Answer:
C. [tex] \dfrac{1}{3x + 1} [/tex]
Step-by-step explanation:
[tex] \dfrac{x + 2}{x^3 + 2x^2 - 9x - 18} \div \dfrac{3x + 1}{x^2 - 9} = [/tex]
Factor all polynomials.
[tex] = \dfrac{x + 2}{x^2(x + 2) -9(x + 2)} \div \dfrac{3x + 1}{(x + 3)(x - 3)} [/tex]
[tex] = \dfrac{x + 2}{(x^2 - 9)(x + 2)} \div \dfrac{3x + 1}{(x + 3)(x - 3)} [/tex]
[tex] = \dfrac{x + 2}{(x + 3)(x - 3)(x + 2)} \div \dfrac{3x + 1}{(x + 3)(x - 3)} [/tex]
To divide by a fraction, multiply by its reciprocal.
[tex] = \dfrac{x + 2}{(x + 3)(x - 3)(x + 2)} \times \dfrac{(x + 3)(x - 3)}{3x + 1} [/tex]
Cancel out common factors in the numerator and denominator.
[tex] = \dfrac{1}{3x + 1} [/tex]
