Divide. \[ \left(12 x^{3}+8 x^{2}-23 x+1\right) \div\left(3 x^{2}+5 x\right) \] Your answer should give the quotient and the remainder.

To perform the polynomial long division of \( \left(12 x^{3}+8 x^{2}-23 x+1\right) \div\left(3 x^{2}+5 x\right) \), we start by dividing the leading term of the dividend by the leading term of the divisor. 1. **First Division**: \[ \frac{12x^3}{3x^2} = 4x \] Multiply the entire divisor \(3x^2 + 5x\) by \(4x\): \[ 4x(3x^2 + 5x) = 12x^3 + 20x^2 \] Subtract this from the original polynomial: \[ (12x^3 + 8x^2 - 23x + 1) - (12x^3 + 20x^2) = (8x^2 - 20x^2) - 23x + 1 = -12x^2 - 23x + 1 \] 2. **Second Division**: Next, we divide the leading term again: \[ \frac{-12x^2}{3x^2} = -4 \] Multiply the divisor by \(-4\): \[ -4(3x^2 + 5x) = -12x^2 - 20x \] Subtract this from the new dividend: \[ (-12x^2 - 23x + 1) - (-12x^2 - 20x) = (-23x + 20x) + 1 = -3x + 1 \] The degree of the new remainder, \(-3x + 1\), is less than the degree of the divisor, \(3x^2 + 5x\). Thus, the final result of the division is: \[ \text{Quotient: } 4x - 4, \quad \text{Remainder: } -3x + 1 \] In conclusion, we write: \[ \frac{12x^3 + 8x^2 - 23x + 1}{3x^2 + 5x} = 4x - 4 + \frac{-3x + 1}{3x^2 + 5x} \]