1. **State the problem:** We need to divide the polynomial $9x^3 + 18x^2 - 4x + 8$ by the binomial $3x + 2$ and express the result in the form of a quotient plus a remainder over the divisor. 2. **Recall the division formula:** For polynomials, dividing $P(x)$ by $D(x)$ gives $$P(x) = Q(x) \cdot D(x) + R(x)$$ where $Q(x)$ is the quotient and $R(x)$ is the remainder with degree less than $D(x)$. 3. **Perform polynomial long division:** - Divide the leading term $9x^3$ by $3x$ to get $3x^2$. - Multiply $3x^2$ by $3x + 2$ to get $9x^3 + 6x^2$. - Subtract this from the original polynomial: $$9x^3 + 18x^2 - 4x + 8 - (9x^3 + 6x^2) = 12x^2 - 4x + 8$$ 4. **Continue division:** - Divide $12x^2$ by $3x$ to get $4x$. - Multiply $4x$ by $3x + 2$ to get $12x^2 + 8x$. - Subtract: $$12x^2 - 4x + 8 - (12x^2 + 8x) = -12x + 8$$ 5. **Continue division:** - Divide $-12x$ by $3x$ to get $-4$. - Multiply $-4$ by $3x + 2$ to get $-12x - 8$. - Subtract: $$-12x + 8 - (-12x - 8) = 16$$ 6. **Write the final result:** $$\frac{9x^3 + 18x^2 - 4x + 8}{3x + 2} = 3x^2 + 4x - 4 + \frac{16}{3x + 2}$$ **Answer:** $3x^2 + 4x - 4 + \frac{16}{3x + 2}$