1. Find all quadratic polynomials $p(x)=a x^{2}+b x+c$, which give a remainder of 2 when divided by the binomial $x+1$ and a remainder of 1 when divided by the binomial $x+2$, given that $p(1)=61$.
SOLUTION. Using the division algorithm twice, we get

$$
\begin{aligned}
& a x^{2}+b x+c=(a x+b-a)(x+1)+c-b+a, \\
& a x^{2}+b x+c=(a x+b-2 a)(x+2)+c-2 b+4 a .
\end{aligned}
$$

Let us add that the found remainders $c-b+a$ and $c-2 b+4 a$ are clearly equal to the values $p(-1)$ and $p(-2)$, respectively, which is consistent with the fact that any polynomial $q(x)$ gives a remainder equal to the number $q\left(x_{0}\right)$ when divided by the binomial $x-x_{0}$.

According to the problem, $c-b+a=2$ and $c-2 b+4 a=1$. The third equation $a+b+c=61$ expresses the condition $p(1)=61$. We will solve the obtained system of three equations using one of many possible methods.

From the first equation, we express $c=b-a+2$, and by substituting into the third equation, we get $a+b+(b-a+2)=61$ or $2 b=59$. Hence, $b=59 / 2$, which, when substituted into the first and second equations, gives $a+c=63 / 2$ and $c+4 a=60$. By subtracting the last two equations from each other, we get $3 a=57 / 2$, from which $a=19 / 2$, so $c=63 / 2-19 / 2=22$. The sought quadratic polynomial is thus unique and has the form

$$
p(x)=\frac{19}{2} \cdot x^{2}+\frac{59}{2} \cdot x+22=\frac{19 x^{2}+59 x+44}{2} .
$$

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