Question

Simplify the sum. d^2-9d+20/d^2-3d-10 plus d^2-2d-8/d^2+4d-32

Answer 1

For this case what you should do is follow the following steps:
 1) factorize numerator and denominator of both expressions.
 2) cancel similar terms
 3) add both fractions by cross product
 4) Rewrite the numerator and denominator.
 Answer:
 See attached image.

Answer 2

Answer:

Hence, the final result after simplification is:

$\dfrac{2d^2+8d-28}{d^2+10d+16}$

Step-by-step explanation:

We have to simplify the sum:

d^2-9d+20/d^2-3d-10 plus d^2-2d-8/d^2+4d-32 i.e.

$\dfrac{d^2-9d+20}{d^2-3d-10}+\dfrac{d^2-2d-8}{d^2+4d-32}$

now we will apply the method of splitting the middle term in each of the polynomial terms in the numerator and denominator to obtain:

$=\dfrac{d^2-5d-4d+20}{d^2-5d+2d-10}+\dfrac{d^2-4d+2d-8}{d^2+8d-4d-32}\\\\\\\\=\dfrac{d(d-5)-4(d-5)}{d(d-5)+2(d-5)}+\dfrac{d(d-4)+2(d-4)}{d(d+8)-4(d+8)}\\\\\\=\dfrac{(d-4)(d-5)}{(d+2)(d-5)}+\dfrac{(d+2)(d-4)}{(d+8)(d-4)}\\\\\\=\dfrac{d-4}{d+2}+\dfrac{d+2}{d+8}\\\\\\=\dfrac{(d-4)(d+8)+(d+2)(d+2)}{(d+2)(d+8)}$

which on solving gives us the final result as:

$\dfrac{2d^2+8d-28}{d^2+10d+16}$