Divide. \[ \left(12 x^{3}+16 x^{2}+20 x+13\right) \div(4 x+4) \] Your answer should give the guotient and the remainder.

To divide \( 12x^{3} + 16x^{2} + 20x + 13 \) by \( 4x + 4 \), we can use polynomial long division. 1. First, simplify the divisor: \[ 4x + 4 = 4(x + 1) \] This gives us: \[ \frac{12x^3 + 16x^2 + 20x + 13}{4(x + 1)} \] We can factor out the 4 from the entire polynomial: \[ 3x^3 + 4x^2 + 5x + \frac{13}{4} \] 2. Now, perform long division: - Divide the leading term \( 3x^3 \) by \( x \) to get \( 3x^2 \). - Multiply \( 3x^2 \) by \( 4(x + 1) \): \[ 3x^2 \cdot (4x + 4) = 12x^3 + 12x^2 \] - Subtract from the original polynomial: \[ (12x^3 + 16x^2 + 20x + 13) - (12x^3 + 12x^2) = 4x^2 + 20x + 13 \] 3. Repeat: - Divide the leading term \( 4x^2 \) by \( x \) to get \( 4x \). - Multiply \( 4x \) by \( 4(x + 1) \): \[ 4x \cdot (4x + 4) = 16x^2 + 16x \] - Subtract: \[ (4x^2 + 20x + 13) - (16x^2 + 16x) = 4x + 13 \] 4. Repeat one more time: - Divide \( 4x \) by \( x \) to get \( 4 \). - Multiply \( 4 \) by \( 4(x + 1) \): \[ 4 \cdot (4x + 4) = 16x + 16 \] - Subtract: \[ (4x + 13) - (16x + 16) = -12x - 3 \] Now, the process is complete since the degree of the remainder (\(-12x - 3\)) is less than the degree of the divisor (\(4x + 4\)). So, the final answer is: \[ 3x^2 + 4x + 4 \quad \text{and the remainder} \quad -12x - 3. \]