It's x power 3 plus 4x power 2 plus x minus %q

Possible rational roots are $\pm1, \pm2, \pm3, \pm6$.

Test $x=1$:

$$\begin{align*} 1^3 +{}& 4(1)^2 + 1 - 6 \\ ={}& 1 + 4 + 1 - 6 \\ ={}& 0 \end{align*}$$

$x=1$ is a root, so $(x-1)$ is a factor.

$$\begin{array}{r|r} 1 & 1 & 4 & 1 & -6 \\ & & 1 & 5 & 6 \\ \hline & 1 & 5 & 6 & 0 \\ \end{array}$$

Quotient is $x^2 + 5x + 6$.

$$x^3 + 4x^2 + x - 6 = (x-1)(x+2)(x+3)$$

$\boxed{(x-1)(x+2)(x+3)}$