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 $(-2 \pm i\sqrt 2)^2 = 2 \mp 4i\sqrt 2$
$(-2 \pm i\sqrt 2)^4 = -28 \mp 16i \sqrt 2$
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, $a+b= 32$
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:
$x_1^2 + x_2^2 = (x_1 + x_2)^2 - 2x_1x_2 = 4$
$x_1^2x_2^2 = 36$
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:
$y_1 + y_2 = -a \implies x_1^2 + x_2^2 = -a$
$y_1y_2 = b \implies x_1^2x_2^2 = b$
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.