Question

What is the difference?

Answer 1

Answer:

  $\dfrac{x^2+3x-12}{(x+3)(x-5)(x+7)}$

Step-by-step explanation:

Factor the denominators, then use a common denominator to add the terms.

   $\dfrac{x}{x^2-2x-15} -\dfrac{4}{x^2+2x-35} =\dfrac{x(x+7)}{(x+3)(x-5)(x+7)}-\dfrac{4(x+3)}{(x+3)(x-5)(x+7)}\\\\=\dfrac{x^2+7x-4x-12}{(x+3)(x-5)(x+7)}=\boxed{\dfrac{x^2+3x-12}{(x+3)(x-5)(x+7)}}$