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))
