1. **State the problem:** Perform corner division (polynomial division) of the polynomial $3x^5 - x^4 - 3x + 1$ by the polynomial $x^2 - 5x^2 + 6x$. 2. **Simplify the divisor:** Note that $x^2 - 5x^2 + 6x = (1 - 5)x^2 + 6x = -4x^2 + 6x$. 3. **Rewrite the division:** We want to divide $$3x^5 - x^4 - 3x + 1$$ by $$-4x^2 + 6x$$. 4. **Set up the division:** The dividend is $3x^5 - x^4 + 0x^3 + 0x^2 - 3x + 1$ (fill missing powers with zero coefficients). 5. **Divide the leading terms:** $$\frac{3x^5}{-4x^2} = -\frac{3}{4}x^3$$. 6. **Multiply divisor by this term:** $$-\frac{3}{4}x^3 \times (-4x^2 + 6x) = 3x^5 - \frac{9}{2}x^4$$. 7. **Subtract:** $$\begin{aligned} &(3x^5 - x^4 + 0x^3 + 0x^2 - 3x + 1) - (3x^5 - \frac{9}{2}x^4) \\ &= 0x^5 + \left(-1 + \frac{9}{2}\right)x^4 + 0x^3 + 0x^2 - 3x + 1 \\ &= \frac{7}{2}x^4 + 0x^3 + 0x^2 - 3x + 1 \end{aligned}$$ 8. **Divide leading terms again:** $$\frac{\frac{7}{2}x^4}{-4x^2} = -\frac{7}{8}x^2$$. 9. **Multiply divisor by this term:** $$-\frac{7}{8}x^2 \times (-4x^2 + 6x) = \frac{7}{2}x^4 - \frac{21}{4}x^3$$. 10. **Subtract:** $$\begin{aligned} &(\frac{7}{2}x^4 + 0x^3 + 0x^2 - 3x + 1) - (\frac{7}{2}x^4 - \frac{21}{4}x^3) \\ &= 0x^4 + \frac{21}{4}x^3 + 0x^2 - 3x + 1 \end{aligned}$$ 11. **Divide leading terms:** $$\frac{\frac{21}{4}x^3}{-4x^2} = -\frac{21}{16}x$$. 12. **Multiply divisor by this term:** $$-\frac{21}{16}x \times (-4x^2 + 6x) = \frac{21}{4}x^3 - \frac{63}{16}x^2$$. 13. **Subtract:** $$\begin{aligned} &(\frac{21}{4}x^3 + 0x^2 - 3x + 1) - (\frac{21}{4}x^3 - \frac{63}{16}x^2) \\ &= 0x^3 + \frac{63}{16}x^2 - 3x + 1 \end{aligned}$$ 14. **Divide leading terms:** $$\frac{\frac{63}{16}x^2}{-4x^2} = -\frac{63}{64}$$. 15. **Multiply divisor by this term:** $$-\frac{63}{64} \times (-4x^2 + 6x) = \frac{63}{16}x^2 - \frac{189}{64}x$$. 16. **Subtract:** $$\begin{aligned} &(\frac{63}{16}x^2 - 3x + 1) - (\frac{63}{16}x^2 - \frac{189}{64}x) \\ &= 0x^2 + \left(-3 + \frac{189}{64}\right)x + 1 \\ &= \left(-\frac{192}{64} + \frac{189}{64}\right)x + 1 = -\frac{3}{64}x + 1 \end{aligned}$$ 17. **Since the degree of the remainder $-\frac{3}{64}x + 1$ is less than the degree of the divisor $-4x^2 + 6x$, the division stops here.** 18. **Final answer:** $$\frac{3x^5 - x^4 - 3x + 1}{-4x^2 + 6x} = -\frac{3}{4}x^3 - \frac{7}{8}x^2 - \frac{21}{16}x - \frac{63}{64} + \frac{-\frac{3}{64}x + 1}{-4x^2 + 6x}$$