B1. If the polynomial $p$ is divided by the polynomial $2-x$, the quotient is $2 x^{2}-x+3$. Determine the remainder of this division, given that the product of all the zeros of the polynomial $p$ is equal to $\frac{11}{2}$.
B1. According to the division theorem, $p(x)=\left(2 x^{2}-x+3\right)(2-x)+o$, where $o$ is the remainder when the polynomial $p(x)$ is divided by the polynomial $2-x$. Since we are dividing by a polynomial of degree 1, the remainder $o$ is some constant. Therefore, the constant term of the polynomial $p(x)=-2 x^{3}+5 x^{2}-5 x+(6+o)$ is equal to $6+o$, and the leading coefficient is -2. According to Vieta's formulas, the product of all the roots of the polynomial $p(x)$ is thus $-\frac{6+o}{-2}=\frac{11}{2}$. From this, we calculate the remainder $o=5$.

The polynomial $p$ written as $\left(2 x^{2}-x+3\right)(2-x)+o \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots$

![](https://cdn.mathpix.com/cropped/2024_06_07_58dcb28c79ea4cc315e1g-23.jpg?height=51&width=1573&top_left_y=1705&top_left_x=276)

The constant term of the polynomial $p$ ..........................................................................

![](https://cdn.mathpix.com/cropped/2024_06_07_58dcb28c79ea4cc315e1g-23.jpg?height=51&width=1573&top_left_y=1805&top_left_x=276)

Calculation of the remainder ...........................................................................................