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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
avatar+274 
+5

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! angel

 

P.S. Suggestions and corrections are welcome.

       

 Apr 26, 2021
 #2
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+1

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

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