1. **State the problem:** We need to perform polynomial division of $$4x^4 - 5x^3 + x^2$$ by $$3x^2 - 4x + 1$$. 2. **Recall the division process:** Polynomial division is similar to long division with numbers. We divide the leading term of the dividend by the leading term of the divisor, multiply the divisor by that result, subtract, and repeat with the remainder. 3. **Divide the leading terms:** $$\frac{4x^4}{3x^2} = \frac{4}{3}x^2$$. 4. **Multiply divisor by $$\frac{4}{3}x^2$$:** $$\frac{4}{3}x^2 \times (3x^2 - 4x + 1) = 4x^4 - \frac{16}{3}x^3 + \frac{4}{3}x^2$$. 5. **Subtract this from the original polynomial:** $$\left(4x^4 - 5x^3 + x^2\right) - \left(4x^4 - \frac{16}{3}x^3 + \frac{4}{3}x^2\right) = 0x^4 - \left(-5x^3 + \frac{16}{3}x^3\right) + \left(x^2 - \frac{4}{3}x^2\right)$$ Simplify: $$-5x^3 + \frac{16}{3}x^3 = -\frac{15}{3}x^3 + \frac{16}{3}x^3 = \frac{1}{3}x^3$$ $$x^2 - \frac{4}{3}x^2 = \frac{3}{3}x^2 - \frac{4}{3}x^2 = -\frac{1}{3}x^2$$ So remainder is: $$\frac{1}{3}x^3 - \frac{1}{3}x^2$$. 6. **Bring down the remainder and repeat:** Divide leading term of remainder by leading term of divisor: $$\frac{\frac{1}{3}x^3}{3x^2} = \frac{1}{9}x$$. 7. **Multiply divisor by $$\frac{1}{9}x$$:** $$\frac{1}{9}x \times (3x^2 - 4x + 1) = \frac{1}{3}x^3 - \frac{4}{9}x^2 + \frac{1}{9}x$$. 8. **Subtract this from the remainder:** $$\left(\frac{1}{3}x^3 - \frac{1}{3}x^2\right) - \left(\frac{1}{3}x^3 - \frac{4}{9}x^2 + \frac{1}{9}x\right) = 0x^3 + \left(-\frac{1}{3}x^2 + \frac{4}{9}x^2\right) - \frac{1}{9}x$$ Simplify: $$-\frac{1}{3}x^2 + \frac{4}{9}x^2 = -\frac{3}{9}x^2 + \frac{4}{9}x^2 = \frac{1}{9}x^2$$ So new remainder: $$\frac{1}{9}x^2 - \frac{1}{9}x$$. 9. **Divide leading term of remainder by leading term of divisor:** $$\frac{\frac{1}{9}x^2}{3x^2} = \frac{1}{27}$$. 10. **Multiply divisor by $$\frac{1}{27}$$:** $$\frac{1}{27} \times (3x^2 - 4x + 1) = \frac{1}{9}x^2 - \frac{4}{27}x + \frac{1}{27}$$. 11. **Subtract this from remainder:** $$\left(\frac{1}{9}x^2 - \frac{1}{9}x\right) - \left(\frac{1}{9}x^2 - \frac{4}{27}x + \frac{1}{27}\right) = 0x^2 + \left(-\frac{1}{9}x + \frac{4}{27}x\right) - \frac{1}{27}$$ Simplify: $$-\frac{1}{9}x + \frac{4}{27}x = -\frac{3}{27}x + \frac{4}{27}x = \frac{1}{27}x$$ So remainder is: $$\frac{1}{27}x - \frac{1}{27}$$. 12. **Since degree of remainder (1) is less than degree of divisor (2), division ends here.** **Final answer:** $$\frac{4x^4 - 5x^3 + x^2}{3x^2 - 4x + 1} = \frac{4}{3}x^2 + \frac{1}{9}x + \frac{1}{27} + \frac{\frac{1}{27}x - \frac{1}{27}}{3x^2 - 4x + 1}$$