# 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!