Solve a quartic function with three unknowns and two given roots.

Question

I'd like to know how I could solve the following quartic function: $p(z)=2z^4+az^3+bz^2+cz+3$ given that it $2$ and $i$ should be part of their roots.

I thought I should maybe be trying to turn this into a quadratic function and then I should be able to solve it for the 2 given roots, but I don't really know how I could get there.

Answer 1

If the aim is to find $a$, $b$ and $c$, then you can substitute the known zeros of $p$ to get a set of linear equations:

$\begin{eqnarray}p(2) & = & 0 & \implies & 2 \times 2^4 + a \times 2^3 + b \times 2^2 + c \times 2 + 3 & = & 0 & \implies 32 + 8a + 4b + 2c + 3 & = & 0\\ p(i) & = & 0 & \implies & 2 \times i^4 + a \times i^3 + b \times i^2 + c \times i + 3 & = & 0 & \implies 2 - ai - b + ci + 3 & = & 0\end{eqnarray}$

Now, you have a few options:

1. Use the knowledge that you have real coefficients to separate the second equation into a real and imaginary part;

2. Use the fact that $p(-i) = 0$, which results from having real coefficients;

3. Get the 4th root using Vieta's formulas, as given in the other answer.

All of these will give you enough information to solve the entire system for the three unknowns.

Answer 2

using Vieta's formulas, and the fact that complex roots occurs in pairs, we get $x_{1}x_{2}x_{3}x_{4}=x_{1}.2.i.(-i)=(-1)^{n}\frac{a_{0}}{a_{n}}=3/2$. So, $x_{1}=3/4$