Question 977233

{{{f(x)=x^4-3x^3+27x-81}}}

{{{f(x)=(x^4-3x^3)+(27x-81)}}}

{{{f(x)=x^3(x-3)+27(x-3)}}}

{{{f(x)=(x-3)(x^3+27)}}}

{{{f(x)=(x-3)(x+3) (x^2-3x+9)}}}

set {{{f(x)=0}}}

{{{0=(x-3)(x+3) (x^2-3x+9)}}}

if {{{0=(x-3)}}}=>{{{x=3}}}
if {{{0=(x+3)}}}=>{{{x=-3}}}
if{{{0=(x^2-3x+9)}}}=>use quadratic formula

{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}} 

{{{x = (-(-3) +- sqrt((-3)^2-4*1*9 ))/(2*1) }}}


{{{x = (3 +- sqrt(9-36 ))/2 }}}
 

{{{x = (3 +- sqrt(-27 ))/2 }}}

{{{x = (3 +- sqrt(-9*3 ))/2 }}}

{{{x = (3 +- 3i*sqrt(3 ))/2 }}}

{{{x = (3/2)(1 +- 3i*sqrt(3 )) }}}

{{{x = (3/2) (1-i*sqrt(3))}}} and {{{x = (3/2) (1+i*sqrt(3))}}}




so, zeros are:({{{x[1]}}},{{{x[2]}}},{{{x[3]}}},{{{x[4]}}})=({{{3}}},{{{-3}}},{{{(3/2) (1-i*sqrt(3))}}},{{{(3/2) (1+i*sqrt(3))}}})