What will be the remainder if we divide the polynomial $x^{2001}$ by $(x+1)^{2}$?
I. solution. According to the binomial theorem,

$$
x^{2001}=((x+1)-1)^{2001}=(x+1)^{2001}-\binom{2001}{1}(x+1)^{2000}+\binom{2001}{2}(x+1)^{1999}-\cdots+\binom{2001}{2000}(x+1)-1
$$

Except for the last two terms, each term is divisible by $(x+1)^{2}$, so the remainder is the sum of the last two terms, 2001(x+ 1) $-1=2001 x+2000$.

Gábor Zavarkó (Miskolc, Földes F. Gymn., 11th grade)

II. solution. The sought remainder is of the form $a x+b$, that is,

$$
x^{2001}=(x+1)^{2} \cdot f+a x+b
$$

with a suitable polynomial $f$. Substituting $x=-1$ on both sides, we get

$$
-1=-a+b, \quad \text { that is } \quad b=a-1
$$

Substituting this into (1), we obtain

$$
x^{2001}+1=(x+1)^{2} \cdot f+a(x+1)
$$

After dividing by $(x+1)$, we get

$$
x^{2000}-x^{1999}+x^{1998}-\cdots+\cdots-x+1=(x+1) f+a
$$

Again, substituting the value $x=-1$:

$$
(-1)^{2000}-(-1)^{1999}+(-1)^{1998}-\cdots+\cdots-(-1)+1=a
$$

Thus, $a=2001$, so the sought remainder is $2001 x+2000$.