Answer:
[tex]\frac{x^2+3x -12}{(x-5) (x +3)(x+7)}[/tex]
Step-by-step explanation:
Given
[tex]\frac{x}{x^2-2x-15} - \frac{4}{x^2+2x-35}[/tex]
Required
Calculate the difference
We start by factorizing the denominator of both fractions
[tex]\frac{x}{x^2-2x-15} - \frac{4}{x^2+2x-35}[/tex]
[tex]\frac{x}{x^2+3x - 5x -15} - \frac{4}{x^2+7x - 5x-35}[/tex]
[tex]\frac{x}{x(x+3) - 5(x +3)} - \frac{4}{x(x+7) - 5(x+7)}[/tex]
[tex]\frac{x}{(x-5) (x +3)} - \frac{4}{(x-5)(x+7)}[/tex]
Take LCM
[tex]\frac{x(x+7) - 4(x +3)}{(x-5) (x +3)(x+7)}[/tex]
Open Brackets
[tex]\frac{x^2+7x - 4x -12}{(x-5) (x +3)(x+7)}[/tex]
[tex]\frac{x^2+3x -12}{(x-5) (x +3)(x+7)}[/tex]
This can't be simplified any further;
