Question

2. Add: Simplify and state the domain.

Answer 1

Answer:

see explanation

Step-by-step explanation:

$\frac{3}{5x+10}$ + $\frac{2x}{x^2-2x-8}$ ← factor the denominators

= $\frac{3}{5(x+2)}$ + $\frac{2x}{(x-4)(x+2)}$

= $\frac{3(x-4)}{5(x+2)(x-4)}$ +$\frac{2x(5)}{5(x+2)(x-4)}$ ← simplify numerator

=$\frac{3x-12+10x}{5(x+2)(x-4)}$

= $\frac{13x-12}{5(x+2)(x-4)}$

The denominator cannot be zero as this would make the rational function undefined

Equating the denominator to zero and solving gives the values of x that cannot be in the domain

5(x + 2)(x - 4) = 0

Equate each factor to zero and solve for x

x + 2 = 0 ⇒ x = - 2

x - 4 = 0 ⇒ x = 4

x = - 2, x = 4 ← excluded values

domain : x ∈ R ( x ≠ - 2, x ≠ 4 )

Answer 2

I think the answer to your question is -2