2
$\begingroup$

Find the real root of the equation $z^3 + z + 10 = 0$ given that one complex root is $1 – 2i$.

I've realized that the roots are $(1-2i), (1+2i)$, and a real number we'll call $a$.

So using the theorem got me $(z-1-2i)(z-1+2i)(z-x)$.

No idea on where to go next.

$\endgroup$
  • $\begingroup$ Multiply out $(z - 1 - 2i)(z - 1 + 2i)$ and divide $z^3 + z + 10$ by the result. $\endgroup$ – user61527 Jan 23 '14 at 23:37
  • 2
    $\begingroup$ The last factor should be $z-a$ so that $a$ is a root. $\endgroup$ – Ross Millikan Jan 23 '14 at 23:41
5
$\begingroup$

The polynomial is monic (lead coefficient $1$). The coefficient of $z^2$ is therefore the negative of the sum of the roots. This coefficient is $0$.

The two known roots have sum $2$, so the missing root must be $-2$.

$\endgroup$
2
$\begingroup$

If the leading term of the polynomial has coefficient $1$, then the product of its roots gives the free term.

Your polynomial has real coefficients; if $1-2i$ is a root, then so is $1+2i$. Thus, we arrive to $10 = (1-2i)(1+2i)a$, where $a$ is the real root. We conclude that $a=2$.

$\endgroup$
1
$\begingroup$

To add to the answers, when you multiply out $(z-1-2i)(z-1+2i)$, observe that it is a difference of two squares, namely, $$((z-1)-2i)((z-1)+2i)=(z-1)^2-(2i)^2=(z-1)^2+4=z^2-2z+5$$ So $z^3+z+10=(z^2-2z+5)(z-a)=z^3-(2+a)z^2+(5+2a)z-5a$

Equating the like terms gives us $$2+a=0$$ $$5+2a=1$$ $$-5a=10$$ They all give the same answer...

$\endgroup$

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.