Question 1004735


{{{ F(x) =x^5 + 5x^3 - 36x}}}....factor completely

{{{ F(x) =x(x^4 + 5x^2 - 36)}}}

{{{ F(x) =x(x^4 + 9x^2-4x^2 - 36)}}}

{{{ F(x) =x((x^4 + 9x^2)-(4x^2+ 36))}}}

{{{ F(x) =x(x^2(x^2 + 9)-4(x^2+ 9))}}}

{{{ F(x) =x (x^2+9) (x^2-4) }}}

{{{ F(x) =x (x^2+9) (x-2) (x+2)}}}

zeros:

{{{ 0 =x (x^2+9) (x-2) (x+2)}}}

 {{{ 0 =x}}}
if  {{{ 0 =x^2+9}}}=>{{{x^2=-9}}}=>{{{x=sqrt(-9)}}}=>{{{x=3i}}} and {{{x=-3i}}}

if {{{ 0 = x-2}}} =>{{{x=2}}}
if {{{ 0 = x+2}}}=>{{{x=-2}}}


so, c) +/- {{{3}}} is not a solution to the equation