## Task 12/85

What is the remainder when the polynomial $P(x)=x^{n}-x^{n-1}$ with $n \in N, n \neq 0$ is divided by the polynomial $Q(x)=(x-1)^{2}$?
Since $Q(x)$ is of the second degree, the remainder $R(x)$ must be linear: $R(x)=a_{1} x+a_{0}$ with real numbers $a_{1}, a_{0}$. Therefore, we have

$$
P(x)=(x-1)^{2} P_{1}(x)+a_{1} x+a_{0}
$$

where $P_{1}(x)$ is a polynomial of degree $(n-2)$ in $\mathrm{x}$. For $x=1$, we get

$$
P(1)=0 \cdot P_{1}(1)+a_{1}+a_{0}=0
$$

thus $a_{0}=-a_{1}$. By differentiating, we also obtain

$$
P^{\prime}(x)=2(x-1) P_{1}(x)+(x-1)^{2} P_{1}^{\prime}(x)+a_{1}=n x^{n-1}-(n-1) x^{n-2}
$$

Substituting $x=1$ into this, we get $P^{\prime}(1)=a_{1}=n-(n-1)=1$. Therefore, $R(x)=x-1$.