1. **State the problem:** Divide the polynomial $3x^{2}+x^{5}+x^{3}+47-28x$ by $x^{3}-1$. 2. **Rewrite the dividend in standard form:** Arrange terms in descending powers of $x$: $$x^{5} + x^{3} + 3x^{2} - 28x + 47$$ 3. **Recall the division formula:** For polynomials $P(x)$ and $D(x)$, dividing $P(x)$ by $D(x)$ gives quotient $Q(x)$ and remainder $R(x)$ such that: $$P(x) = Q(x) \cdot D(x) + R(x)$$ where degree of $R(x)$ is less than degree of $D(x)$. 4. **Start polynomial long division:** - Divide leading term $x^{5}$ by $x^{3}$ to get $x^{2}$. - Multiply divisor by $x^{2}$: $x^{2}(x^{3}-1) = x^{5} - x^{2}$. - Subtract this from dividend: $$\left(x^{5} + x^{3} + 3x^{2} - 28x + 47\right) - \left(x^{5} - x^{2}\right) = x^{3} + 4x^{2} - 28x + 47$$ 5. **Next step:** - Divide leading term $x^{3}$ by $x^{3}$ to get $1$. - Multiply divisor by $1$: $x^{3} - 1$. - Subtract: $$\left(x^{3} + 4x^{2} - 28x + 47\right) - \left(x^{3} - 1\right) = 4x^{2} - 28x + 48$$ 6. **Remainder check:** Degree of remainder $4x^{2} - 28x + 48$ is 2, which is less than degree 3 of divisor, so division ends here. 7. **Final answer:** $$\frac{3x^{2}+x^{5}+x^{3}+47-28x}{x^{3}-1} = x^{2} + 1 + \frac{4x^{2} - 28x + 48}{x^{3} - 1}$$