When working with algebraic fractions, try to factorise first:
#(x^3 - 4x^2+2x - 8)/(2x^2+4x-2)" "(larr "try grouping")/(larr"HCF, quadratic trinomial")#
#=(x^2(x-4)+2(x-4))/(2(x^2+2x+1))#
#= ((x-4)(x^2+2))/(2(x+1)(x+1))#
This cannot be simplified, there are no common factors to cancel
Let's do long division:
#(x^3 - 4x^2+2x - 8)/(2x^2+4x-2) = (x^3 - 4x^2+2x - 8)/(2(x^2+2x-1)#
Let's divide by #(x^2+2x-1)# to start, and divide by #2# later
#color(white)(......................................)x-6#
#x^2+2x-1 )bar(x^3 - 4x^2+2x - 8)#
#color(white)(.....................)ul(x^3 +2x^2-x)" "larr# subtract
#color(white)(........................)-6x^2+3x-8#
#color(white)(.........................)ul(-6x^2-12x-6)" "larr# subtract
#color(white)(.......................................)15x-14" "larr# remainder
We still need to divide by #2#. Therefore we have:
#(x^3 - 4x^2+2x - 8)/(2x^2+4x-2) = (x-6)/2 +(15x-4)/(2(x^2+4x-2))#
