3. Simplify the fraction $\frac{2 x^{6}+5 x^{4}-3 x^{3}+2 x^{2}-12 x-14}{4 x^{6}-4 x^{4}-6 x^{3}-3 x^{2}+25 x-28}$. After simplification, the degrees of the polynomials in the numerator and the denominator should be reduced.

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# Solution:

Let's find the greatest common divisor of the polynomials in the numerator and the denominator using the Euclidean algorithm.

For this, we will divide the denominator by the numerator with a remainder:

$$
\begin{gathered}
4 x^{6}-4 x^{4}-6 x^{3}-3 x^{2}+25 x-28= \\
=2 \cdot\left(2 x^{6}+5 x^{4}-3 x^{3}+2 x^{2}-12 x-14\right)-14 x^{4}-7 x^{2}+49
\end{gathered}
$$

As a result of the division, we obtained the remainder $-14 x^{4}-7 x^{2}+49$. Now, we will divide the numerator (which now acts as the divisor) by the remainder (for example, using long division):

$$
2 x^{6}+5 x^{4}-3 x^{3}+2 x^{2}-12 x-14=\left(-14 x^{4}-7 x^{2}+49\right) \cdot\left(-\frac{x^{2}}{2}-\frac{2}{7}\right)+4 x^{3}+2 x-14
$$

Next, we need to divide the divisor by the remainder again. This time, the remainder from the division turns out to be zero:

$$
-14 x^{4}-7 x^{2}+49=-\frac{7 x}{2} \cdot\left(4 x^{3}+2 x-14\right)
$$

This means that the polynomial $4 x^{3}+2 x-14$ is the greatest common divisor we are looking for.

The Interregional Olympiad for Schoolchildren based on Departmental Educational Organizations in Mathematics

of the numerator and the denominator of the original fraction, and it can be "factored out" (to avoid fractional coefficients, it will be more convenient to use the polynomial $2 x^{3}+x-7$):

Thus,

$$
\frac{2 x^{6}+5 x^{4}-3 x^{3}+2 x^{2}-12 x-14}{4 x^{6}-4 x^{4}-6 x^{3}-3 x^{2}+25 x-28}=\frac{\left(2 x^{3}+x-7\right) \cdot\left(x^{3}+2 x+2\right)}{\left(2 x^{3}+x-7\right) \cdot\left(2 x^{3}-3 x+4\right)}
$$

Answer: For example, $\frac{x^{3}+2 x+2}{2 x^{3}-3 x+4}$.