Precalculus
posted by Kate .
Hello,
Thanks for all of the help guys. I really appreciate it.
I was woudnering if you could show me how to do this step by step and the thought process that goes through your head while you do this. This way I will be able to try to solve some of these problems on my own becasue I really have no idea how to do them.
Perform the indicated operation and simplify the ersult. Leave your anser in factored form.
1/x  2/(x^2 + x) + 3/(x^3  x^2)
THANK YOU!!!

remember to add/subtract fractions we need a common denominator.
let's write our expression in factored form
1/x  2/(x^2 + x) + 3/(x^3  x^2)
= 1/x  2/[x(x+1)] + 3/[x^2(x1)]
so the common denominator is x^2(x+1)(x1)
first term: x(x+1)(x1)/[x^2(x+1)(x1)]
notice if we cancel we get 1/x
second term: 2x(x1)/[x^2(x+1)(x1)]
again, see what you get when you cancel
third term: 3(x+1)/[x^2(x+1)(x1)]
so we get
[x(x+1)(x1)  2x(x1) + 3(x+1)]/[x^2(x+1)(x1)]
I will let you finish it,
(I got (x^3  2x^2 + 4x + 3)/[x^2(x^2  1)] 
I bet you have a sign wrong but will do as shown.
1/x
2/[x(x+1)]
+3/[x^2(x1)]
we need a common denominator
need LCD = x^2 (x1)(x+1) = x^2(x^21)
so
1[(x)(x+1)(x1)]/LCD
2 [ (x)(x1) ] /LCD
+3 [ (x+1 ] / LCD
then
[1 (x(x^21) 2 (x^2x) +3 (x+1) ]/LCD
[ x^3x 2x^2 +2x +3x+3 ]/LCD
(x^3 2x^2 +4x +3)/[x^2(x^21)]
I could not simplify, suspct mistake but that is the idea. 
Oh, I see Reiny got the same answer. If you typed it right the answer is right.