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?


There is some quadratic $x^2 + cx + d$ such that:
$$(x^2 + 4x + 6)(x^2 + cx + d) = x^4 + ax^2 + b$$
Multiply it out:
$$x^4 + (c + 4)x^3 + (d + 4c + 6)x^2 + (6c + 4d)x + 6d = x^4 + ax^2 + b$$
Equate coefficients:
$$\begin{cases}
c + 4 = 0 \\
d + 4c + 6 = a \\
4d + 6c = 0 \\
6d = b
\end{cases}$$
Use the first equation to find $c$.  Use the third to find $d$.  Use those to find $a$ and $b$.
