Given the two functions $f$ and $h$ such that $f ( x ) = x ^ { 3 } - 3 x ^ { 2 } - 4 x + 12$ and $h ( x ) = \left\{ \begin{array} { l } \frac { f ( x ) } { x - 3 } , \text { for } x \neq 3 \\ p , \text { for } x = 3 . \end{array} \right.$ (a) Find all zeros of the function $f$. (b) Find the value of $p$ so that the function $h$ is continuous at $x = 3$. Justify your answer. (c) Using the value of $p$ found in (b), determine whether $h$ is an even function. Justify your answer.

When does the polynomial $1 + x + \cdots + x ^ { n }$ have $x - a$ as a factor? Here $n$ is a positive integer greater than 1000 and $a$ is a real number.
(A) if and only if $a = - 1$
(B) if and only if $a = - 1$ and $n$ is odd
(C) if and only if $a = - 1$ and $n$ is even
(D) We cannot decide unless $n$ is known.

Let $f ( x ) = 7 x ^ { 32 } + 5 x ^ { 22 } + 3 x ^ { 12 } + x ^ { 2 }$. (i) Find the remainder when $f ( x )$ is divided by $x ^ { 2 } + 1$. (ii) Find the remainder when $x f ( x )$ is divided by $x ^ { 2 } + 1$. In each case your answer should be a polynomial of the form $a x + b$, where $a$ and $b$ are constants.

Find a polynomial $p(x)$ that simultaneously has both the following properties.
(i) When $p(x)$ is divided by $x^{100}$ the remainder is the constant polynomial 1.
(ii) When $p(x)$ is divided by $(x-2)^{3}$ the remainder is the constant polynomial 2.

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.

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.

Statements
(25) To divide an integer $b$ by a nonzero integer $d$, define a quotient $q$ and a remainder $r$ to be integers such that $b = q d + r$ and $| r | < | d |$. Such integers $q$ and $r$ always exist and are both unique for given $b$ and $d$. (26) To divide a polynomial $b ( x )$ by a nonzero polynomial $d ( x )$, define a quotient $q ( x )$ and a remainder $r ( x )$ to be polynomials such that $b = q d + r$ and $\deg ( r ) < \deg ( d )$. (Here $b ( x )$ and $d ( x )$ have real coefficients and the 0 polynomial is taken to have negative degree by convention.) Such polynomials $q ( x )$ and $r ( x )$ always exist and are both unique for given $b ( x )$ and $d ( x )$. (27) Suppose that in the preceding question $b ( x )$ and $d ( x )$ have rational coefficients. Then $q ( x )$ and $r ( x )$, if they exist, must also have rational coefficients. (28) The least positive number in the set $$\left\{ \left( a \times 2023 ^ { 2020 } \right) + \left( b \times 2020 ^ { 2023 } \right) \right\}$$ as $a$ and $b$ range over all integers is 3.

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 $$p(x) = x^6 + 10x^5 + 11x^4 + 12x^3 + 13x^2 - 12x - 11.$$ Find the remainder when $p(x)$ is divided by $x+1$. [1 point]

For a natural number $n$, let $a _ { n }$ be the remainder when the polynomial $2 x ^ { 2 } - 3 x + 1$ is divided by $x - n$. Find the value of $\sum _ { n = 1 } ^ { 7 } \left( a _ { n } - n ^ { 2 } + n \right)$. [3 points]

Show that, for all $x \in \mathbb { R }$ and for all $n \in \mathbb { N } ^ { * }$, $$\left( x ^ { n } - 1 \right) ^ { 2 } = \prod _ { k = 1 } ^ { n } \left( x ^ { 2 } - 2 x \cos \frac { 2 k \pi } { n } + 1 \right)$$

Throughout this subsection II.B, we fix two natural integers $m$ and $n$. The Chebyshev polynomials of the second kind $U_n$ have roots $\cos\left(\frac{k\pi}{n+2}\right)$ for $k = 1, \ldots, n+1$.
Let $h$ be the $\gcd$ in $\mathbb{N}$ of $m+1$ and $n+1$. By examining the common roots of $U_n$ and $U_m$, show that $U_{h-1}$ is a $\gcd$ in $\mathbb{R}[X]$ of $U_n$ and $U_m$.

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.

Let $\widehat { \mu } = \left( \mu _ { 1 } > \cdots > \mu _ { n } \right) \in \mathbb { R } ^ { n }$. We define the polynomials $$Q _ { 0 } = \prod _ { k = 1 } ^ { n } \left( X - \mu _ { k } \right) \quad \text { and } \quad \forall j \in \{ 1 , \ldots , n \} , \quad P _ { j } = \frac { Q _ { 0 } } { \left( X - \mu _ { j } \right) } .$$ Let $P \in \mathbb { R } [ X ]$ be a monic polynomial of degree $n + 1$.
(a) Show that there exists a unique vector $\left( a , \alpha _ { 1 } , \alpha _ { 2 } , \ldots , \alpha _ { n } \right) \in \mathbb { R } ^ { n + 1 }$ such that $$P = ( X - a ) Q _ { 0 } - \sum _ { j = 1 } ^ { n } \alpha _ { j } P _ { j }$$ (b) Assume that the real numbers $\alpha _ { 1 } , \ldots , \alpha _ { n }$ are all strictly positive. Show that $P$ has $n + 1$ distinct real roots $\lambda _ { 1 } > \cdots > \lambda _ { n + 1 }$, and that $\hat { \lambda } = \left( \lambda _ { 1 } > \cdots > \lambda _ { n + 1 } \right)$ and $\widehat { \mu }$ are strictly interlaced.
(c) Conversely, assume that $P$ has $n + 1$ distinct real roots $\lambda _ { 1 } > \cdots > \lambda _ { n + 1 }$, and that $\widehat { \lambda } = \left( \lambda _ { 1 } > \cdots > \lambda _ { n + 1 } \right)$ and $\widehat { \mu }$ are strictly interlaced. Show that, for all $j \in \{ 1 , \ldots , n \} , \alpha _ { j } > 0$.

Let $\widehat { \mu } = \left( \mu _ { 1 } > \cdots > \mu _ { n } \right) \in \mathbb { R } ^ { n }$. We are given integers $m _ { k } \geqslant 1$ for $k = 1 , \ldots , n$. We set $$Q _ { 1 } = \prod _ { k = 1 } ^ { n } \left( X - \mu _ { k } \right) ^ { m _ { k } } \quad \text { and, this time, } \quad P _ { j } = \frac { Q _ { 1 } } { X - \mu _ { j } } .$$ Show that $$Q _ { 1 } \wedge Q _ { 1 } ^ { \prime } = \prod _ { k = 1 } ^ { n } \left( X - \mu _ { k } \right) ^ { m _ { k } - 1 }$$

Let $\widehat { \mu } = \left( \mu _ { 1 } > \cdots > \mu _ { n } \right) \in \mathbb { R } ^ { n }$. We are given integers $m _ { k } \geqslant 1$ for $k = 1 , \ldots , n$. We set $$Q _ { 1 } = \prod _ { k = 1 } ^ { n } \left( X - \mu _ { k } \right) ^ { m _ { k } } \quad \text { and } \quad P _ { j } = \frac { Q _ { 1 } } { X - \mu _ { j } } .$$ Let $\left( a , \alpha _ { 1 } , \alpha _ { 2 } , \ldots , \alpha _ { n } \right) \in \mathbb { R } ^ { n + 1 }$ and let $P \in \mathbb { R } [ X ]$ be defined by the formula $$P = ( X - a ) Q _ { 1 } - \sum _ { j = 1 } ^ { n } \alpha _ { j } P _ { j }$$ (a) Give an expression of $P \wedge Q _ { 1 }$ in terms of the $\mu _ { j }$, the $m _ { j }$ and the set $J$ of indices for which $\alpha _ { j } = 0$.
(b) Assume that the numbers $\alpha _ { 1 } , \ldots , \alpha _ { n }$ are non-negative. Show that all roots of $P$ are real.

Let $n$ be a non-zero natural number. Let $A \in \mathbb{C}_{2n}[X]$, split with simple roots, and $(\alpha_1, \ldots, \alpha_{2n})$ its roots. Show that $$\forall B \in \mathbb{C}_{2n-1}[X], \quad B(X) = \sum_{k=1}^{2n} B(\alpha_k) \frac{A(X)}{(X - \alpha_k) A'(\alpha_k)}$$

Let $n$ be a non-zero natural number. Let $P$ be in $\mathbb{C}_{2n}[X]$, and, for all $\lambda \in \mathbb{C}$, $P_\lambda(X) = P(\lambda X) - P(\lambda)$. If $\lambda \in \mathbb{C}$, verify that $X - 1$ divides $P_\lambda$.

Let $n$ be a non-zero natural number. Let $P$ be in $\mathbb{C}_{2n}[X]$, and for all $\lambda$ in $\mathbb{C}$, let $Q_\lambda$ be the quotient of $P_\lambda = P(\lambda X) - P(\lambda)$ by $X-1$: $$Q_\lambda(X) = \frac{P(\lambda X) - P(\lambda)}{X - 1} \in \mathbb{C}_{2n-1}[X]$$ Show that, for all $\lambda$ in $\mathbb{C}$, $Q_\lambda(1) = \lambda P'(\lambda)$.

Let $n$ be a non-zero natural number. We consider the polynomial $R(X) = X^{2n} + 1$. For $k$ in $\llbracket 1, 2n \rrbracket$, we denote $\varphi_k = \frac{\pi}{2n} + \frac{k\pi}{n}$ and $\omega_k = \mathrm{e}^{\mathrm{i}\varphi_k}$. Show that $$R(X) = \prod_{k=1}^{2n}(X - \omega_k)$$

Let $n$ be a non-zero natural number. Let $P$ be in $\mathbb{C}_{2n}[X]$, and for all $\lambda$ in $\mathbb{C}$, let $Q_\lambda(X) = \frac{P(\lambda X) - P(\lambda)}{X-1} \in \mathbb{C}_{2n-1}[X]$. We consider the polynomial $R(X) = X^{2n} + 1$. For $k$ in $\llbracket 1, 2n \rrbracket$, we denote $\varphi_k = \frac{\pi}{2n} + \frac{k\pi}{n}$ and $\omega_k = \mathrm{e}^{\mathrm{i}\varphi_k}$.
Using formula $$\forall B \in \mathbb{C}_{2n-1}[X], \quad B(X) = \sum_{k=1}^{2n} B(\alpha_k) \frac{A(X)}{(X - \alpha_k) A'(\alpha_k)} \tag{I.1}$$ show that $$\forall \lambda \in \mathbb{C}, \quad Q_\lambda(X) = -\frac{1}{2n} \sum_{k=1}^{2n} \frac{P(\lambda\omega_k) - P(\lambda)}{\omega_k - 1} \frac{X^{2n}+1}{X - \omega_k} \omega_k$$ then deduce that $$\forall \lambda \in \mathbb{C}, \quad \lambda P'(\lambda) = \frac{1}{2n} \sum_{k=1}^{2n} P(\lambda\omega_k) \frac{2\omega_k}{(1-\omega_k)^2} - \frac{P(\lambda)}{2n} \sum_{k=1}^{2n} \frac{2\omega_k}{(1-\omega_k)^2}.$$

Let $n$ be a non-zero natural number. Let $P$ be in $\mathbb{C}_{2n}[X]$. For $k$ in $\llbracket 1, 2n \rrbracket$, we denote $\varphi_k = \frac{\pi}{2n} + \frac{k\pi}{n}$ and $\omega_k = \mathrm{e}^{\mathrm{i}\varphi_k}$. Show that $$\forall \lambda \in \mathbb{C}, \quad \lambda P'(\lambda) = \frac{1}{2n} \sum_{k=1}^{2n} P(\lambda\omega_k) \frac{2\omega_k}{(1-\omega_k)^2} + nP(\lambda) \tag{I.2}$$ One may apply equality (I.2) to the polynomial $X^{2n}$.

Let $n$ be a non-zero natural number. Let $f$ be in $\mathcal{S}_n$, where $\mathcal{S}_n$ is the $\mathbb{C}$-vector space of functions $f : \mathbb{R} \rightarrow \mathbb{C}$ satisfying $$\exists (a_0, \ldots, a_n) \in \mathbb{C}^{n+1}, \quad \exists (b_1, \ldots, b_n) \in \mathbb{C}^n, \quad \forall t \in \mathbb{R}, \quad f(t) = a_0 + \sum_{k=1}^{n}\left(a_k \cos(kt) + b_k \sin(kt)\right)$$ Show that there exists $U \in \mathbb{C}_{2n}[X]$ such that, for all $\theta \in \mathbb{R}$, $f(\theta) = \mathrm{e}^{-\mathrm{i}n\theta} U(\mathrm{e}^{\mathrm{i}\theta})$.

Let $n$ be a non-zero natural number. For $k$ in $\llbracket 1, 2n \rrbracket$, we denote $\varphi_k = \frac{\pi}{2n} + \frac{k\pi}{n}$ and $\omega_k = \mathrm{e}^{\mathrm{i}\varphi_k}$. Let $f \in \mathcal{S}_n$.
Verify that, for all $k \in \llbracket 1, 2n \rrbracket$, $\frac{2\omega_k}{(1-\omega_k)^2} = \frac{-1}{2\sin(\varphi_k/2)^2}$ and deduce from questions 11 and 12 that $$\forall \theta \in \mathbb{R}, \quad f'(\theta) = \frac{1}{2n} \sum_{k=1}^{2n} f(\theta + \varphi_k) \frac{(-1)^k}{2\sin(\varphi_k/2)^2}.$$