Question 1190368


{{{ x^3-4x^2+x+6}}}

if given that {{{x=-2}}} is one of the solutions, means {{{(x-(-2))}}}={{{x+2}}} is one factor

so, use long division to see if you can factor it completely


---------({{{x^2-6x+13}}}
{{{(x+2)}}}|{{{ x^3-4x^2+x+6}}}
----------{{{ x^3+2x^2}}}.........subtract
-------------{{{ -6x^2}}}....bring {{{x}}} down
-------------{{{ -6x^2+x}}}
-------------{{{ -6x^2-12x}}}........subtract
-------------------{{{ 13x}}}...bring {{{6}}} down
-------------------{{{ 13x+6}}}
-------------------{{{ 13x+26}}}.......subtract
----------------------{{{ -20}}}->reminder


{{{ x^3-4x^2+x+6=(x^2-6x+13)(x+2)+(-20)}}}