Homework Help: Polynomial Help

1. Sep 19, 2010

chillfactor

1. The problem statement, all variables and given/known data

find a polynomial P(x) which has nonnegative coefficients. If P(1)=1 and P(5)= 426, then wast is p(3)

2. Relevant equations
P(1)= 6
P(5)= 426
P(3)= x

3. The attempt at a solution
I have tried to use guess and check. I cant find a way to solve algebraically.

Last edited: Sep 19, 2010
2. Sep 19, 2010

rock.freak667

Do you have a specified degree for p(x) ?

3. Sep 19, 2010

chillfactor

There is no specified degree for p(X). It just has to have nonnegative coefficents

4. Sep 19, 2010

rock.freak667

Then it is best to choose a degree and work with that.

5. Sep 19, 2010

chillfactor

There is only one polynomial that works. One hint I received was that it had to be less than 6.

6. Sep 19, 2010

Mentallic

I have no idea how to approach it either. Can't quite gather what to do with the fact that all the coefficients aren't negative. Either way, it must be at least a polynomial of degree 4. But what suggests that it must be less than 6?

7. Sep 19, 2010

chillfactor

how do you know it must be at least degree 4?

8. Sep 19, 2010

Office_Shredder

Staff Emeritus
If you try to fit for example a quadratic

$$ax^2+bx+c$$ to the conditions, then you get two conditions:

a+b+c=1

25a+5b+c=426

We can subtract one from the other to get

24a+4b=425

Now we know that a+b+c=1 must be satisfied, with all non-negative numbers, so none of a, b or c can be larger than 1. That's clearly not possible if 24a+4b=425. A similar argument kills cubic polynomials

9. Sep 19, 2010

chillfactor

i am afraid i made a mistake when I posted the question. Actually P(1)= 6. So, could you try again. I would appreciate it.

10. Sep 19, 2010

Office_Shredder

Staff Emeritus
There are lots of choices of polynomials actually, just consider a degree n polynomial of the form axn+b. We know that a+b=6, and that 5na+b=426. Two equations, two unknowns, start trying to find solutions!

If you know a little linear algebra/convex geometry there's a neat explanation for how you can decide exactly which coefficients are able to be non-zero

Last edited: Sep 19, 2010
11. Sep 20, 2010

tiny-tim

welcome to pf!

hi chillfactor! welcome to pf!
i assume all the coefficients must be whole numbers?

then ∑an = 6 and ∑an5n = 426

obviously a0 = 1,

and by subtracting we have 4a1 + 24a2 + 124a3 = 420 …

carry on from there

12. Sep 20, 2010

chillfactor

Re: welcome to pf!

thanks!