Question 107107
find the zeros of

 {{{(2x-7)^2(x^2-9)}}}

First, you can write {{{(2x-7)^2}}} as {{{(2x-7)(2x-7)}}}

And {{{(x^2-9)}}} as{{{ (x-3)(x+3)}}}

Then you will have:

{{{(2x-7)(2x-7) (x-3)(x+3)}}}
Zeros: product is equal to {{{0}}} if one factor is equal to {{{0}}}
So,
{{{2x-7=0}}}……… => …{{{2x = 7}}}… => {{{x = 7/2}}}....first and secon  zero ( we have {{{2x -7}}} two times
Or
{{{x-3 = 0}}}……… =>…{{{x = 3}}}....third zero
Or
{{{x + 3 = 0}}}……..=> …{{{x = -3}}}....fourth zero

check:

{{{(2x-7)^2(x^2-9)}}}

plug in zeros:

if {{{x=7/2}}}
{{{(2*(7/2)-7)^2((7/2)^2-9) =  (7-7)^2(49/4-9)=0(49/4-9)=0}}} (since one factor is {{{0}}}
Or
{{{x=3}}}
{{{(2*3-7)^2(3^2-9) = (6-7)^2(9-9) =1*0 = 0}}}
Or
{{{x = -3}}}
{{{(2*(-3)-7)^2((-3)^2-9) = (-6-7)^2(9-9) =1*0 = 0}}}