# Show that $c^2 + a^2d=abc$ for a monic quartic polynomial

I apologize in advance for asking a homework question, but I have genuinely no idea on how to approach part b). The question is as follows:

Consider the polynomial equation $$\rm{P}(x)=x^4 + ax^3 + bx^2 + cx+d$$, with a root $$ki$$. In addition, $$a,b,c,d\in Z$$ and $$k\in R,\; k\neq 0$$

(a) Show that $$c=k^2a$$

(b) Show that $$c^2 + a^2d=abc$$

Part (a) is quite trivial and can be done by knowing that $$ki$$ and $$\bar{ki}$$ are roots since all the coefficients are real. However I am unsure of how to do part (b), do I square the expression in part (a). I tried working backwards from the solution to a true statement here is what I had:

\begin{align} c^2 + a^2d &=abc \\ k^4a^2+a^2d &=abc \\ a^2\left(k^4 +d\right) &=abc \\ a\left(k^4 +d\right) &=bc \tag{assuming a\neq 0, though this may not be true} \\ \dots &\dots \\ k^4 &=\frac{bc-ad}{a} \end{align}

This gets me nowhere

• Have you tried using the fact that $x^2+k^2$ is a factor? I don't know if this works, but it's the fist thing that occurs to me. Try dividing and see what you need to make the remainder $0$. – saulspatz Apr 20 at 16:34

We know that $$ki$$ is a solution to $$P(x)=0$$ so in particular,
$$\begin{split} (ki)^4+a(ki)^3+b(ki)^2+c(ki)+d&=0 \\ \Leftrightarrow k^4-ak^3i-bk^2+cki+d&=0 \\ \Leftrightarrow k^4-bk^2+d+ki(c-ak^2)&=0 \end{split}$$
Using part (a), this gives $$k^4-bk^2+d=0$$. Now add $$bk^2$$ on both sides and multiply through with $$a^2$$ to get $$a^2k^4+a^2d=a^2bk^2=abak^2.$$ Due to part (a) we can substitute $$c^2$$ for $$a^2k^4$$ as well as $$c$$ for $$ak^2$$ and obtain $$c^2+a^2d=abc.$$
Hint: Both the real and imaginary parts of $$P(ki)$$ are $$0$$. Consider the real part and use (a).