# Thread: Real and non-real roots to a quintic equation

1. ## 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?

2. 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.

3. 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?

4. 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:

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

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

What are a and c?

5. Try multiplying $\displaystyle (z-[2-3i])(z-[2+3i])$.

6. Originally Posted by roninpro
Try multiplying $\displaystyle (z-[2-3i])(z-[2+3i])$.
I got z^2 - 4z + 13 can someone verify please?

I'll crack on with the question now anyway!

7. 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.

8. 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)

9. Originally Posted by roninpro
Try multiplying $\displaystyle (z-[2-3i])(z-[2+3i])$.
Easier to expand $\displaystyle ([z - 2] + 3i)([z - 2] - 3i)$.