MHB Adding binomials and trinomials

[CSCI MARIO]

( x - 3/ x^2- 7x + 10 ) + ( x + 4 / x ^ 2 - 9 )
i have to add these and put the denominator in a factored form

(x-5)(x-2)(x-3)(x+3)

but the example in my book came up with

2x^3 - 6x^2 - 27x + 67 as the numerator.
This is what i don't understand.
can someone explain how the numerator was solved.

 

[evinda]

( x - 3/ x^2- 7x + 10 ) + ( x + 4 / x ^ 2 - 9 )
i have to add these and put the denominator in a factored form

(x-5)(x-2)(x-3)(x+3)

but the example in my book came up with

2x^3 - 6x^2 - 27x + 67 as the numerator.
This is what i don't understand.
can someone explain how the numerator was solved.

$$x^2- 7x + 10=0 \Leftrightarrow (x-2) (x-5)=0$$

$$\frac{ x - 3}{ x^2- 7x + 10} + \frac{ x + 4 }{ x^2 - 9}= \frac{x-3}{(x-2)(x-5)}+\frac{x+4}{(x-3)(x+3)} \\ =\frac{(x-3)(x-3)(x+3)}{(x-2)(x-5)(x-3)(x+3)}+\frac{(x+4)(x-2)(x-5) }{(x-3)(x+3) (x-2)(x-5)} \\ =\frac{(x-3)(x-3)(x+3)+(x+4)(x-2)(x-5) }{(x-3)(x+3) (x-2)(x-5)} \\ =\frac{(x^2-9)(x-3)+ (x^2+2x-8)(x-5)}{(x-3)(x+3) (x-2)(x-5)}\\ =\frac{x^3-3x^2-9x+27+x^3-3x^2-18x+40}{(x-3)(x+3) (x-2)(x-5)}$$