What are the roots of the polynomial equation x^3-3x^2-6x+8

Ve must factor

factor by grouping
hmm
erm
(trial and error)

if we group it like this:
(-3x^2-6x)+(x^3+8)
we can factor each group
the first is a -3x
the 2nd is a sum of 2 perfect cubes
(-3x)(x+2)+(x+2)(x²-2x+4)
undistribute (x+2) from each
(x+2)(-3x+x²-2x+4)
(x+2)(x²-5x+4)
can we factor more?
what 2 numbers multiply to get 4 and add to get -5?, -1 and -4
(x+2)(x-1)(x-4)

set eqch to zero (assuming it is equal to 0)

0=(x+2)(x-1)(x-4)

0=x+2
-2=x

0=x-1
1=x

0=x-4
x=4

x=-2, 1, 4

$7.50x+$4.50y=midk it's been awhile since i've done one of these