Solve 3x^3 - 21x^2 + 36 x =0

To solve 3x^3-21x^2+36x =
0

We factor the
left.

3x(x^2-7x+12) =
0............(1)

We take
x^2-7x+12

We split the middle tern -7x = -4x and -3x  such
that their product is equal to the product o first and last term , that is, x^2 and
12.

x^2-7x+12

x^2-7x+12=x^2-4x-3x+12

x^2-7x+12=x(x-4)-3(x-4)

x^2-7x+12=(x-4)(x-3)

Therefore 
3x^3-21x^2+36x = 0 could be rewritten as:

3x(x-4)(x-3) =
0.

Therefore by zero product rule we
get:

x = 0 , x-4 = 0 , x -3 =
0.

x = 0 , or x = 4 , or x = 3.