Finding a Polynomial with Nonnegative Coefficients

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chillfactor
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Homework Statement



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

Homework Equations


P(1)= 6
P(5)= 426
P(3)= x

The Attempt at a Solution


I have tried to use guess and check. I can't find a way to solve algebraically.
 
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There is no specified degree for p(X). It just has to have nonnegative coefficents
 
chillfactor said:
There is no specified degree for p(X). It just has to have nonnegative coefficents

Then it is best to choose a degree and work with that.
 
There is only one polynomial that works. One hint I received was that it had to be less than 6.
 
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?
 
If you try to fit for example a quadratic

[tex]ax^2+bx+c[/tex] 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
 
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.
 
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
 
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welcome to pf!

hi chillfactor! welcome to pf! :wink:
chillfactor said:
find a polynomial P(x) which has nonnegative coefficients. If P(1)=6 and P(5)= 426, then wast is p(3)

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 :smile: