12. Let $p(x)$ be a polynomial of degree 2010. What is the maximum degree that the polynomial $p(x-1)-3 p(x)+3 p(x+1)-p(x+2)$ can have?
(A) It is always the zero polynomial
(B) 0
(C) 1
(D) 2007
(E) 2010

## Numerical answer problems -5 points
12. The answer is $(\mathbf{D})$. First, we note that if $p(x)$ has degree $d$, then $p(x+1)-p(x)$ has degree exactly $d-1$. For simplicity, we denote with $\star$ a sum of terms of degree less than $d-1$. Setting $p(x)=a x^{d}+b x^{d+1}+\star$, we have

$$
\begin{aligned}
p(x+1)-p(x) & =a(x+1)^{d}-a x^{d}+b(x+1)^{d-1}-b x^{d-1}+\star \\
& =\left(a x^{d}+a d x^{d-1}+\star\right)-a x^{d}+\left(b x^{d-1}+\star\right)-b x^{d-1}+\boldsymbol{t} \\
& =a d x^{d-1}+\star .
\end{aligned}
$$

Thus, $p(x+1)-p(x)$ has degree exactly $d-1$. Now, let $r(x):=p(x+1)-p(x)$ and $s(x):=r(x+1)-r(x)$; we have

$$
\begin{aligned}
s(x) & =(p(x+2)-p(x+1))-(p(x+1)-p(x)) \\
& =p(x+2)-2 p(x+1)+p(x), \\
s(x)-s(x-1) & =(p(x+2)-2 p(x+1)+p(x))-(p(x+1)-2 p(x)+p(x-1)) \\
& =p(x+2)-3 p(x+1)+3 p(x)-p(x-1) .
\end{aligned}
$$

Applying the result proven multiple times, we then have that the degree of $r(x)$ is $2010-1=2009$, the degree of $s(x)$ is 2008, the degree of $s(x)-s(x-1)$ is 2007 (to obtain this last one, we must set $y:=x-1$ and apply the result to $s(y+1)-s(y))$.

Second Solution

For every positive integer $i$, I define the polynomial

$$
\left(\begin{array}{c}
x \\
i
\end{array}\right)=\frac{x \cdot(x-1) \cdots(x-i+1)}{i !}
$$

I set $\left(\begin{array}{l}x \\ 0\end{array}\right)=1$ and $\left(\begin{array}{l}x \\ i\end{array}\right)=0$ for every integer $i<0$. The facts I need about these polynomials are two:

the first is that, for positive $i$, $\left(\begin{array}{c}x \\ i\end{array}\right)$ has degree $i$, and, in particular, the coefficient of $x^{i}$ in $\left(\begin{array}{c}x \\ i\end{array}\right)$ is $\frac{1}{i !}$ (it suffices to observe that the coefficient of $x$ in each of the $i$ linear factors that compose the numerator is 1)

the second is the equation $\left(\begin{array}{c}x+1 \\ i+1\end{array}\right)-\left(\begin{array}{c}x \\ i+1\end{array}\right)=\left(\begin{array}{c}x \\ i\end{array}\right)$, which holds for every integer $i$: for positive values of $i$ we have

$$
\begin{aligned}
\left(\begin{array}{c}
x+1 \\
i+1
\end{array}\right)-\left(\begin{array}{c}
x \\
i+1
\end{array}\right)= & \\
& =\frac{(x+1) \cdot x \cdots(x-i+1)}{(i+1) !}-\frac{x \cdots(x-i+1) \cdot(x-i)}{(i+1) !} \\
& =\frac{x+1}{i+1}\left(\begin{array}{c}
x \\
i
\end{array}\right)-\frac{x-i}{i+1}\left(\begin{array}{c}
x \\
i
\end{array}\right)=\left(\begin{array}{c}
x \\
i
\end{array}\right)
\end{aligned}
$$

for $i=0$ the equation reduces to $x+1-x=1$, and for $i<0$ it is an immediate consequence of the definition.

Applying this second fact, I obtain

$$
\begin{aligned}
& \left(\begin{array}{c}
x-1 \\
i
\end{array}\right)-3\left(\begin{array}{c}
x \\
i
\end{array}\right)+3\left(\begin{array}{c}
x+1 \\
i
\end{array}\right)-\left(\begin{array}{c}
x+2 \\
i
\end{array}\right)= \\
& =-\left(\left(\begin{array}{c}
x \\
i
\end{array}\right)-\left(\begin{array}{c}
x-1 \\
i
\end{array}\right)\right)+2\left(\left(\begin{array}{c}
x+1 \\
i
\end{array}\right)-\left(\begin{array}{c}
x \\
i
\end{array}\right)\right)-\left(\left(\begin{array}{c}
x+2 \\
i
\end{array}\right)-\left(\begin{array}{c}
x+1 \\
i
\end{array}\right)\right) \\
& =-\begin{array}{ccc}
\left(\begin{array}{c}
x \\
i-1
\end{array}\right) & +2 & \left(\begin{array}{c}
x+1 \\
i-1
\end{array}\right)
\end{array} \quad-\quad\left(\begin{array}{c}
x+2 \\
i-1
\end{array}\right) \\
& =\left(\left(\begin{array}{c}
x+1 \\
i-1
\end{array}\right)-\left(\begin{array}{c}
x \\
i-1
\end{array}\right)\right)-\left(\left(\begin{array}{c}
x+2 \\
i-1
\end{array}\right)-\left(\begin{array}{c}
x+1 \\
i-1
\end{array}\right)\right) \\
& =\left(\begin{array}{l}
x+1 \\
i-2
\end{array}\right) \quad-\quad\left(\begin{array}{c}
x+2 \\
i-2
\end{array}\right) \\
& =-\left(\begin{array}{c}
x+2 \\
i-3
\end{array}\right)
\end{aligned}
$$

where in the first and third steps I simply reordered the sum.

Now, consider a polynomial $p(x)$ of degree $d$. I claim that it can be written in the form

$$
p(x)=a_{d}\left(\begin{array}{l}
x \\
d
\end{array}\right)+\cdots+a_{1}\left(\begin{array}{l}
x \\
1
\end{array}\right)+a_{0}
$$

for a suitable choice of real numbers $a_{0}, \ldots, a_{d}$. Indeed, calling $k$ the leading coefficient of $p(x)$ and setting $a_{d}=k d$!, I obtain that $p(x)-a_{d}\left(\begin{array}{l}x \\ d\end{array}\right)$ has degree less than $d$, since the monomials of degree $d$ cancel out. Repeating the process on this new polynomial, I obtain an $a_{d-1}$ such that the degree of $p(x)-a_{d}\left(\begin{array}{l}x \\ d\end{array}\right)-a_{d-1}\left(\begin{array}{c}x \\ d-1\end{array}\right)$ is less than $d-1$. Continuing to lower the degree of the polynomial, I reach the point where $p(x)-a_{d}\left(\begin{array}{l}x \\ d\end{array}\right)-a_{d-1}\left(\begin{array}{c}x \\ d-1\end{array}\right)-\cdots-a_{1}\left(\begin{array}{l}x \\ 1\end{array}\right)$ has degree 0, i.e., it is a constant, and I set $a_{0}$ equal to this constant.

Now it is easy to compute the polynomial requested by the problem, collecting the coefficients $a_{i}$

$$
\begin{aligned}
p(x-1)-3 p(x)+3 p(x+1)-p(x+2) & = \\
& =a_{d}\left(\left(\begin{array}{c}
x-1 \\
d
\end{array}\right)-3\left(\begin{array}{c}
x \\
d
\end{array}\right)+3\left(\begin{array}{c}
x+1 \\
d
\end{array}\right)-\left(\begin{array}{c}
x+2 \\
d
\end{array}\right)\right)+\cdots \\
& =-a_{d}\left(\begin{array}{c}
x+2 \\
d-3
\end{array}\right)-a_{d-1}\left(\begin{array}{c}
x+2 \\
d-4
\end{array}\right)-\cdots-a_{3}
\end{aligned}
$$

and this is a polynomial of degree $d-3$, since the degree of $\left(\begin{array}{l}x+2 \\ d-3\end{array}\right)$, like that of $\left(\begin{array}{c}x \\ d-3\end{array}\right)$, is $d-3$, while the degree of the other terms is less. The correct answer is therefore 2007.