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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?

 

f(1) = 9 means ax^3 + bx^2 + cx + d = 9. But we can have coefficients of 0. I am really confused.

 Nov 18, 2022
 #1
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By stars and bars, the number of such polynomials is C(12,4) = 495.

 Nov 18, 2022
 #3
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Thanks!

Guest Nov 20, 2022
 #2
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Guest, you are super close, but note that a cannot be 0 as that will violate the definition of a cubic polynomial. 

 

By f(1) = 9, you will get that:

a + b + c + d = 9

Since a has to be greater than or equal to 1, we can set another variable f that is equivalent to a - 1.

 

f + b + c + d = 8

 

We can use stars and bars from here, as f can take a value of 0 as well. Hence, the answer is actually 

 

C(11, 3) = 11 x 10 x 9/6 = 15 x 11 = 165.

 Nov 19, 2022
 #4
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Actually, 495 is correct, but thanks for your help!

Guest Nov 20, 2022

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