Polynomial Problem of the Week #194: Find $f(2008)$ for a Degree 2008 Polynomial

[anemone]

Here is this week's POTW:

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Let $f(x)$ be a polynomial with degree $2008$ and leading coefficient $1$ such that

$$f(0)=2007,\,f(1)=2006,\,f(2)=2005,\,\cdots\,f(2007)=0$$

Determine the value of $f(2008)$.

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[anemone]

Congratulations to the following members for their correct solution::)

1. kaliprasad
2. MarkFL

Solution from MarkFL:

Spoiler

Let:

\(\displaystyle g(x)=f(x)+x-2007\)

Clearly, $g(x)$ has the roots $x\in\{0,1,2,\,\cdots\,2007\}$, hence:

\(\displaystyle g(x)=\prod_{k=0}^{2007}(x-k)\)

And so we may state:

\(\displaystyle \prod_{k=0}^{2007}(x-k)=f(x)+x-2007\)

Solving for $f(x)$, we obtain:

\(\displaystyle f(x)=\prod_{k=0}^{2007}(x-k)-x+2007\)

Hence:

\(\displaystyle f(2008)=\prod_{k=0}^{2007}(2008-k)-2008+2007=2008!-1\)