To divide \(6x^4 + x^3 - x^2 + 5x - 6\) by \(2x^2 - x + 1\), we perform polynomial long division.

1. Divide the leading term \(6x^4\) by \(2x^2\) to get \(3x^2\).

2. Multiply \(3x^2\) by \(2x^2 - x + 1\) to get \(6x^4 - 3x^3 + 3x^2\).

3. Subtract \(6x^4 - 3x^3 + 3x^2\) from \(6x^4 + x^3 - x^2 + 5x - 6\) to get \(4x^3 - 4x^2 + 5x - 6\).

4. Divide \(4x^3\) by \(2x^2\) to get \(2x\).

5. Multiply \(2x\) by \(2x^2 - x + 1\) to get \(4x^3 - 2x^2 + 2x\).

6. Subtract \(4x^3 - 2x^2 + 2x\) from \(4x^3 - 4x^2 + 5x - 6\) to get \(-2x^2 + 3x - 6\).

7. Divide \(-2x^2\) by \(2x^2\) to get \(-1\).

8. Multiply \(-1\) by \(2x^2 - x + 1\) to get \(-2x^2 + x - 1\).

9. Subtract \(-2x^2 + x - 1\) from \(-2x^2 + 3x - 6\) to get \(2x - 5\).

The quotient is \(3x^2 + 2x - 1\) and the remainder is \(2x - 5\).