Answer
The solutions are 0 and 4 and 8.
Work Step by Step
$x^{3}$-12$x^{2}$+32x=0
x($x^{2}$-12$x^{}$+32)=0
x($x^{2}$-4$x^{}$-8x+32)=0
x[x(x-4)-8(x-4)]=0
x(x-8)(x-4)=0
x=0 or x-8=0 or x-4=0
x=0 or x=8 or x=4
The solutions are 0 and 4 and 8.
Check
Let x=0
$x^{3}$-12$x^{2}$+32x=0
$0^{3}$-12$0^{2}$+32*0=0
0-0+0=0
0=0
Let x=4
$x^{3}$-12$x^{2}$+32x=0
$4^{3}$-12$4^{2}$+32*4=0
64-12*16+128=0
64-192+128=0
0=0
Let x=8
$x^{3}$-12$x^{2}$+32x=0
$8^{3}$-12$8^{2}$+32*8=0
512-12*64+256=0
512-768+256=0
768-768=0
0=0
