Given $x^2 + 4x + 6$ as factor of $x^4 + ax^2 + b$, then $a + b$ is I got this task two days ago, quite difficult for me, since I have not done applications of Vieta's formulas and Bezout's Theorem for a while. If can someone solve this and add exactly how I am supposed to use these two theorem's on this task, I would be thankful.

Given $x^2 + 4x + 6$ as factor of $x^4 + ax^2 + b$, then $a + b$ is equal to?


Let $r_1,r_2$ be the roots of $x^2+4x+6$. By Vieta's $r_1+r_2 = -4$ and $r_1r_2 = 6$.
Since $x^2+4x+6$ is a factor of $x^4+ax^2+b$, we have that $r_1,r_2$ are also roots of $x^4+ax^2+b$. 
Then, since $x^4+ax^2+b$ is an even polynomial, $-r_1,-r_2$ are also roots of $x^4+ax^2+b$. 
It is easy to see that $r_1,r_2,-r_1,-r_2$ are distinct, so we've found all four roots of $x^4+ax^2+b$.
Therefore, $x^4+ax^2+b$ $= (x-r_1)(x-r_2)(x+r_1)(x+r_2)$ $= \left[(x^2-(r_1+r_2)x+r_1r_2\right]\left[(x^2+(r_1+r_2)x+r_1r_2\right]$ $= (x^2+4x+6)(x^2-4x+6)$ $= x^4 - 4x^2 + 36$.
Hence, $a = -4$, $b = 36$, and thus, $a+b = 32$. 
Note: I'm not sure how Bezout's Theorem is needed here.
EDIT: Little Bezout's Theorem is also known as the polynomial remainder theorem, which was used above.
