<s>[INST] 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.
 [/INST] Lets try it like this:


*

*Suppose $(x+ \sqrt3)(x-\sqrt3)$ are both factors.

*Expand the expression above to get $x^2 - 3$.

*Perform polynomial long division with $x^2 -3$.

*You will get $x^2 + x + 2$ as the second quadratic factor.

*Re-expanding the quadratic factors $(x^2 -3)(x^2 +x+2)$ gives the     original quartic polynomial. I uploaded a picture of my work with this reply.



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