10.9. When dividing the polynomial $x^{1951}-1$ by $x^{4}+x^{3}+2 x^{2}+$ $+x+1$, a quotient and a remainder are obtained. Find the coefficient of $x^{14}$ in the quotient.

## 10.4. Vieta's Theorem
10.9. Answer: -1. The equalities $x^{4}+x^{3}+2 x^{2}+x+1=\left(x^{2}+1\right) \times$ $\times\left(x^{2}+x+1\right)$ and $x^{12}-1=(x-1)\left(x^{2}+x+1\right)\left(x^{3}+1\right)\left(x^{2}+1\right)\left(x^{4}-x^{2}+1\right)$ show that

$$
\begin{aligned}
& x^{4}+x^{3}+2 x^{2}+x+1=\frac{x^{12}-1}{(x-1)\left(x^{3}+1\right)\left(x^{4}-x^{2}+1\right)}= \\
& =\frac{x^{12}-1}{x^{8}-x^{7}-x^{6}+2 x^{5}-2 x^{3}+x^{2}+x-1}
\end{aligned}
$$

Therefore, dividing the polynomial $x^{1951}-1$ by $x^{4}+x^{3}+2 x^{2}+x+1$ is the same as first dividing it by $x^{12}-1$, and then multiplying by $x^{8}-x^{7}-x^{6}+2 x^{5}-2 x^{3}+x^{2}+x-1$. But

$$
\frac{x^{1951}-1}{x^{12}-1}=x^{1939}+x^{1927}+x^{1915}+\ldots+x^{19}+x^{7}+\frac{x^{7}-1}{x^{12}-1}
$$

so the desired coefficient is the coefficient of $x^{14}$ in the product

$$
\left(x^{1939}+\ldots+x^{19}+x^{7}+\frac{x^{7}-1}{x^{12}-1}\right)\left(x^{8}-x^{7}-x^{6}+2 x^{5}-2 x^{3}+x^{2}+x-1\right)
$$