O produto das raízes da equação $2x^2 - 5x + 3 = 0$ é
(A) $\dfrac{2}{3}$ (B) $\dfrac{3}{2}$ (C) $\dfrac{5}{2}$ (D) $2$ (E) $3$

The function $f(x) = x^2 - 4x + 3$ has roots $x_1$ and $x_2$. What is the value of $x_1 + x_2$?
(A) $-4$
(B) $-3$
(C) $3$
(D) $4$
(E) $7$

a) Find a polynomial $p ( x )$ with real coefficients such that $p ( \sqrt { 2 } + i ) = 0$. b) Find a polynomial $q ( x )$ with rational coefficients and having the smallest possible degree such that $q ( \sqrt { 2 } + i ) = 0$. Show that any other polynomial with rational coefficients and having $\sqrt { 2 } + i$ as a root has $q ( x )$ as a factor.

Let $x^{3} + ax^{2} + bx + 8 = 0$ be a cubic equation with integer coefficients. Suppose both $r$ and $-r$ are roots of this equation, where $r > 0$ is a real number. List all possible pairs of values $(a, b)$.

The polynomial $p(x) = 10x^{400} + ax^{399} + bx^{398} + 3x + 15$, where $a, b$ are real constants, is given to be divisible by $x^{2}-1$.
(i) If you can, find the values of $a$ and $b$. Write your answers as $a =$ $\_\_\_\_$, $b =$ $\_\_\_\_$. If it is not possible to decide, state so.
(ii) If you can, find the sum of reciprocals of all 400 (complex) roots of $p(x)$. Write your answer as sum $=$ $\_\_\_\_$. If it is not possible to decide, state so.

[12 points] Consider polynomials $p(x)$ with the following property, called $(\dagger)$. $(\dagger)$ If $r$ is a root of $p(x)$, then $r^{2} - 4$ is also a root of $p(x)$.
(i) We want to find every quadratic polynomial of the form $p(x) = x^{2} + bx + c$ such that $p(x)$ has two distinct roots, has integer coefficients and has property $(\dagger)$. Prove that there are exactly two such polynomials and list them.
(ii) It is also true that there are exactly two cubic polynomials of the form $p(x) = x^{3} + ax^{2} + bx + c$ with the property $(\dagger)$ such that $p(x)$ shares no root with the polynomials you found in part (i). Explain fully how you will prove this along with the method to find the polynomials, but do not try to explicitly find the polynomials.

Consider polynomials of the form $f ( x ) = x ^ { 3 } + a x ^ { 2 } + b x + c$ where $a , b , c$ are integers. Name the three (possibly non-real) roots of $f ( x )$ to be $p , q , r$.
(a) If $f ( 1 ) = 2021$, then $f ( x ) = ( x - 1 ) \left( x ^ { 2 } + s x + t \right) + 2021$ where $s , t$ must be integers.
(b) There is such a polynomial $f ( x )$ with $c = 2021$ and $p = 2$.
(c) There is such a polynomial $f ( x )$ with $r = \frac { 1 } { 2 }$.
(d) The value of $p ^ { 2 } + q ^ { 2 } + r ^ { 2 }$ does not depend on the value of $c$.

[14 points] We want to find a nonzero polynomial $p ( x )$ with integer coefficients having the following property.
$$\text { Letting } q ( x ) : = \frac { p ( x ) } { x ( 1 - x ) } , \quad q ( x ) = q \left( \frac { 1 } { 1 - x } \right) \text { for all } x \notin \{ 0,1 \}$$
(i) Find one such polynomial with the smallest possible degree.
(ii) Find one such polynomial with the largest possible degree OR show that the degree of such polynomials is unbounded.

Suppose that for a given polynomial $p ( x ) = x ^ { 4 } + a x ^ { 3 } + b x ^ { 2 } + c x + d$, there is exactly one real number $r$ such that $p ( r ) = 0$.
(a) If $a, b, c, d$ are rational, show that $r$ must be rational.
(b) If $a, b, c, d$ are integers, show that $r$ must be an integer.
Possible hint: Also consider the roots of the derivative $p ^ { \prime } ( x )$.

Consider the polynomial $$p(x) = x^6 + 10x^5 + 11x^4 + 12x^3 + 13x^2 - 12x - 11.$$ Let $z_1, z_2, z_3, z_4, z_5, z_6$ be the six complex roots of $p(x)$. Evaluate $\sum_{i=1}^{6} z_i^2$. [2 points]

A function $f(x) = x^{3} + ax^{2} + bx + 4$ satisfies the following condition for two integers $a$ and $b$. What is the maximum value of $f(1)$? [4 points] For all real numbers $\alpha$, the limit $\lim_{x \rightarrow \alpha} \frac{f(2x+1)}{f(x)}$ exists.

For $n \in \mathbb{N}^*$, $T_n$ denotes the polynomial function satisfying $T_n(x) = 2^{1-n} F_n(x)$ where $F_n(x) = \cos(n \arccos x)$.
Let $n \in \mathbb{N}^*$. Show that the polynomial function $T_n$ has exactly $n$ distinct zeros all belonging to $]-1,1[$. For $j \in \{1, 2, \ldots, n\}$, we denote by $x_{n,j}$ the $j$-th zero of $T_n$ in increasing order. Give the value of $x_{n,j}$.

For $n \in \mathbb{N}^*$, $T_n$ denotes the polynomial function satisfying $T_n(x) = 2^{1-n} F_n(x)$ where $F_n(x) = \cos(n \arccos x)$, and $x_{n,j}$ denotes the $j$-th zero of $T_n$ in increasing order.
Let $n \in \mathbb{N}^*$ and $x \in \mathbb{R} \setminus \{x_{n,j},\, 1 \leqslant j \leqslant n\}$. Show that: $$\frac{T_n'(x)}{T_n(x)} = \sum_{j=1}^{n} \frac{1}{x - x_{n,j}}$$

For $n \in \mathbb{N}^*$, $T_n$ denotes the polynomial function satisfying $T_n(x) = 2^{1-n} F_n(x)$ where $F_n(x) = \cos(n \arccos x)$, and $x_{n,j}$ denotes the $j$-th zero of $T_n$ in increasing order. We denote by $E_{n-1}$ the vector subspace of polynomial functions of degree at most $n-1$.
Let $n \in \mathbb{N}^*$, $x \in \mathbb{R} \setminus \{x_{n,j},\, 1 \leqslant j \leqslant n\}$ and $P \in E_{n-1}$.
a) Show that: $$P(x) = \sum_{j=1}^{n} \frac{P(x_{n,j})}{T_n'(x_{n,j})} \frac{T_n(x)}{x - x_{n,j}}$$
b) Deduce that: $$P(x) = \frac{2^{n-1}}{n} \sum_{j=1}^{n} (-1)^{n-j} \sqrt{1 - x_{n,j}^2}\, P(x_{n,j}) \frac{T_n(x)}{x - x_{n,j}}$$

We assume $\alpha = 1$. We denote $T_n$ the unique polynomial eigenvector of $\varphi_1$ of degree $n$, of norm 1 (with respect to $S_1$) and with positive leading coefficient. For $n \in \mathbb{N}^*$, determine the roots of $T_n$.

Determine $T_0, T_1, T_2$ and $T_3$, where the Chebyshev polynomials of the first kind $(T_n)_{n \in \mathbb{N}}$ are defined by $$\forall n \in \mathbb{N}, \quad \forall \theta \in \mathbb{R}, \quad T_n(\cos\theta) = \cos(n\theta)$$

The Chebyshev polynomials of the first kind $(T_n)_{n \in \mathbb{N}}$ are defined by $T_n(\cos\theta) = \cos(n\theta)$.
By noting that for every real $\theta$, we have $e^{in\theta} = \left(e^{i\theta}\right)^n$, show that $$\forall n \in \mathbb{N}, \quad T_n = \sum_{0 \leqslant k \leqslant n/2} \binom{n}{2k} \left(X^2 - 1\right)^k X^{n-2k}$$
Write in Maple or Mathematica language a function $T$ taking as argument a natural integer $n$ and returning the expanded expression of the polynomial $T_n$.

The Chebyshev polynomials of the first kind $(T_n)_{n \in \mathbb{N}}$ are defined by $T_n(\cos\theta) = \cos(n\theta)$.
Show that the sequence $(T_n)_{n \in \mathbb{N}}$ satisfies the recurrence relation $$\forall n \in \mathbb{N}, \quad T_{n+2} = 2X T_{n+1} - T_n$$
Deduce, for every natural integer $n$, the degree and the leading coefficient of $T_n$. Find this result again with the expression from question I.A.2.

The Chebyshev polynomials of the first kind $(T_n)_{n \in \mathbb{N}}$ are defined by $T_n(\cos\theta) = \cos(n\theta)$.
Show that, for every natural integer $n$, the polynomial $T_n$ is split over $\mathbb{R}$, with simple roots belonging to $]-1,1[$. Determine the roots of $T_n$.

The Chebyshev polynomials of the second kind $(U_n)_{n \in \mathbb{N}}$ are defined by $$\forall n \in \mathbb{N}, \quad U_n = \frac{1}{n+1} T_{n+1}'$$ where $(T_n)_{n \in \mathbb{N}}$ are the Chebyshev polynomials of the first kind defined by $T_n(\cos\theta) = \cos(n\theta)$.
Show that $$\forall n \in \mathbb{N}, \quad \forall \theta \in \mathbb{R} \backslash \pi\mathbb{Z}, \quad U_n(\cos\theta) = \frac{\sin((n+1)\theta)}{\sin\theta}$$

The Chebyshev polynomials of the second kind $(U_n)_{n \in \mathbb{N}}$ are defined by $U_n = \frac{1}{n+1} T_{n+1}'$, and satisfy $U_n(\cos\theta) = \frac{\sin((n+1)\theta)}{\sin\theta}$ for $\theta \notin \pi\mathbb{Z}$.
Deduce the following properties:
a) The sequence $(U_n)_{n \in \mathbb{N}}$ satisfies the same recurrence relation $T_{n+2} = 2X T_{n+1} - T_n$ as the sequence $(T_n)_{n \in \mathbb{N}}$.
b) For every natural integer $n$, the polynomial $U_n$ is split over $\mathbb{R}$ with simple roots belonging to $]-1,1[$. Determine the roots of $U_n$.

The Chebyshev polynomials of the first and second kind are defined by $T_n(\cos\theta) = \cos(n\theta)$ and $U_n(\cos\theta) = \frac{\sin((n+1)\theta)}{\sin\theta}$ for $\theta \notin \pi\mathbb{Z}$.
Show that $$\begin{cases} T_m \cdot T_n = \frac{1}{2}\left(T_{n+m} + T_{n-m}\right) & \text{for all integers } 0 \leqslant m \leqslant n \\ T_m \cdot U_{n-1} = \frac{1}{2}\left(U_{n+m-1} + U_{n-m-1}\right) & \text{for all integers } 0 \leqslant m < n \end{cases}$$

For $\alpha \in \mathbb{C}$, we set $P_\alpha = X^2 + \alpha$. We denote by $\mathcal{C}(P)$ the set of complex polynomials that commute with the polynomial $P$ under composition.
Let $\alpha \in \mathbb{C}$ and let $Q$ be a non-constant complex polynomial that commutes with $P_\alpha$. Show that $Q$ is monic.

For $\alpha \in \mathbb{C}$, we set $P_\alpha = X^2 + \alpha$. We denote by $\mathcal{C}(P)$ the set of complex polynomials that commute with the polynomial $P$ under composition. Every non-constant polynomial commuting with $P_\alpha$ is monic.
Deduce that, for every integer $n \geqslant 1$, there exists at most one polynomial of degree $n$ that commutes with $P_\alpha$. Determine $\mathcal{C}(X^2)$.

For $\alpha \in \mathbb{C}$, we set $P_\alpha = X^2 + \alpha$. We denote by $G$ the set of complex polynomials of degree 1, and the inverse of $U \in G$ under composition is denoted $U^{-1}$.
Let $P$ be a complex polynomial of degree 2. Justify the existence and uniqueness of $U \in G$ and $\alpha \in \mathbb{C}$ such that $U \circ P \circ U^{-1} = P_\alpha$. Determine these two elements when $P = T_2$.