Question

add.
\\(\dfrac{2}{3x^2 + 20x + 32} + \dfrac{1}{x + 4}\\)
simplify your answer as much as possible.

Explanation:

Step1: Factor the quadratic denominator

Factor $3x^2+20x+32$:
We find two numbers that multiply to $3\times32=96$ and add to 20, which are 12 and 8.
$$3x^2+12x+8x+32=3x(x+4)+8(x+4)=(3x+8)(x+4)$$

Step2: Rewrite fractions with common denominator

The common denominator is $(3x+8)(x+4)$. Adjust the second fraction:
$$\frac{2}{(3x+8)(x+4)} + \frac{1}{x+4} = \frac{2}{(3x+8)(x+4)} + \frac{3x+8}{(3x+8)(x+4)}$$

Step3: Add the numerators

Combine the fractions over the common denominator:
$$\frac{2 + 3x + 8}{(3x+8)(x+4)}$$

Step4: Simplify the numerator

Combine like terms in the numerator:
$$\frac{3x + 10}{(3x+8)(x+4)}$$

Answer:

$\frac{3x + 10}{(3x+8)(x+4)}$