Find the quotient and remainder using long division. \( \frac{3 x^{3}+5 x^{2}-24 x+16}{3 x-4} \) The quotient is The remainder is

To find the quotient and remainder using long division for the expression \( \frac{3 x^{3}+5 x^{2}-24 x+16}{3 x-4} \), follow the long division steps: 1. Divide the leading term of the numerator \(3x^3\) by the leading term of the denominator \(3x\), which gives \(x^2\). 2. Multiply \(x^2\) by \(3x - 4\) to get \(3x^3 - 4x^2\). 3. Subtract this from the original numerator: \((3x^3 + 5x^2 - 24x + 16) - (3x^3 - 4x^2) = 9x^2 - 24x + 16\). 4. Now, divide the leading term \(9x^2\) by \(3x\), which gives \(3x\). 5. Multiply \(3x\) by \(3x - 4\) to get \(9x^2 - 12x\). 6. Subtract this from what you have: \((9x^2 - 24x + 16) - (9x^2 - 12x) = -12x + 16\). 7. Divide \(-12x\) by \(3x\) which gives \(-4\). 8. Multiply \(-4\) by \(3x - 4\) to get \(-12x + 16\). 9. Subtract this from what you have: \((-12x + 16) - (-12x + 16) = 0\). At the end of the long division, there's no remainder. Thus, the quotient is \(x^2 + 3x - 4\) and the remainder is \(0\).