Finding real coefficients of polynomial with complex roots

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I have been tasked with finding the just real coefficients of a polynomial with the roots:

$z_{0}=10$

$z_{1}=3-i$

$z_{2}=-8+2i$

Writing the polynomial as factors...

$(z-10)(z-3+i)(z+8-2i)$

..obviously does not work since this results in complex coefficients.

Can anyone help me out here?

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2 Answers

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There is not enough information, say the degree of a given polynomial.

Anyway, since it has (only) real coeficients if $a+bi$ is one root, then $a-bi$ is also a root of the polynomial.

So this polynomial is divisible with:

$$q(z)=(z-10)\color{red}{(z-3+i)(z-3-i)}\color{blue}{(z+8-2i)(z+8+2i)} $$$$= (z-10)(z^2-6z+10)(z^2+16z+68)$$

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Polynomial with only real coefficients is$$( z-10) \, ( z-2i+8)\,(z+2 i+8) \, ( z-i-3) \, ( z+i-3)\\= {{z}^{5}}-118 {{z}^{3}}-68 {{z}^{2}}+3160 z-6800 $$

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