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.

You can solve all the questions in one formula: since you correctly identified $x^2-3$ as the quadratic factor, try polynomial long division by $x^2-3$. This will
a. prove that $\pm \sqrt{3}$ are roots of the polynomials (since the remainder is zero),
b. give the other quadratic factor (as the quotient).
