Finding the 5th root.

1. Jul 17, 2007

Mr. Mathmatics

1. The problem statement, all variables and given/known data
Find a Polynomial function of degree 5 whose zeros are: 2, radical 7, and 5i

2. Relevant equations
Multiplying roots

3. The attempt at a solution

Hello, well it's my first time at these forums so just following the template given. Thanks for any help by the way I'm just so confused.

I know that being given an imaginary number makes another root -5i automatically but the polynomial is a degree 5 with 5 roots. So my roots are (2, radical 7, 5i and -5i) I only have 4 and I don't really know any way to find the 5th.

I've already multiplied 5i and -5i to get one part of the polynomial but without the 5th root, I just can't finish. Anyone have any suggestions?

Thanks so much again.

2. Jul 17, 2007

Hurkyl

Staff Emeritus
Just because a polynomial has 5i as a root doesn't mean that it must also have -5i as a root.

e.g. f(x) = x - 5i is a perfectly good polynomial function.

I don't see anything in your statement of the problem that requires those to be the only roots of your polynomial, or that they have to have multiplicity one.

3. Jul 17, 2007

chanvincent

Did you miss something in the question? i.e. the coefficient of the polynomial is restricted to real interger ...

4. Jul 17, 2007

Mr. Mathmatics

chanvincent: I'm pretty sure she wants to keep the polynomial away from imaginary numbers.

Ahhh, good point I guess I can find any polynomial with those roots giving it any multiplicity I want since it says "a" polynomial function. Thanks for the speedy replies both of you.

But now I have a new problem I have x^4-6x^3+12x^2-8x and I have to multiply that by -5i and (x-radical 7) When I tried it, I was welcomed with a new imaginary polynomial. So is there any better way to multiplying this out, or am I just doing it wrong?

Thanks again

Last edited: Jul 17, 2007
5. Jul 17, 2007

Dick

If your polynomial needs to have real coefficients then -5i does need to be a root. That's what Hurkyl and chanvincent were pointing out. I don't know where you got that quartic polynomial (0 is also a root) but you should probably pose this question a little more clearly.

6. Jul 18, 2007

lalbatros

From how you stated the question, I would say that more than one polynomial are solutions to this question.

"Find a polynomial" does not mean "Find the polynomial".

7. Jul 18, 2007

chanvincent

Ha ha... a very good point... then the general solution would be:

$$(x-2)(x-\sqrt{7})(x-5i)(x-c_1)(x-c_2)$$

but then the question would become meaningless...

Ha ha :tongue:

8. Jul 18, 2007

derekjn

Since you need to come up with two more zeroes, and the conjugate of (x - 5i) needs to be one of them (assuming your polynomial is to have real coefficients), use the conjugate of (x - sqrt(7)) as the other zero and it should be a little cleaner when you multiply.

9. Jul 18, 2007

HallsofIvy

Staff Emeritus
Of course, if you multiply a polynomial with real coefficients by -5i, you will get a polynomial with imaginary coefficients. I doubt that you were told to "multiply by -5i". What really was the problem? Is this a completely new problem or more on the original?

As far as the first problem is concerned, yes, you must have $(x-2)(x-\sqrt{7})(x-5i)(x-c_1)(x-c_2)$ but that certainly does not make the problem "meaningless"- choose c1 and c2 to be whatever you want. As you yourself said, the problem asks for "a" polynomial, not "the" polynomial. What others have been saying is that to make the polynomial as simple as possible, having not only real but integer coefficients, choose $c_1= -5i$ and $c_2= -\sqrt{7}$ so that your polynomial is $(x-2)(x-\sqrt{7})(x+\sqrt{7})(x-5i)(x+5i)$.

10. Jul 18, 2007

Gib Z

Of course, $(x-2)(x-\sqrt{7})(x-5i)(x-c_1)(x-c_2)$ is a perfectly good polynomial anyway, and also quite a good solution to the question. It is a polynomial, regardless that you have not defined c_1 or c_2. Also, you lose no generality.