Question 1136714

{{{g(x) = x^3- 2x^2 -6x + 72}}} 

Zero 
if you have {{{3 + 3i }}}, you also have {{{3 - 3i }}}; complex zeros always coming in pairs

find zero product

{{{(x-(3 + 3i) )(x-(3 - 3i) )}}}
={{{(x-3 - 3i )(x-3 +3i )}}}
={{{x^2 - 6 x + 18}}}

 use long division,  divide {{{x^3- 2x^2 -6x + 72}}}  by {{{x^2 - 6 x + 18}}}
...................................({{{x+4}}}
 {{{x^2 - 6 x + 18}}}|{{{x^3- 2x^2 -6x + 72}}}
...................................{{{x^3 - 6 x^2 + 18x}}}.....subtract
.........................................{{{4x^2-24x}}}
.........................................{{{4x^2-24x+72}}}
..............................................{{{4x^2-24x+72}}}....subtract
.................................................................{{{0}}}

 third zero is {{{x+4}}}


so, your answer is: {{{x+4}}},{{{3 + 3i }}},{{{3 - 3i }}}