Question 756798
{{{fx=-9(x-8)(x-1)}}} .....to find the real zero of a polynomial function set {{{f(x)=0}}}

{{{-9(x-8)(x-1)=0}}}...since you have {{{f(x)}}} completely factored, use zero product rule to find solutions

{{{-9(x-8)(x-1)=0}}}...since {{{-9<>0}}}, the product will be equal to zero if {{{(x-8)=0}}} or {{{(x-1)=0}}}

if {{{(x-8)=0}}} => {{{highlight(x=8)}}}

if {{{(x-1)=0}}} => {{{highlight(x=1)}}}

so, you have two real zeros

let see them on a graph:

{{{ drawing( 600, 600, -10, 10, -10,115,circle(8,0,0.2),circle(1,0,0.2),graph( 600, 600, -10, 10, -10, 115, -9(x-8)(x-1))) }}}