How do you show that (x+2) is a factor of f(x) = x^3 - 19x - 30, and then factorise f(x) completely?

A1) Showing that (x+2) is a factor of f(x).         General Method:  If A is a number, then (x- A) is a factor of f(x) if when x=A,  f(x=A)=0    <-- This is what we want to show                                              This is because if (x-A) is a factor of f(x) then we can write:  f(x) = (x - A) x (some other function)      [ in the same way as 8= 2 x 4 =  2 x (another number)]                                      Therefore if x = A, then f(x=A) = (A-A) x (some other function) = 0 x (some other function) = 0.        Our Question:   We have to show (x + 2) is a factor, and our general method uses (x - A).  We can still use our method as (x - A) = (x + {-A}). Therefore comparing the formula,  -A = 2    =>   A = -2                                f(x=A = -2) = (-2)^3  - 19(-2) - 30 = -8 + 38 - 30 = 38 - 38 =0.    A2)  To factorise completely:   We know that f(x) = (x+2) x (some other function)                                                The largest term in f(x) = x^3, so f(x) = (x+2)(ax^2 + bx + c) = x^3 - 19x - 30                                                Multiplying out & comparing terms:    ax^3 + bx^2 +cx + 2ax^2 +2bx + 2c  = ax^3+x^2(b+2a) + x(c+2b) + 2c = x^3 - 19x - 30                                                   a=1 ,   b+2a= b+2=0 => b = -2   ,    c+2b = c + 2(-2) = c - 4 = -19 => c= -15                          Therefore: f(x) = (x+2)(x^2 - 2x - 15)                        We can factorise further (x^2 - 2x - 15) = (x+d)(x+e)  comparing terms in a similar way as before leads to final factorisation of       f(x) = (x+2)(x-5)(x+3)