1. **State the problem:** Divide the polynomial $6x^{3}+x^{2}-x+1$ by the binomial $4x-1$. 2. **Recall the division formula:** 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 this quotient, subtract, and repeat with the remainder. 3. **Divide the leading terms:** $\frac{6x^{3}}{4x} = \frac{6}{4}x^{2} = \frac{3}{2}x^{2}$. 4. **Multiply divisor by $\frac{3}{2}x^{2}$:** $$\left(4x - 1\right) \times \frac{3}{2}x^{2} = 4x \times \frac{3}{2}x^{2} - 1 \times \frac{3}{2}x^{2} = 6x^{3} - \frac{3}{2}x^{2}$$ 5. **Subtract this from the original polynomial:** $$\left(6x^{3} + x^{2} - x + 1\right) - \left(6x^{3} - \frac{3}{2}x^{2}\right) = \cancel{6x^{3}} + x^{2} - x + 1 - \cancel{6x^{3}} + \frac{3}{2}x^{2} = \left(x^{2} + \frac{3}{2}x^{2}\right) - x + 1 = \frac{5}{2}x^{2} - x + 1$$ 6. **Repeat the division with the new polynomial:** Divide leading term $\frac{5}{2}x^{2}$ by $4x$: $$\frac{\frac{5}{2}x^{2}}{4x} = \frac{5}{2} \times \frac{1}{4} x = \frac{5}{8}x$$ 7. **Multiply divisor by $\frac{5}{8}x$:** $$\left(4x - 1\right) \times \frac{5}{8}x = 4x \times \frac{5}{8}x - 1 \times \frac{5}{8}x = \frac{20}{8}x^{2} - \frac{5}{8}x = \frac{5}{2}x^{2} - \frac{5}{8}x$$ 8. **Subtract this from the current polynomial:** $$\left(\frac{5}{2}x^{2} - x + 1\right) - \left(\frac{5}{2}x^{2} - \frac{5}{8}x\right) = \cancel{\frac{5}{2}x^{2}} - x + 1 - \cancel{\frac{5}{2}x^{2}} + \frac{5}{8}x = \left(-x + \frac{5}{8}x\right) + 1 = -\frac{3}{8}x + 1$$ 9. **Divide leading term $-\frac{3}{8}x$ by $4x$:** $$\frac{-\frac{3}{8}x}{4x} = -\frac{3}{8} \times \frac{1}{4} = -\frac{3}{32}$$ 10. **Multiply divisor by $-\frac{3}{32}$:** $$\left(4x - 1\right) \times -\frac{3}{32} = 4x \times -\frac{3}{32} - 1 \times -\frac{3}{32} = -\frac{12}{32}x + \frac{3}{32} = -\frac{3}{8}x + \frac{3}{32}$$ 11. **Subtract this from the current polynomial:** $$\left(-\frac{3}{8}x + 1\right) - \left(-\frac{3}{8}x + \frac{3}{32}\right) = \cancel{-\frac{3}{8}x} + 1 - \cancel{-\frac{3}{8}x} - \frac{3}{32} = 1 - \frac{3}{32} = \frac{32}{32} - \frac{3}{32} = \frac{29}{32}$$ 12. **Final result:** Quotient is $$\frac{3}{2}x^{2} + \frac{5}{8}x - \frac{3}{32}$$ and remainder is $$\frac{29}{32}$$ So, $$6x^{3} + x^{2} - x + 1 = \left(4x - 1\right) \left(\frac{3}{2}x^{2} + \frac{5}{8}x - \frac{3}{32}\right) + \frac{29}{32}$$