Question 960679
Conjugate Zeros Theorem states: 

if {{{1 + i}}} is zero, then {{{1 - i}}}  is zero too

so,  the product {{{(1 + i)(1 - i)=1^2-i^2=1-(-1)=1+1=2}}} is a factor of

{{{f(x) = 12x^4 -23x^3 + 21x^2 + 4x -2}}} ......so,divide by {{{2}}}

{{{f(x) = (12x^4-23x^3 + 21x^2 + 4x -2)/2}}}

{{{f(x) =  (1/2)(12x^4 -23x^3 + 21x^2 + 4x -2)}}}.......factor {{{(12x^4 -23x^3 + 21x^2 + 4x -2)}}}

{{{f(x) = (1/2)(12x^4 -24x^3 +x^3+ 24x^2 -2x^2+ 2x -x^2+2x-2)}}}......group


{{{f(x) = (1/2)((12x^4 -24x^3+ 24x^2) +(x^3-2x^2 + 2x) -(x^2-2x+2))}}}


{{{f(x) = (1/2)(12x^2(x^2 -2x+ 2) +x(x^2-2x + 2) -(x^2-2x+2))}}}


{{{f(x) =(1/2) (12x^2+4x-3x-1) (x^2-2x+2)}}}

{{{f(x) =(1/2) ((12x^2+4x)-(3x+1)) (x^2-2x+2)}}}

{{{f(x) =(1/2) (4x(3x+1)-(3x+1)) (x^2-2x+2)}}}

{{{f(x) =(1/2) (4x-1) (3x+1) (x^2-2x+2)}}}

zeros:

{{{(4x-1) =0}}} =>{{{highlight(x=1/4) }}}

{{{(3x+1)=0}}} =>{{{highlight(x=-1/3)}}}


{{{x^2-2x+2=0 }}}=>use quadratic formula

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

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


{{{x = (2 +- sqrt( 4-8 ))/2 }}} 

{{{x = (2 +- sqrt( -4 ))/2 }}} 

{{{x = (2 +- 2i)/2 }}} 

{{{x = (1 +- i) }}}


 =>{{{highlight(x = 1+i)}}} and {{{highlight(x = 1-i)}}}