# Cubic equation with two complex roots and one real root?

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.

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

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 $-2$.
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 $a=2$.
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...