The first real zero is easy. It's zero. We can factor out an x.
x^5-x^4-3x^3+5x^2-2x=x(x^4-x^3-3x^2+5x-2)
Now we note that when x=-2,
x^4-x^3-3x^2+5x-2
=(-2)^4-(-2)^3-3(-2)^2+5(-2)-2
=16-(-8)-3(4)-10-2=0.
This means that x+2 is a factor of x^4-x^3-3x^2+5x-2.
Let's factor x+2 from x^4-x^3-3x^2+5x-2.
x^4-x^3-3x^2+5x-2
= x^4+2x^3-3x^3-6x^2+3x^2+6x-x-2
=x^3(x+2)-3x^2(x+2)+3x(x+2)-(x+2)
=(x^3-3x^2+3x-1)(x+2)
Now we see that when x=1
x^3-3x^2+3x-1
=1^3-3(1)^2+3(1)-1=1-3+3-1=0.
This means that x-1 is a factor of x^3-3x^2+3x-1. Let's factor x-1 from x^3-3x^2+3x-1.
x^3-3x^2+3x-1
=x^3-x^2-2x^2+2x+x-1
=x^2(x-1)-2x(x-1)+(x-1)=(x^2-2x+1)(x-1)
Finally we recognize that
x^2-2x+1=(x-1)(x-1)
So putting it all together, we have
x^5-x^4-3x^3+5x^5-2x=x(x+2)(x-1)(x-1)(x-1)
=x(x+2)(x-1)^3.
