Question 50682
This is factoring by grouping.  Group the first two and last two terms together.


{{{x^3-x^2+4*x-4=0}}}
{{{x^2(x-1) + 4(x-1) = 0}}}


Notice that there is a common factor of (x-1):
{{{ (x-1) (x^2+4) = 0}}}

{{{x=1}}} or {{{x^2 = -4}}}


Are you solving with complex numbers or real numbers only?  If real numbers only, then the only answer you have to worry about is x=1.  End of problem, since real numbers squared cannot equal a negative.


However, if complex numbers are allowed, then 
{{{x^2 = -4}}} gives two solutions:
{{{x= sqrt(-4)}}}, {{{x=-sqrt(-4) }}}
{{{x= 2i}}}, {{{x= -2i}}} in addition to {{{x= 1}}}.


R^2 at SCC