Cubic equation with cubic

Question

I have a question that I am finding difficult to answer.

$a = 1 + 4i$ satifies the cubic equation:$$z³+5z²+kz+m=0$$ where $k$ and $m$ are real constants

I believe using $a^2$ and $a^3$ will help to find what $k$ and $m$ and also the other roots of the equation but I am not sure where to go.

Thank you for your help

Answer 1

Hint: Expand the cubic $(z-\alpha)(z-1-4i)(z-1+4i)$ and then choose $\alpha$ such that the coefficient of $z^2$ is $5$.

Answer 2

Hint : if $\alpha=1+4i$ is a root (of a real polynomial) then $\beta=1-4i$ is also a root. We also have \begin{eqnarray*} \alpha+\beta+\gamma=-5. \end{eqnarray*} Now calculate $(z^2-2z+17)(z-\gamma)=0$.

Answer 3

Option 1: Plug in the root and retrieve the real and imaginary parts and put them to zero :

$$(1+4i)^3+5\,(1+4i)^2+(1+4i)k+m=0$$ $$(k+m-122)+(4k-12)i=0$$

The imaginary part equal to zero gives $k=3$ and this in the real part leads to $m=119$

Option 2: You have the root, so you know that $f(z)$ is divisible by $z-1-4i$. By polynomial long-division, you find :

$$z^3+5\,z^2+k\,z+m = (z-1-4\,i)\,(z^2+(6+4\,i)\,z+k - 10 + 28\,i)$$ $$ + (k+m-122)+(4k-12)\,i$$

The remainder of this division should be zero, thus :

$$(k+m-122)+(4k-12)i=0$$