How do you factorise a quadratic with a co-efficient in front of the x^2 - e.g: 3x^2 + 14x + 8

Ok so to factorise a quadratic with a co-efficient in front of the x^2 you need to follow a couple of steps: Step 1 - multiple the co-efficient by the number at the end of the expression - e.g. 3^8 = 24 Step 2 - we now need to find two numbers that multiply to make the number we generated (24) and add to make the number in front of the x (14) - e.g. in this example the numbers would be 2, 12 (2^12 = 24, 2+12 = 14) Step 3 - Break up the x into those 2 numbers - e.g. 3x^2 + 12x + 2x + 8 Step 4 - Factorise each half of the expression - e.g.: 3x^2 + 12x => 3x(x + 4)         +2x + 8 => +2(x + 4) NOTE: In each half of the expression once you have factorised there should be an element in both halves that is the same. In this case (x + 4) is in both halves. Step 5 - Put both halves back together again - e.g. (3x + 2)(x + 4)