Answered

2/x-3 + x-5/9-x^2 - 3x-5/9+6x+x^2


I understand how to factor to get the denominator but the numerator when I foil and add I keep getting wrong.

Verified Answer :

LammettHash

I'm taking a wild guess on what the first expression is:

\(\dfrac2{x-3}+\dfrac{x-5}{9-x^2}-\dfrac{3x-5}{9+6x+x^2}\)

First, some factoring in the denominators:

\(\dfrac2{x-3}+\dfrac{x-5}{(3-x)(3+x)}-\dfrac{3x-5}{(3+x)^2}\)
\(\dfrac2{x-3}-\dfrac{x-5}{(x-3)(x+3)}-\dfrac{3x-5}{(x+3)^2}\)

Find the common denominator:

\(\dfrac{2(x+3)^2}{(x-3)(x+3)^2}-\dfrac{(x-5)(x+3)}{(x-3)(x+3)^2}-\dfrac{(3x-5)(x-3)}{(x-3)(x+3)^2}\)

Combine the numerators:

\(\dfrac{2(x+3)^2-(x-5)(x+3)-(3x-5)(x-3)}{(x-3)(x+3)^2}\)

Expand the numerator:

\(\dfrac{(2x^2+12x+18)-(x^2-2x-15)-(3x^2-14x+15)}{(x-3)(x+3)^2}\)

Combine like terms:

\(\dfrac{-2x^2+28x+18}{(x-3)(x+3)^2}\)

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