# Real and non-real roots to a quintic equation

• Feb 13th 2011, 10:25 AM
MickHarford
Real and non-real roots to a quintic equation
The question is:

Establish that alpha=2+3i is a root of the quintic equation

3z^5 + 2z^4 - 64z^3 +384z^2 - 667z +182 =0

and find the remaining roots.

Now, I know that as it's a quintic there will be five roots. As I'm told that 2+3i is a root, I can know that 2-3i is also a root. And once I get so far I will be able to divide, take the one real root and the other two complex roots.

But I'm falling at the first hurdle, how do I show that 2+3i is a root?
• Feb 13th 2011, 10:39 AM
dwsmith
Quote:

Originally Posted by MickHarford
The question is:

Establish that alpha=2+3i is a root of the quintic equation

3z^5 + 2z^4 - 64z^3 +384z^2 - 667z +182 =0

and find the remaining roots.

Now, I know that as it's a quintic there will be five roots. As I'm told that 2+3i is a root, I can know that 2-3i is also a root. And once I get so far I will be able to divide, take the one real root and the other two complex roots.

But I'm falling at the first hurdle, how do I show that 2+3i is a root?

If 2 + 3i and 2 - 3i are two solutions, what quadratic do they represent?

Once you obtain that and factor it out, the cubic will factor nicely.

To show 2 + 3i is a root, plug it in.
• Feb 13th 2011, 10:47 AM
MickHarford
Quote:

Originally Posted by dwsmith
If 2 + 3i and 2 - 3i are two solutions, what quadratic do they represent?

Once you obtain that and factor it out, the cubic will factor nicely.

To show 2 + 3i is a root, plug it in.

This is the part I don't get, or at least I'm being thick.

I've got an example written down:
p(z) = z^4 - 4z^3 +55z^2 -8z +106
2+7i and 2-7i are roots.
So p(z)=(z^2-4z+53)(z^2+2)

I understand how the division and eventual solutions are obtained, I jsut don't quite understand the step to what the quadratic from the solutions is?

It's probably something incredibly simple isn't it?
• Feb 13th 2011, 11:02 AM
dwsmith
Quote:

Originally Posted by MickHarford
This is the part I don't get, or at least I'm being thick.

I've got an example written down:
p(z) = z^4 - 4z^3 +55z^2 -8z +106
2+7i and 2-7i are roots.
So p(z)=(z^2-4z+53)(z^2+2)

I understand how the division and eventual solutions are obtained, I jsut don't quite understand the step to what the quadratic from the solutions is?

It's probably something incredibly simple isn't it?

What would z need to be is the question we are asking:

$\displaystyle z=\frac{-b\pm\sqrt{b^2-4ac}}{2a}=\cdots$

$\displaystyle 2\pm 3i=\frac{-(-4)\pm\sqrt{-36}}{2}=\frac{-(-4)\pm\sqrt{16-4ac}}{2(1)}$

What are a and c?
• Feb 13th 2011, 11:23 AM
roninpro
Try multiplying $(z-[2-3i])(z-[2+3i])$.
• Feb 13th 2011, 12:21 PM
MickHarford
Quote:

Originally Posted by roninpro
Try multiplying $(z-[2-3i])(z-[2+3i])$.

I got z^2 - 4z + 13 can someone verify please?

I'll crack on with the question now anyway!
• Feb 13th 2011, 01:17 PM
skeeter
Quote:

Originally Posted by MickHarford
I got z^2 - 4z + 13 can someone verify please?

I'll crack on with the question now anyway!

that's fine.
• Feb 14th 2011, 12:44 PM
MickHarford
Thanks guys. Solved the problem, after my initial struggles it just required two divisions through.

Solution set I got was 2+3i, 2-3i, 2, 7 and -1/3 (going off memory right now)
• Feb 14th 2011, 09:42 PM
mr fantastic
Quote:

Originally Posted by roninpro
Try multiplying $(z-[2-3i])(z-[2+3i])$.

Easier to expand $([z - 2] + 3i)([z - 2] - 3i)$.