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I have been presented with the following question:

given that $(2-i)$ and $(1+3i)$ are roots of the equation $$ax^4 + bx^3 + cx^2 + dx + e = 0$$ find the other two roots.


I am not sure where to begin on this question, any help with the methodology to find the solution would be greatly appreciated.

here is the exact question:

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    $\begingroup$ Assuming there is a comma between $(2-i)$ and $(3-i)$, there is no way to determine the other two roots, without some constraints on the complex values of $a,b,c,d,e$. $\endgroup$ – user228113 Jan 3 '17 at 15:00
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    $\begingroup$ Same as above, with "$(1+3i)$" instead of "$(3-i)$". Pick your two favourite complex numbers $\alpha$ and $\beta$, and consider the polynomial $(x-\alpha)(x-\beta)(x-2+i)(x-1-3i)$. $\endgroup$ – user228113 Jan 3 '17 at 15:14
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    $\begingroup$ $a,b,c,d,e\in\mathbb R$ or not? $\endgroup$ – MattG88 Jan 3 '17 at 15:14
  • $\begingroup$ very sorry I had switched the polarities by mistake, the question is now correct I have got the answer which is $(2+i)$ and $(1-3i)$ but I need to know how to do the workings out, e.g when do I look out for a complex conjugate? how do I show on paper that I have established that there are complex conjugates to be found $\endgroup$ – Flewitt Connor Jan 3 '17 at 15:18
  • $\begingroup$ @FlewittConnor I have edited your question to state that you mean real coefficients as that is a necessary condition for the roots to be as you say there are. $\endgroup$ – gowrath Jan 3 '17 at 15:41
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Hint: If $a, b, c, d, e \in \mathbb{R}$, then the complex roots of $$ax^4 + bx^3 + cx^2 + dx + e = 0$$ come in conjugate pairs.

Can you take it from here?

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The other two solutions are: $2+i$ and $1-3i$. This a consequence of the fundamental theorem of algebra when $a,b,c,d,e\in\mathbb R$.

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  • $\begingroup$ Why should they? $(x-i)(x+i)(x-2+i)(x-3+i)$ is as good as any other polynomial of degree $4$. $\endgroup$ – user228113 Jan 3 '17 at 15:02
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    $\begingroup$ Doesn't this assume real coefficients? $\endgroup$ – John Jan 3 '17 at 15:03
  • $\begingroup$ You are right:) $\endgroup$ – MattG88 Jan 3 '17 at 15:04
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This question doesn't seem complete because you need to have some additional constraints on the coefficients of the polynomial for this to have one answer.

I am going to assume your question means $a, b, c, d, e \in \mathbb{R}$. In this case, since it is a 4th degree polynomial, it is of the form

$$ ax^4 + bx^3 + cx^2 + dx + e = (x-z_1)(x-z_2)(x - z_3)(x - z_4) $$

where $z_1, z_2, z_3, z_4$ are the roots of the polynomial. We already know two of these so we have:

$$ \begin{align} z_1 &= 2 - i \\ z_3 &= 1 + 3i \end{align} $$

Now since the coefficients are real, the roots must come in conjugate pairs i.e. for every complex root $z = \alpha+\beta i$, there must be another root $\overline{z} = \alpha - \beta i$ so that $(x-z)(x-\overline{z})$ will cancel out their imaginary parts, resulting in real coefficients. Thus:

$$ \begin{align} z_2 &= \overline{z_1} = 2 + i \\ z_4 &= \overline{z_3} = 1 - 3i \\ \end{align} $$

So your polynomial is:

$$ \begin{align} ax^4 + bx^3 + cx^2 + dx + e &= (x-z_1)(x-z_2)(x - z_3)(x - z_4) \\ &= (x- 2 + i)(x- 2 - i)(x - 1 - 3i)(x - 1 + 3i) \\ &= x^4 - 6 x^3 + 23 x^2 - 50 x + 50 \end{align} $$

Matching coefficients on the LHS and RHS, we get:

$$ \begin{align} a &= 1 \\ b &= -6\\ c &= 23\\ d &= -50 \\ e &= 50 \end{align} $$

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