# Finding Real and Complex Roots of a Polynomial Equation

• May 6th 2009, 12:41 PM
ft_fan
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!

• May 6th 2009, 01:01 PM
TheAbstractionist
Quote:

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. (Wink)
• May 6th 2009, 01:03 PM
Plato
Quote:

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?
• May 6th 2009, 04:58 PM
ft_fan
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!
• May 6th 2009, 05:42 PM
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.
• May 6th 2009, 05:50 PM
TheEmptySet
Quote:

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 (Clapping)