# Thread: Finding Real and Complex Roots of a Polynomial Equation

1. ## Finding Real and Complex Roots of a Polynomial Equation

Hi there,

I have to answer to following question, and am having issues:

Given that $\displaystyle 2-i$ is a root of the polynomial equation

$\displaystyle z^6-5z^5+8z^4-2z^3-3z^2+3z-10=0$,

find all other roots.

I can see that $\displaystyle i$ is a root, and that because of conjugate pairs, $\displaystyle -2-i$ must be, and I understand that there must be 6 roots due to the degree of the polynomial, but I don't know how to find the other 3. Any advice or pointers would be hugely appreciated!

2. Originally Posted by ft_fan
I can see that $\displaystyle i$ is a root, and that because of conjugate pairs, $\displaystyle -2-i$ must be, and I understand that there must be 6 roots due to the degree of the polynomial, but I don't know how to find the other 3. Any advice or pointers would be hugely appreciated!
Hi ft_fan.

If $\displaystyle i$ is a root, then $\displaystyle -i$ must also be a root. So you know four of the six roots of the equation already. You should be able to find the remaining two roots by solving a quadratic equation.

3. Originally Posted by ft_fan
Hi there,

I have to answer to following question, and am having issues:

Given that $\displaystyle 2-i$ is a root of the polynomial equation

$\displaystyle z^6-5z^5+8z^4-2z^3-3z^2+3z-10=0$,

find all other roots.

I can see that $\displaystyle i$ is a root, and that because of conjugate pairs, $\displaystyle \color{red}-2-i$ must be, and I understand that there must be 6 roots due to the degree of the polynomial, but I don't know how to find the other 3.
The conjugate of $\displaystyle 2+i$ is $\displaystyle 2\color{blue}-i$ not $\displaystyle \color{red}-2-i$.

You know four roots $\displaystyle 2+i,~2-i,~i,~-i$.
So here are four factors $\displaystyle [z-(2+i)][z-(2-i)][z-i][z+i]$
Can you finish?

4. Ok, getting there, and thanks for picking up my mistake in typing above. I am still having a slight issue though;
I have taken the 4 roots,

$\displaystyle 2-i$, $\displaystyle 2+i$, $\displaystyle i$ and $\displaystyle -i$,

and multiplied them together as follows,

$\displaystyle (z-(2-i))(z-(2+i))(z-(i))(z-(-i))=0$,

giving the following polynomial,

$\displaystyle z^4-4z^3+6z^2-4z+5=0$.

I have divided this by the original polynomial,

$\displaystyle z^6-5z^5+8z^4-2z^3-3z^2+3z-10=0$,

but each time I get the same thing,
$\displaystyle z^2+z+1$ remainder $\displaystyle -10z^2-z-5$,

which clearly isn't correct.

Am I doing something wrong in thr division of polynomials, or have I perhaps multiplied the roots together incorrectly? I've reached the point where I have retried the problem on so many occasions I cant work out what's going wrong.

Any help much appreciated!

5. when you divide $\displaystyle z^6-5z^5+8z^4-2z^3-3z^2+3z-10=0$by $\displaystyle z^4-4z^3+6z^2-4z+5$ you will get $\displaystyle z^2-z-2$. Check it.

6. Originally Posted by chakravarthiponmudi
when you divide $\displaystyle z^6-5z^5+8z^4-2z^3-3z^2+3z-10=0$by $\displaystyle z^4-4z^3+6z^2-4z+5$ you will get $\displaystyle z^2-z-2$. Check it.

Right....

but $\displaystyle z^2-z-2=(z-2)(z+1)$

so both $\displaystyle z=-1,2$ are both roots