Pls solve in English not French

Step 1: Use the Remainder Theorem to form an equation for a and b. Given that when P(x) is divided by (x-1), the remainder is -6, we have P(1) = -6. P(1) = a(1)^3 - 3(1)^2 + b(1) + 6 = -6 a - 3 + b + 6 = -6 a + b + 3 = -6 a + b = -9 (*) Step 2: Use the Factor Theorem to form a second equation for a and b. Given that (x+2) is a factor of P(x), we have P(-2) = 0. P(-2) = a(-2)^3 - 3(-2)^2 + b(-2) + 6 = 0 -8a - 3(4) - 2b + 6 = 0 -8a - 12 - 2b + 6 = 0 -8a - 2b - 6 = 0 -8a - 2b = 6 Dividing by -2: 4a + b = -3 (**) Step 3: Solve the system of equations to find a and b. Subtract equation () from equation (*): (4a + b) - (a + b) = -3 - (-9) 3a = 6 a = 2 Substitute a=2 into equation (*): 2 + b = -9 b = -11 So, the polynomial is P(x) = 2x^3 - 3x^2 - 11x + 6. The values are a=2, b=-11. Step 4: Solve the equation P(x) = 0. Since (x+2) is a factor, x=-2 is a root. We can use synthetic division to find the quadratic factor. ` -2 | 2 -3 -11 6 | -4 14 -6 ------------------ 2 -7 3 0 ` The quotient is 2x^2 - 7x + 3. So, P(x) = (x+2)(2x^2 - 7x + 3) = 0. Now, we solve the quadratic equation 2x^2 - 7x + 3 = 0. We can factor this quadratic: 2x^2 - 6x - x + 3 = 0 2x(x - 3) - 1(x - 3) = 0 (2x - 1)(x - 3) = 0 This gives us two more roots: 2x - 1 = 0 x = (1)/(2) x - 3 = 0 x = 3 The solutions to P(x)=0 are x = -2, x = (1)/(2), and x = 3. The solutions are x = -2, x = (1)/(2), x = 3. Send me the next one 📸