Question 134721
Everything you have done so far is absolutely correct, but you stopped short of saying that:


{{{x^4-6x^3+71x^2-146x+530=(x^2-4x+53)(x^2-2x+10)=0}}} =>  {{{x^2-4x+53=0}}} or {{{x^2-2x+10=0}}}.


You already have the roots for {{{x^2-4x+53=0}}} since you were given one of them and used it and its conjugate to derive the trinomial in the first place.  Now all you need is the roots of the quotient polynomial that came from division of the quartic by {{{x^2-4x+53}}}.  In other words, just solve


{{{x^2-2x+10=0}}}


This one adapts readily to completing the square:


{{{x^2-2x=-10}}}


{{{x^2-2x+1=-10+1}}}


{{{(x-1)^2=-9}}}


{{{x-1=3i}}} or {{{x-1=-3i}}}


{{{x=1+3i}}} or {{{x=1-3i}}}


In sum, the four roots of {{{x^4-6x^3+71x^2-146x+530=0}}} are:


{{{2+7i}}}
{{{2-7i}}}
{{{1+3i}}}
{{{1-3i}}}