Solve this math equation? polynomials?

Question

x^5 - 5x^3 + 4x

Answer 1

x^5 - 5x^3+4x = 0

x[x^4-5x^2+4]=0

x=0  or  x^4-5x^2+4=0

assume  x^2 = t

so  t^2 - 5t + 4 =0  

t^2 - 4t -t + 4 =0

t(t-4) - 1(t-4) = 0

(t-4) (t-1) =0

t=4 and t=1

but t= x^2

so  x^2 =4 and x^2 = 1

    x= 2, -2   and x= 1, -1

so x=0, 1, -1, 2, -2  

(x)(x+1)(x-1)(x+2)(x-2)      

Answer 2

x^5 - 5x^3 + 4x = 0

x(x^4 - 5x^2 + 4) = 0

then x = 0 or x^4 - 5x^2 + 4 = 0

assume y = x^2

then y^2 - 5y + 4 = 0 (we have a+b+c = 1+(-5) + 4 = 0)

y1 = 1 or y2 = 4/1 = 4

as x^2 = y, then

x1= +-1, x2=+-2

finally:

we have: -2, -1, 0, 1, and 2