How many cubic (i.e., third-degree) polynomials \(f(x)\) are there such that \(f(x)\) has nonnegative integer coefficients and \(f(1)=9\)?

Hint 1: Keep in mind that these are nonnegative integers not just postive integers.

Hint 2: Try to think about the problem by using "sticks and stones"

 Apr 26, 2021

Consider a third-degree polynomial \(f(x)\) such that

        \(f(x) =ax^3+bx^2+cx+d\)

⇒    \(f(1) = a+b+c+d\)


Now, since  \(f(1)=9\)

∴ \(a+b+c+d=9\)


So, you're looking for the number of ways to write 9 as the sum of four non-negative integers.

Here I'm applying "sticks and stones" method as you suggested.


⇒No. of solutions \(=\binom{n+r-1}{r-1}\)

Here, \(n=9\)  and  \(r=4\)


Total no. of possible solutions  \(=\binom{9+4-1}{4}\)


                                                 \(={12! \over (12-4)!4!}\)

                                                 \(={12! \over 8!4!}\)



∴ There are 495 possible third-degree polynomials that fulfills the given condition. 




Phew! This question blew my mind! angel


P.S. Suggestions and corrections are welcome.


 Apr 26, 2021

Nice solution, but I think that you have to consider that a cannot equal 0. (otherwise it wouldn't be cubic


Guest Apr 27, 2021

Yeah you're right, in that case lets eliminate one possibility and make it 494 possible polynomials.

amygdaleon305  Apr 27, 2021

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