1. The problem is to verify the correctness of the polynomial long division shown. 2. The division is: $$\frac{15x^4 - x^3 + 7x^2 + 0x + 5}{3x^2 + 1}$$. 3. The quotient and remainder are found by dividing step-by-step: - Divide the leading term $15x^4$ by $3x^2$ to get $5x^2$. - Multiply $5x^2$ by the divisor $3x^2 + 1$ to get $15x^4 + 5x^2$. - Subtract this from the dividend: $(15x^4 - x^3 + 7x^2) - (15x^4 + 5x^2) = -x^3 + 2x^2$. 4. Next, divide $-x^3$ by $3x^2$ to get $-\frac{1}{3}x$. - Multiply $-\frac{1}{3}x$ by $3x^2 + 1$ to get $-x^3 - \frac{1}{3}x$. - Subtract: $(-x^3 + 2x^2) - (-x^3 - \frac{1}{3}x) = 2x^2 + \frac{1}{3}x$. 5. Divide $2x^2$ by $3x^2$ to get $\frac{2}{3}$. - Multiply $\frac{2}{3}$ by $3x^2 + 1$ to get $2x^2 + \frac{2}{3}$. - Subtract: $(2x^2 + \frac{1}{3}x) - (2x^2 + \frac{2}{3}) = \frac{1}{3}x - \frac{2}{3}$. 6. Since the degree of the remainder $\frac{1}{3}x - \frac{2}{3}$ is less than the divisor degree 2, the division ends here. 7. The quotient is $$5x^2 - \frac{1}{3}x + \frac{2}{3}$$ and the remainder is $$\frac{1}{3}x - \frac{2}{3}$$. 8. Comparing with the user's work, their quotient terms and remainder differ, indicating the division is incorrect. Final answer: The polynomial division shown is not correct.