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?

 A: 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 $b$.
A: 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.
A: \begin{align*}
x^4+ax^2+b & = x^2\color{blue}{(x^2+4x+6)}-4x\color{blue}{(x^2+4x+6)}+\\
&(a+10)\color{blue}{(x^2+4x+6)}-\color{red}{(4a+16)(x)+(b-6a-60)}
\end{align*}
For $x^2+4x+6$ to divide the given polynomial.We need $4a+16=0$ and $b-6a-60=0$. Thus $a=-4$ and $b=36$.
A: The first solution I came up with involves actually finding the roots of the quadratic, then employing Factor/Remainder theorem, so while it is a tad more tedious, it's also very direct. I will follow with an alternative solution that is more elegant.
By the quadratic formula, roots of $x^2 + 4x + 6$ are the complex conjugate pair $-2 \pm i\sqrt 2$.
By Factor Theorem, those will also be roots of the quartic (biquadratic), giving the linear simultaneous equations:
$(-2 + i\sqrt 2)^4 + a(-2 + i\sqrt 2)^2 + b = 0$
$(-2 - i\sqrt 2)^4 + a(-2 - i\sqrt 2)^2 + b = 0$
Now $(-2 \pm i\sqrt 2)^2 = 2 \mp 4i\sqrt 2$
Squaring again,
$(-2 \pm i\sqrt 2)^4 = -28 \mp 16i \sqrt 2$
which allows you to express the simultaneous equations as:
$(-28 - 16i\sqrt 2) + a(2 - 4i\sqrt 2) + b = 0$
$(-28 + 16i\sqrt 2) + a(2 + 4i\sqrt 2) + b = 0$
by subtraction, we almost immediately have $a(8i\sqrt 2) = -32i\sqrt 2 \implies a = -4$
By adding and back substitution, we get: $-56 + 4(-4) + 2b = 0 \implies b = 36$
And of course, to answer the original question, $a+b= 32$
Alternative solution:
Let the roots of the original quadratic be $x_1, x_2$.
Vieta's formulas give:
$x_1 + x_2 = -4$
$x_1x_2 = 6$
from which we can deduce that:
$x_1^2 + x_2^2 = (x_1 + x_2)^2 - 2x_1x_2 = 4$
and
$x_1^2x_2^2 = 36$
Now, by Factor Theorem, $x_1$ and $x_2$ are also roots of the biquadratic.
If we let $y = x^2$, the biquadratic can be written as:
$y^2 + ay + b = 0$
which will have roots $y_1$ and $y_2$ where $y_1 = x_1^2$ and $y_2 = x_2^2$
Applying Vieta's formula to the quadratic in $y$ allows us to deduce that:
$y_1 + y_2 = -a \implies x_1^2 + x_2^2 = -a$
and
$y_1y_2 = b \implies x_1^2x_2^2 = b$
and by reference to the above results, that allows us to immediately conclude that $a = -4, b = 36$ and $a+b = 32$.
This solution involves Vieta's formulas, and is less tedious than my original method.
A: Since $x^4 + a x^2 + b$ is even (i.e. invariant under $x \to -x$), so is its factorization.  If $x^2 + 4 x + 6$ is one factor, the other must be .... 
A: 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 $a+b=32$
