Question 137324
{{{f(x)=x^3-6x^2+9x}}}
You can factor out an x to get:
{{{f(x)=x(x^2-6x+9)}}}
Then factor the perfect square quadratic.
{{{f(x)=x(x-3)^2}}}
Zeros are at x=0 and x=3.
{{{ graph( 300, 300, -5, 5, -5, 5, x^3-6x^2+9x) }}}