Question 936332

{{{x^3-2x^2-x-6=0}}}

{{{x^3-3x^2+x^2-3x+2x-6=0}}}...group

{{{(x^3-3x^2)+(x^2-3x)+(2x-6)=0}}}..factor out

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

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

one solution is: if {{{x-3=0}}} => {{{x=3}}}

use quadratic formula to find other two solutions:

{{{x^2+x+2 = 0}}}

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

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

{{{x = (-1 +- sqrt( 1-8 ))/2 }}}

{{{x = (-1 +- sqrt( -7 ))/2 }}}

{{{x = (-1 +- sqrt( 7 )i)/2 }}}

so, we have complex solutions:

{{{x = (-1 + sqrt( 7 )i)/2 }}} or {{{x = -1/2 + sqrt( 7 )i)/2 }}}

and

{{{x = (-1 - sqrt( 7 )i)/2 }}} or {{{x = -1/2- sqrt( 7 )i)/2 }}}


{{{ graph( 600, 600, -10, 10, -10, 10, x^3-2x^2-x-6) }}}