Question 142421
factor by grouping: {{{ x^3 - x^2 }}} is {{{ (x^2)(x-1) }}} and
{{{ -4x + 4 }}} is {{{ (-4)(x - 1) }}}} then combine like terms in a way: {{{ (x^2)(x-1) }}} and {{{ (-4)(x - 1)}}}}  have {{{ (x-1) }}} in common take that and mult by whats left: {{{ (x-1)(x^2-4) }}}. to finish solving set (each quantity in brackets) equal to zero: {{{ x-1 = 0 }}} so add one to both sides and you get {{{ x= 1 }}} and {{{ (x^2)-4 =0 }}} so add 4 to both sides and you get {{{ (x^2)=4 }}}, take the square root of both sides which cancels the 2nd power and you get {{{ x= square root of 4}}} which is {{{ x=2 }}} so your solution is x= 1 and x=2.