Because this is a third degree polynomial, we have 3 solutions. There's a cubic equation (just like the quadratic equation) but it's really hard so let's try and be smart about this!
We can use the rational root theorem, which states that any solution will be a factor of the #x^0# coefficient divided by a factor of the highest power (here, #x^3#). That means that any rational solution will be a factor of 8 divided by a factor of 1. Since the only factor of 1 is 1, this makes our job a little easier.
We therefore should try out the following numbers to be roots:
# x = pm 1, pm 2, pm 4, pm 8#.
By rational roots theorem, all rational roots are going to be in that list, which makes our job a lot easier.
At this point, we sort of have to plug and chug.
Let's try them in order:
x = -1
#-1^3 - 2 (-1)^2 + 4 (-1) - 8 = -1 - 2 - 4 - 8 neq 0 #
x = 1
#1^3 - 2 (1)^2 + 4 (1) - 8 = 1 - 2 + 4 - 8 neq 0 #
x = -2
#-2^3 - 2 (-2)^2 + 4 (-2) - 8 = -8 - 8 - 8 - 8 neq 0 #
x = 2
#2^3 - 2 (2)^2 + 4 (2) - 8 = 8 - 8 + 8 - 8 = 0 #
So #x=2# is a root! This means that the polynomial divides #x-2#. We can now use "synthetic division" in order to factor the polynomial factors.
#underline2| 1\ \ (-2)\ \ 4 \ \ (-8) #
#\ \ \ \ \ \ downarrow \ \ \ \ underline 2\ \ \ \ underline 0 \ \ \ \ \ \ \ \ underline 8 #
#\ \ \ \ \ \ \ \ 1\ \ \\ \ \ \ 0 \ \ \ \ \ 4 \ \ \ \ \ | underline 0 #
i.e. the polynomial factors to
#(x-2)(x^2 + 4) #
Depending on what you want, this may be the last step. However, if you understand complex numbers, you may want to go one more step. The second half of the polynomial has roots at #pm 2i#. Therefore, we could write the polynomial fully factored into its roots as
#(x-2)(x-2i)(x+2i)#
