Question 174076
Note: {{{10x^2-30x+20}}} factors to {{{10x^2-30x+20=10(x^2-3x+2)=10(x-1)(x-2)}}} and {{{x^2-1}}} factors to {{{x^2-1=(x+1)(x-1)}}}



{{{(10x^2-30x+20)/(x^2-1)=(10(x-1)(x-2))/((x+1)(x-1))=(10*cross((x-1))(x-2))/((x+1)cross((x-1)))=(10(x-2))/(x+1)}}}



So {{{(10x^2-30x+20)/(x^2-1)=(10(x-2))/(x+1)}}} where {{{x<>-1}}} or {{{x<>1}}}