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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

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  • $\begingroup$ 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 $0$. $\endgroup$ – saulspatz Apr 20 at 16:34
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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 $$c^2+a^2d=abc.$$

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Hint: Both the real and imaginary parts of $P(ki)$ are $0$. Consider the real part and use (a).

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