293. Find all integers $k$ for which the value of the fraction

$$
\frac{2 k^{2}+k-8}{k-1}
$$

is an integer.
$\triangle$ To isolate the integer part of the fraction, we will divide its numerator by the denominator using the "corner" method, similar to dividing a multi-digit natural number by another natural number. We get:

![](https://cdn.mathpix.com/cropped/2024_05_21_475fdfbb473422773b9eg-063.jpg?height=139&width=368&top_left_y=1828&top_left_x=848)

$$
\begin{aligned}
& \frac{-3 k-3}{-5}
\end{aligned}
$$

This notation should be understood as follows: we divide the leading term of the dividend by the leading term of the divisor; the resulting quotient $2 k$ is multiplied by the divisor, and the product

$$
2 k(k-1)=2 k^{2}-2 k
$$

is written under the dividend; we subtract this product from the dividend and divide the leading term of the resulting difference $3 k-8$ by the leading term of the divisor, and so on.

The quotient $2 k+3$ is the integer part of the given fraction. Therefore,

$$
\frac{2 k^{2}+k-8}{k-1}=2 k+3-\frac{5}{k-1}
$$

Now, the usual approach: we list all the divisors of the number 5.

Answer: $6, 2, 0, -4$.