(d) \( \left(x^{2}+4 x+4\right) \div(x+2) \) (c) \( \left(15 x^{2}+5 x y-4 y^{2}\right) \div(3 x+2 y) \) (f) \( \left(35 a^{2}-a b-12 b^{2}\right) \div(5 a-3 b) \) (g) \( \left(x^{3}-27 y^{3}\right) \div(x-3 y) \) (h) \( \left(8 x^{3}+27 y^{3}\right) \div(2 x+3 y) \)

Divide the polynomials by following steps: - step0: Using Long Division to Divide Polynomials: \(\left(x^{2}+4x+4\right)\div \left(x+2\right)\) - step1: Calculate: \(\left(x+2\right)\text{ }\textrm{remainder}\text{ }0\) Let's go through the polynomial long division problems step by step. ### (d) \( \left(x^{2}+4 x+4\right) \div(x+2) \) 1. **Identify the dividend and divisor**: - Dividend: \( x^{2} + 4x + 4 \) - Divisor: \( x + 2 \) 2. **Perform the division**: - The first term of the dividend \( x^{2} \) divided by the first term of the divisor \( x \) gives \( x \). - Multiply \( x \) by the divisor \( (x + 2) \) to get \( x^{2} + 2x \). - Subtract \( (x^{2} + 2x) \) from \( (x^{2} + 4x + 4) \): \[ (x^{2} + 4x + 4) - (x^{2} + 2x) = 2x + 4 \] - Now, divide \( 2x \) by \( x \) to get \( 2 \). - Multiply \( 2 \) by the divisor \( (x + 2) \) to get \( 2x + 4 \). - Subtract \( (2x + 4) \) from \( (2x + 4) \): \[ (2x + 4) - (2x + 4) = 0 \] 3. **Conclusion**: - The result of the division is \( x + 2 \) with a remainder of \( 0 \). - Final result: \( (x + 2) \) remainder \( 0 \). ### (c) \( \left(15 x^{2}+5 x y-4 y^{2}\right) \div(3 x+2 y) \) I encountered an error while trying to compute this division. Let's analyze it manually: 1. **Identify the dividend and divisor**: - Dividend: \( 15x^{2} + 5xy - 4y^{2} \) - Divisor: \( 3x + 2y \) 2. **Perform the division**: - The first term of the dividend \( 15x^{2} \) divided by the first term of the divisor \( 3x \) gives \( 5x \). - Multiply \( 5x \) by the divisor \( (3x + 2y) \) to get \( 15x^{2} + 10xy \). - Subtract \( (15x^{2} + 10xy) \) from \( (15x^{2} + 5xy - 4y^{2}) \): \[ (15x^{2} + 5xy - 4y^{2}) - (15x^{2} + 10xy) = -5xy - 4y^{2} \] - Now, divide \( -5xy \) by \( 3x \) to get \( -\frac{5}{3}y \). - Multiply \( -\frac{5}{3}y \) by the divisor \( (3x + 2y) \) to get \( -5xy - \frac{10}{3}y^{2} \). - Subtract \( (-5xy - \frac{10}{3}y^{2}) \) from \( (-5xy - 4y^{2}) \): \[ (-5xy - 4y^{2}) - (-5xy - \frac{10}{3}y^{2}) = -4y^{2} + \frac{10}{3}y^{2} = -\frac{2}{3}y^{2} \] 3. **Conclusion**: - The result of the division is \( 5x - \frac{5}{3}y \) with a remainder of \( -\frac{2}{3}y^{2} \). ### (f) \( \left(35 a^{2}-a b-12 b^{2}\right) \div(5 a-3 b) \) I encountered an error while trying to compute this division. Let's analyze it manually: 1. **Identify the dividend and divisor**: - Dividend: \( 35a^{2} - ab - 12b^{2} \) - Divisor: \( 5a - 3b \) 2. **Perform the division**: - The first term of the dividend \( 35a^{2} \) divided by the first term of the divisor \( 5a \) gives \( 7a \). - Multiply \( 7a \) by the divisor \( (5a - 3b) \) to get \( 35a^{2} - 21ab \). - Subtract \( (35a^{2} - 21ab) \) from \( (35a^{2} - ab - 12b^{2}) \): \[ (35a^{2} - ab - 12b^{2}) - (35a^{2} - 21ab) = 20ab - 12b^{2} \] - Now, divide \( 20ab \) by \( 5a \) to get \( 4b \). - Multiply \( 4b \) by the divisor \( (5a - 3b) \) to get \( 20ab - 12b^{2} \). - Subtract \( (20ab - 12b^{2}) \) from \( (20ab - 12b^{2}) \): \[ (20ab - 12b^{2}) - (20ab - 12b^{2}) = 0 \] 3. **Conclusion**: - The result of the division is \( 7a + 4b \) with a remainder of \( 0 \). ### (g) \( \left(x^{3}-27 y^{3}\right) \div(x-3 y) \) I encountered an error while trying to compute this division. Let's analyze it manually: 1. **Identify the dividend and divisor**: - Dividend: \( x^{3} - 27y^{3} \) - Divisor: \( x - 3y \) 2. **Perform the division**: - The expression \( x^{3} - 27y^{3} \) can be factored as \( (x - 3y)(x^{2} + 3xy + 9y^{2}) \). - Therefore, dividing \( x^{3} - 27y^{3} \) by \( x - 3y \) gives \( x^{2} + 3xy + 9y^{2} \). 3. **Conclusion**: - The result of