First, let's change our expression to have a leading coeffiecient lower than #1#, which means factoring out a #2#:
#2x^3+8x^2-4x#
#2(x^3+4x-2x)#
and while we're at it, let's factor out an #x# too
#2x(x^2+4x-2)#
Now, we need to factor the expression within the parentheses. That means we must find two numbers that add to 4 and multiply to #-2#. Because the product is a negative number, that means one of the numbers must be negative. We can also tell that the smallest number is the negative one, since the negative #+# positive gives us a positive number. #-4+9# gives us #5#, whereas #-9+4# leaves us with #-5#.
That's all very useful information, but we ought to just find two numbers that give us add to 4 and multiply to #2#:
#color(white)(.)+4#
#color(white)(.)xx2#
...........
#1xx2#
#2xx2#
That didn't work, so let's use the quadratic formula: #(-b+-sqrt(b^2-4xxaxxc))/(2a)#
#(-(4)+-sqrt((4)^2-4xx(1)xx(-2)))/(2(1))#
#(-4+-sqrt(16+8))/(2)#
#(-4+-sqrt(24))/(2)#
#(-4+-sqrt(2xx2xx3xx2))/(2)#
#(-4+-2sqrt(6))/(2)#
#(-2xxcancel(2)+-cancel(2)sqrt(6))/(cancel(2))#
#-2+-sqrt(6)# or about #-4.449# and #0.449#
