## Task A-4.1.

Determine all polynomials $P$ of the third degree that have the following three properties:

(i) $P(x)$ when divided by $x^{2}-1$ gives a remainder of $2 x+1$,

(ii) the sum of the roots of the polynomial $P$ is -2,

(iii) the graph of the polynomial $P$ passes through the point $(0,3)$.
## First solution.

From property (i) it follows that $P(x)$ can be written as

$$
P(x)=\left(x^{2}-1\right) Q(x)+(2 x+1)
$$

Since $P(x)$ is a polynomial of the third degree, $Q(x)$ must obviously be a polynomial of the first degree, i.e., $Q(x)=a x+b$, so we have

$$
P(x)=\left(x^{2}-1\right)(a x+b)+(2 x+1)=a x^{3}+b x^{2}+(2-a) x+(1-b). \quad 2 \text{ points}
$$

From Viète's formulas, it follows that the sum of the roots of the polynomial is equal to $-\frac{b}{a}$. Therefore, $\frac{b}{a}=2. \quad 2$ points From property (iii) it follows that $P(0)=3$, which, when substituted into the expression for $P(x)$, gives $(1-b)=3$, hence $b=-2$.

Finally, from $\frac{b}{a}=2$ and $b=-2$, it follows that $a=-1$, so the desired polynomial is

$$
P(x)=-x^{3}-2 x^{2}+3 x+3
$$

2 points