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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

#1
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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  $$=\binom{9+4-1}{4}$$

$$=\binom{12}{4}$$

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

$$={12! \over 8!4!}$$

$$=495$$

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

Phew! This question blew my mind!

P.S. Suggestions and corrections are welcome.

Apr 26, 2021
#2
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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
#3
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Yeah you're right, in that case lets eliminate one possibility and make it 494 possible polynomials.

amygdaleon305  Apr 27, 2021