Use polynomial division or synthetic division to divide f(x) by (x+1).

Okay, LA, let's tackle question 1. 1. (i) Given that (x+1) is a factor of f(x), where f(x) = x^3 + 6x^2 + 11x + 6, factorise f(x) completely. Step 1: Use polynomial division or synthetic division to divide f(x) by (x+1). Using synthetic division with root -1: ` -1 | 1 6 11 6 | -1 -5 -6 ---------------- 1 5 6 0 ` The quotient is x^2 + 5x + 6. Step 2: Factor the resulting quadratic expression. We need two numbers that multiply to 6 and add to 5. These are 2 and 3. x^2 + 5x + 6 = (x+2)(x+3) Step 3: Write the complete factorization of f(x). f(x) = (x+1)(x+2)(x+3) The final answer is (x+1)(x+2)(x+3). 1. (ii) Let be a real constant. Show that the roots of the quadratic equation 3x^2 + (-4-2)x = 0 are always real. Step 1: Identify the coefficients a, b, and c from the quadratic equation ax^2+bx+c=0. For 3x^2 + (-4-2)x + 0 = 0: a = 3 b = -4-2 c = 0 Step 2: Calculate the discriminant = b^2 - 4ac. = (-4-2)^2 - 4(3)(0) = (-4-2)^2 - 0 = (-4-2)^2 Step 3: Determine if the roots are always real. For roots to be real, the discriminant must be greater than or equal to zero ( 0). Since (-4-2)^2 is a square of a real number, it will always be greater than or equal to zero for any real value of . (-4-2)^2 0 Therefore, the roots of the quadratic equation are always real. That's 2 down. 3 left today — send the next one.