Question

simplify the expression given below.\\(dfrac{x + 2}{x^3 + 2x^2 - 9x - 18}divdfrac{3x + 1}{x^2 - 9}\\)\\(\text{a. } dfrac{1}{(x + 3)(x - 3)}\\)\\(\text{b. } dfrac{1}{3x + 1}\\)\\(\text{c. } dfrac{3x + 1}{(x + 3)(x - 3)}\\)\\(\text{d. } 3x + 1\\)

Explanation:

Step1: Rewrite division as multiplication

$\frac{x+2}{x^3+2x^2-9x-18} \times \frac{x^2-9}{3x+1}$

Step2: Factor cubic denominator

Factor by grouping:
$x^3+2x^2-9x-18 = x^2(x+2)-9(x+2) = (x+2)(x^2-9)$
Substitute back:
$\frac{x+2}{(x+2)(x^2-9)} \times \frac{x^2-9}{3x+1}$

Step3: Cancel common factors

Cancel $(x+2)$ and $(x^2-9)$:
$\frac{1}{1} \times \frac{1}{3x+1}$

Step4: Simplify the product

$\frac{1}{3x+1}$

Answer:

B. $\frac{1}{3x+1}$