Difficulty finding complex roots of an equation

Question

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.

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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:

Answer 1

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?

Answer 2

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

Answer 3

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:

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