3
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ “the real coefficients of a polynomial” or “a polynomial with only real coefficients”? $\endgroup$ – Martin R Oct 9 '18 at 20:22
  • $\begingroup$ Only real coefficients. $\endgroup$ – Boris Grunwald Oct 9 '18 at 20:24
5
$\begingroup$

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:

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

$\endgroup$
  • $\begingroup$ Thanks, I think this makes sense. $\endgroup$ – Boris Grunwald Oct 9 '18 at 20:34
  • $\begingroup$ And sorry, it was the smallest possible degree $\endgroup$ – Boris Grunwald Oct 9 '18 at 20:39
3
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.