# A very very hard college algebra problem

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1. Aug 11, 2016

### nmego12345

1. The problem statement, all variables and given/known data
Note: I'm saying it's very very hard because I still couldn't solve it and I've posted it in stackexchange and no answer till now.

I'm posting here the problem statement, all variables and known data in addition to my solving attempts. Because I'm posting an image of my question and that it would be hard to seperate my solving attempts from the problem statement etc.., but they are obvious in the image. I'm posting an image because i'm not familiar to the formatting here yet, and I don't have time to format all this problem again ( took me 1.5 hours to format)

https://s3.amazonaws.com/diigo/thum...[/B][/B] [B]3. The attempt at a solution[/B]

2. Aug 11, 2016

### Stephen Tashi

We need to state the problem correctly. Isn't the problem to find a polynomial $f_k (x)$ such that $f_k(x)- f_k(x-1) = x^k$ ?

3. Aug 11, 2016

### nmego12345

I'm not sure If I understand the problem correctly, I found it on a textbook, I didn't invent it, but I think
that you're correct

4. Aug 11, 2016

### Stephen Tashi

Let $p(x) = x^{k+1} + ...$ other terms. Suppose, for example, that $p(x) - p(x-1)$ has the term $Cx^3$.

By part 2) , if you form the new polynomial $q(x) = p(x) - C f_3(x)$ then

$q(x) - q(x-1) = p(x) - p(x-1) + C (f_3(x) - f_3(x-1)) = x^{k+1}+ ... Cx^3 + .... - C x^3$ so you can eliminate the $Cx^3$ term by subtracting a multiple of $f_3(x)$.

I think what the problem wants you to do is express $f_k(x)$ as $x^k$ minus multiples of $f_{k-1}, f_{k-2},...$.

After you do that, you may be able to find the numerical values of the coefficients, but perhaps the problem only wants you to write the "recursive" relation between $f_k$ and $f_{k-1}, f_{k-2} ...$.

5. Aug 11, 2016

### nmego12345

First of all, what is Cx3?, is it combinations? but combinations are defined over 2 numbers not 1 number. Is it an arbitary constant?

6. Aug 11, 2016

### Stephen Tashi

C is an arbitrary constant.

7. Aug 11, 2016