4/b+1 + 1/b2-4 + 5/b+2

Answer 1

the common denominator is (b+1)(b+2)(b-2), so when you place all over the common denominator, you have

[4(b^2-4) + 1(b+1) + 5(b+1)(b-2)]/[(b+1)(b^2-4)]

(9b^2-4b-25)/(b^3+b^2-4b-4)