Finding quadratic factors Show that $(x-√3)$ and $(x+√3)$ are factors of $x^4+x^3-x^2-3x-6$. Hence write down one quadratic factor of $x^4+x^3-x^2-3x-6$, and find a second quadratic factor of this polynomial.  
My attempt:
$f(x)=x^4+x^3-x^2-3x-6$
$f(3^\frac12)=(3^\frac12)^4+(3^\frac12)^3-(3^\frac12)^2-3(3^\frac12)-6=0$
$f(-3^\frac12)=(-3^\frac12)^4+(-3^\frac12)^3-(-3^\frac12)^2-3(-3^\frac12)-6=0$  
I know the first quadratic factor is $x^2-3$ but I don't know how to arrive to the second quadratic factor.

We know that there exist $a,b$ with
$$
x^4+x^3-x^2-3x-6=(x^2-3)(x^2+ax+b),
$$
and comparison of the coefficients on both sides immediately gives $a=1$ and $b=2$.
