Finding real coefficients of polynomial with complex roots

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?

• “the real coefficients of a polynomial” or “a polynomial with only real coefficients”? – Martin R Oct 9 '18 at 20:22
• Only real coefficients. – Boris Grunwald Oct 9 '18 at 20:24

Anyway, since it has (only) real coeficients if $$a+bi$$ is one root, then $$a-bi$$ is also a root of the polynomial.
$$q(z)=(z-10)\color{red}{(z-3+i)(z-3-i)}\color{blue}{(z+8-2i)(z+8+2i)}$$ $$= (z-10)(z^2-6z+10)(z^2+16z+68)$$
Polynomial with only real coefficients is $$( z-10) \, ( z-2i+8)\,(z+2 i+8) \, ( z-i-3) \, ( z+i-3)\\= {{z}^{5}}-118 {{z}^{3}}-68 {{z}^{2}}+3160 z-6800$$