Question 976974
{{{f(x) = x^3 -2x^2-16x + 32}}}......set {{{f(x) =0}}}, then group

{{{0 = (x^3 -2x^2)-(16x - 32)}}}

{{{0 = x^2(x -2)-16(x - 2)}}}

{{{0 = (x^2-16)(x - 2)}}}

{{{0 = (x^2-4^2)(x - 2)}}}

{{{0 =(x-4)(x+4)(x-2)}}}

zeros:

{{{0 =(x-4)}}}=>{{{x=4}}}
{{{0 =(x+4)}}}=>{{{x=-4}}}
{{{0 =(x-2)}}}=>{{{x=2}}}

the multiplicity of each zero is {{{1}}}

the number of turning points of the graph of the function is {{{2}}}


{{{ graph( 600, 600, -60, 60, -60, 60, (x-4)(x+4)(x-2)) }}}