1. **State the problem:** Divide the polynomial $x^4 - 2x^3 - 4x + 5$ by $3x^3 + 2$. 2. **Set up the division:** We want to find the quotient $Q(x)$ and remainder $R(x)$ such that: $$x^4 - 2x^3 - 4x + 5 = (3x^3 + 2)Q(x) + R(x)$$ where the degree of $R(x)$ is less than the degree of $3x^3 + 2$ (which is 3). 3. **Divide the leading terms:** Divide $x^4$ by $3x^3$: $$\frac{x^4}{3x^3} = \frac{1}{3}x$$ This is the first term of the quotient $Q(x)$. 4. **Multiply and subtract:** Multiply $3x^3 + 2$ by $\frac{1}{3}x$: $$\frac{1}{3}x \times (3x^3 + 2) = x^4 + \frac{2}{3}x$$ Subtract this from the dividend: $$\left(x^4 - 2x^3 - 4x + 5\right) - \left(x^4 + \frac{2}{3}x\right) = -2x^3 - 4x + 5 - \frac{2}{3}x = -2x^3 - \left(4 + \frac{2}{3}\right)x + 5 = -2x^3 - \frac{14}{3}x + 5$$ 5. **Repeat division with new polynomial:** Divide the leading term $-2x^3$ by $3x^3$: $$\frac{-2x^3}{3x^3} = -\frac{2}{3}$$ This is the next term of the quotient. 6. **Multiply and subtract again:** Multiply $3x^3 + 2$ by $-\frac{2}{3}$: $$-\frac{2}{3} \times (3x^3 + 2) = -2x^3 - \frac{4}{3}$$ Subtract this from the current polynomial: $$\left(-2x^3 - \frac{14}{3}x + 5\right) - \left(-2x^3 - \frac{4}{3}\right) = -2x^3 - \frac{14}{3}x + 5 + 2x^3 + \frac{4}{3} = -\frac{14}{3}x + \left(5 + \frac{4}{3}\right) = -\frac{14}{3}x + \frac{19}{3}$$ 7. **Check remainder degree:** The remainder is $-\frac{14}{3}x + \frac{19}{3}$, which is degree 1, less than degree 3 of divisor, so division stops. 8. **Write final answer:** $$Q(x) = \frac{1}{3}x - \frac{2}{3}, \quad R(x) = -\frac{14}{3}x + \frac{19}{3}$$ **Therefore,** $$x^4 - 2x^3 - 4x + 5 = (3x^3 + 2)\left(\frac{1}{3}x - \frac{2}{3}\right) + \left(-\frac{14}{3}x + \frac{19}{3}\right)$$