients
$$a=1,2a+b=4,2b+c=1,2c=-6.$$
Solve and get
$$a=1,b=2,c=-3,$$
so that
$$x^3+4x^2+x-6=(x+2)(x^2+2x-3).$$
A: I think it may be done the folowing way:
$x^3+4x^2+x−6=x^3+2x^2+2x^2+4x-3x-6=x^2(x+2) + 2x(x+2)-3(x+2)=(x+2)(x^2+2x-3)$
