299. When dividing the polynomial $x^{1051}-1$ 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.
299. The polynomial $x^{4}+x^{8}+2 x^{2}+x+1$ can be factored; it equals $\left(x^{2}+1\right)\left(x^{2}+x+1\right)$. From this, it is easy to see that this polynomial is a divisor of the polynomial

$$
x^{12}-1=\left(x^{6}-1\right)\left(x^{6}+1\right)=\left(x^{3}-1\right)\left(x^{3}+1\right)\left(x^{2}+1\right)\left(x^{4}-x^{2}+1\right)
$$

specifically:

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

Dividing $x^{1951}-1$ by $x^{4}+x^{3}+2 x^{2}+x+1$ is the same as dividing $x^{1951}-1$ by $x^{12}-1$, and then multiplying the result by $x^{8}-x^{7}-x^{6}+2 x^{5}-2 x^{3}+x^{2}+x-1$. But it is easy to see that

$$
\frac{x^{1951}-1}{x^{12}-1}=x^{1939}+x^{1927}+x^{1915}+x^{1903}+\ldots+x^{19}+x^{7}+\frac{x^{7}-1}{x^{12}-1}
$$

(it is easy to verify this by performing polynomial division or by noting that $x^{1951}-1=x^{7}\left[\left(x^{12}\right)^{162}-1\right]+x^{7}-1$ and using the known formula for dividing the difference of even powers of two monomials by the difference of their bases). From this, it follows that the desired coefficient coincides with the coefficient of $x^{14}$ in the product

$$
\begin{aligned}
\left(x^{1939}+x^{1927}+\ldots+x^{31}+\right. & \left.x^{19}+x^{7}+\frac{x^{7}-1}{x^{12}-1}\right) \times \\
& \times\left(x^{8}-x^{7}-x^{6}+2 x^{5}-2 x^{3}+x^{2}+x-1\right)
\end{aligned}
$$

which is equal to -1.