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. 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 $$R(X) = \prod_{k=1}^{2n}(X - \omega_k)$$

Consider the polynomial
$$P = x^2 + 2(a-1)x - 8a - 8.$$
(1) Let $a$ be a rational number. If the value of $P$ is a rational number when $x = 1 - \sqrt{2}$, then $a =$ $\mathbf{K}$ and in this case the value of $P$ is $P =$ $\mathbf{LM}$.
(2) Let $x$ and $a$ be positive integers. We are to investigate $x$ and $a$ which are such that the value of $P$ is a prime number.
When we factorize $P$, we have
$$P = (x - \mathbf{N}\mathbf{N})(x + \mathbf{O}a + \mathbf{P}).$$
Hence $x$ must be $\mathbf{Q}$.
Furthermore, the smallest possible $a$ is $\mathbf{R}$, and in this case the value of $P$ is $P = \mathbf{ST}$.

Consider the polynomial
$$P = x^2 + 2(a-1)x - 8a - 8.$$
(1) Let $a$ be a rational number. If the value of $P$ is a rational number when $x = 1 - \sqrt{2}$, then $a =$ $\mathbf{K}$ and in this case the value of $P$ is $P =$ $\mathbf{LM}$.
(2) Let $x$ and $a$ be positive integers. We are to investigate $x$ and $a$ which are such that the value of $P$ is a prime number.
When we factorize $P$, we have
$$P = (x - \mathbf{N})(x + \mathbf{O}a + \mathbf{P}).$$
Hence $x$ must be $\mathbf{Q}$.
Furthermore, the smallest possible $a$ is $\mathbf{R}$, and in this case the value of $P$ is $P = \mathbf{ST}$.

Let $a$ and $b$ be integers,
$$P(x) = x^{3} + ax^{2} + bx - 2$$
It is known that the polynomial has exactly one real root.
If $\mathbf{P}(1) = 0$, what is the smallest value that the integer $a$ can take?
A) $-6$ B) $-5$ C) $-4$ D) $-3$ E) $-2$