Finding $b$ and $d$ such that $2x^4+ax^3+bx^2+cx+d $ has roots $x=\sqrt2-3$ and $x=3i+2$ 
I've got the following function:
$$f(x)=2x^4+ax^3+bx^2+cx+d$$
Given the two roots $x=(\sqrt2-3)$ and $x=(3i+2)$, I need to compute values $b$ and $d$.

Can anyone point me in the right direction of how to start this?
Thanks in advance.
EDIT: Just to clarify, that's the information I've got. No more, no less.
 A: This is a basic application of Vieta's formulas. Here is one way we can approach the problem:
Assuming rational coefficients, there are two conjugate pairs we are dealing with, namely:

*

*the surd conjugate pair: $\left(\sqrt 2-3\right)$ and $\left(-\sqrt 2 - 3\right)$, whose sum is $-6$ and product is $7$

*the complex conjugate pair: $(3i+2)$ and $(-3i+2)$, whose sum is $4$ and product is $13$
Using the sum and product of roots, a quadratic function that has the surd conjugate pair as zeros must thus have the form $A_1(x^2+6x+7)$; a quadratic function that has the complex conjugate pair as zeros must have the form $A_2(x^2-4x+13)$, where $A_1,A_2\in\mathbb Q$.
We can now think of the original quartic $f(x)$ as a product of two quadratic functions with these respective forms:
$$2x^4+ax^3+bx^2+cx+d=A_1A_2\left(x^4+2x^3-4x^2+50x+91\right)$$
Noticing from the coefficient of $x^4$ that this can only hold if $A_1A_2=2$, we have:
$$2x^4+ax^3+bx^2+cx+d=2x^4+4x^3-8x^2+100x+182$$
From there, we can just compare coefficients to get the values of $a$, $b$, $c$ and $d$.
