Question

D^2+d-30/d^2+3d-40 + d^2+14d+48/d^2-2d-48

Answer 1

Answer: option B.

   2d^2  + 14d + 16        
----------------------------
        (d+8)(d-8)              


Explanation:

The question is:

$\frac{d^2+d-30}{d^2+3d-40} + \frac{d^2+14d+48}{d^2-2d-48}$

1) Start factoring all the polynomials to rewrite the fractions.

2) d^2 + d - 30 = (d + 6)(d - 6)

3) d^2 + 3d - 40 = (d + 8)(d - 5)

4) d^2 + 14d + 48 = (d + 6) (d + 8)

5) d^2 - 2d - 48 = (d - 8)(d + 6)

6) rewrite the fractions:

  (d+6)(d-5)         (d+8)(d+6)
----------------- +  -----------------
  (d+8)(d-5)          (d-8)(d+6)

7) simplify the fractions cancelling the factors that are equal in the numerator and the denominator:

  d+6       d+8
-------- + -------
  d+8       d-8

8) take least common denominatior: (d+8)(d-8), and sum the fractions:

   (d-8)(d+6) + (d+8)^2
--------------------------------
          (d+8)(d-8)

9) expand the parenthesis in the numerator and combine like terms:

 d^2 - 2d - 48 + d^2 + 16d + 64
------------------------------------------- =
               (d+8)(d-8)

        2d^2  + 14d + 16        
=   ----------------------------
              (d+8)(d-8)              

And that is the option B.

Answer 2

I think D.2d^2+15d+18/(d+8)(d-8)