<s>[INST] 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?

 [/INST] If $x^2 + 4x + 6$ is factor  of $x^4 + ax^2 + b$, then it must satify
$$x^4 + ax^2 + b=(x^2 + 4x + 6)Q(x)$$  where $Q(x)$ is a second degree polynominal. You do not need to find $Q(x)$
$x^2 + 4x + 6=0$
$x^2=-(4x + 6)$
$$x^4 + ax^2 + b=(x^2 + 4x + 6)Q(x)$$  
$$(-(4x + 6))^2 - a(4x + 6)+ b=0$$
$$16x^2+48x+36 - 4ax -6a+ b=0$$
$$-16(4x + 6))+48x+36 - 4ax -6a+ b=0$$
$$-64x -96+48x+36 - 4ax -6a+ b=0$$
$$-16x -60 - 4ax -6a+ b=0$$
$$-16-4a=0 $$
$$a=-4$$
$$-60 -6a+ b=0$$
$$-60 +24+ b=0$$
$$b=36$$
then $a+b=32$
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