Adding polynomial fractions and subtracting check

[kpx001]

Did I make an error? I assumed i would be able to factor the answer and cancel numerator to denominator

1/(x+2) + 1/(x^2-4) - 2/(x^2 - x -2)

(x-2)(x+1) / (x+2)(x-2)(x+1) + (x+1) / (x+2)(x-2)(x+1) - 2x+4 / (x+2)(x-2)(x+1)

(x^2 - x - 2) / (x+2)(x-2)(x+1) +(x+1) / (x+2)(x-2)(x+1) - 2x+4 / (x+2)(x-2)(x+1)


then

(x^2 - 2x - 5) / (x+2)(x-2)(x+1)

 

[Deleted member 4993]

Looks okay - you have parentheses missing in your work.

 

[kpx001]

should i foil out the denominator to simplify it better, or keep it the way it is?

 

[Mrspi]

should i foil out the denominator to simplify it better, or keep it the way it is?

It's generally better to leave the denominator in factored form; if you were going to use this answer in another problem, you'd probably want the denominator to be factored.

Just my opinion of course....others may disagree.

 

[Denis]

Did I make an error? I assumed i would be able to factor the answer and cancel numerator to denominator
1/(x+2) + 1/(x^2-4) - 2/(x^2 - x -2)
(x-2)(x+1) / (x+2)(x-2)(x+1) + (x+1) / (x+2)(x-2)(x+1) - 2x+4 / (x+2)(x-2)(x+1)
(x^2 - x - 2) / (x+2)(x-2)(x+1) +(x+1) / (x+2)(x-2)(x+1) - 2x+4 / (x+2)(x-2)(x+1)
then
(x^2 - 2x - 5) / (x+2)(x-2)(x+1)

As "opinion", I (being lazy) like to keep "writing to a minimum":

Let a = x + 2, b = x - 2, c = x + 1

(a b^2 c + a b c - 2 a^2 b) / (a^2 b^2 c)

= (b c + c - 2 a) / (a b c)