Our strategy will aim to factor the polynomial as much as possible: once completely factored, the polynomial will become a multiplication of polynomials of lower degree:
[tex] f(x) = f_1(x)\cdot f_2(x) \cdot f_3(x) [/tex]
and its zeroes will be the ones of its factors.
Since the polynomial has no constant term, you can factor it as follows:
[tex] x^3 - 2x^2 - 8x = x(x^2 - 2x - 8) [/tex]
To continue, we must factor the quadratic expression in the parenthesis. A common way to factor expressions like [tex] ax^2+bx+c [/tex] is to find the two solutions [tex] x_1 [/tex] and [tex] x_2 [/tex] and write the polynomial as [tex] ax^2+bx+c=a(x-x_1)(x-x_2) [/tex].
To find the solutions, we can use the quadratic formula
[tex] x_{1,2} = \frac{-b\pm\sqrt{b^2-4ac}}{2a} [/tex]
and since in our case [tex] a=1, b=-2,c=-8 [/tex], the solving formula becomes
[tex] x_{1,2} = \frac{2\pm\sqrt{4+32}}{2} = \frac{2\pm6}{2} = 1 \pm 3 [/tex]
So, the two solutions are [tex] x_1 = -2 [/tex] and [tex] x_2 = 4 [/tex] and we write the polynomial as [tex] (x+2)(x-4) [/tex].
So, the complete factorization is
[tex] x^3 - 2x^2 - 8x = x(x+2)(x-4) [/tex]
So, the zeroes of the cubic polynomial we started with are the zeroes of the three polynomials in the factorization: [tex] f_1(x)=0 [/tex] yields a solution for [tex] x=0 [/tex], [tex] f_2(x)=x+2 [/tex] yields a solution for [tex] x=-2 [/tex] and [tex] f_3(x)=x-4 [/tex] yields a solution for [tex] x=4 [/tex].
