What are the solutions to these polynomial division problems?

Welcome back atuahenegodlove21 — been a while! Let's pick up where you left off. Here are the solutions to your polynomial division problems: 1. Divide x^3 - 5x^2 + x + 42 by (x-7) Step 1: Set up synthetic division. The divisor is (x-7), so we use k=7. The coefficients of the dividend are 1, -5, 1, 42. c|cccc 7 & 1 & -5 & 1 & 42 \\ & & 7 & 14 & 105 \\ & 1 & 2 & 15 & 147 \\ Step 2: Interpret the result. The last number, 147, is the remainder. The other numbers, 1, 2, 15, are the coefficients of the quotient, starting with x^2. The quotient is x^2 + 2x + 15 with a remainder of 147. So, (x^3 - 5x^2 + x + 42)/(x-7) = x^2 + 2x + 15 + (147)/(x-7). The result of the division is x^2 + 2x + 15 with a remainder of 147. 2. Divide 2x^3 - 15x^2 - 2x + 120 by (x-6) and (x-4) This means dividing by (x-6)(x-4). We can perform two successive synthetic divisions. Step 1: Divide 2x^3 - 15x^2 - 2x + 120 by (x-6). Use k=6. c|cccc 6 & 2 & -15 & -2 & 120 \\ & & 12 & -18 & -120 \\ & 2 & -3 & -20 & 0 \\ The remainder is 0. The quotient is 2x^2 - 3x - 20. Step 2: Divide the new quotient 2x^2 - 3x - 20 by (x-4). Use k=4. c|ccc 4 & 2 & -3 & -20 \\ & & 8 & 20 \\ & 2 & 5 & 0 \\ The remainder is 0. The final quotient is 2x + 5. The result of the division is 2x + 5. 3. Divide 6x^4 + 13x^3 - 36x^2 - 43x + 30 by (x-2) Step 1: Set up synthetic division. The divisor is (x-2), so we use k=2. The coefficients of the dividend are 6, 13, -36, -43, 30. c|ccccc 2 & 6 & 13 & -36 & -43 & 30 \\ & & 12 & 50 & 28 & -30 \\ & 6 & 25 & 14 & -15 & 0 \\ Step 2: Interpret the result. The last number, 0, is the remainder. The other numbers, 6, 25, 14, -15, are the coefficients of the quotient, starting with x^3. The quotient is 6x^3 + 25x^2 + 14x - 15 with a remainder of 0. So, (6x^4 + 13x^3 - 36x^2 - 43x + 30)/(x-2) = 6x^3 + 25x^2 + 14x - 15. The result of the division is 6x^3 + 25x^2 + 14x - 15. Send me the next one 📸