Question

divide: \frac{3 - \frac{3}{a + 2}}{\frac{1}{a + 2} + \frac{a}{3}}. write numerator and denominator in

provide your answer below:

Explanation:

Simplify the numerator

Using the Complex Rational Expressions knowledge point

$$ 3 - \frac{3}{a+2} = \frac{3(a+2) - 3}{a+2} = \frac{3a + 6 - 3}{a+2} = \frac{3a + 3}{a+2} = \frac{3(a+1)}{a+2} $$

Simplify the denominator

Using the Complex Rational Expressions knowledge point

$$ \frac{1}{a+2} + \frac{a}{3} = \frac{3(1) + a(a+2)}{3(a+2)} = \frac{3 + a^2 + 2a}{3(a+2)} = \frac{a^2 + 2a + 3}{3(a+2)} $$

Divide the simplified expressions

Using the Complex Rational Expressions knowledge point

$$ \frac{\frac{3(a+1)}{a+2}}{\frac{a^2+2a+3}{3(a+2)}} = \frac{3(a+1)}{a+2} \cdot \frac{3(a+2)}{a^2+2a+3} $$

Simplify the product

Using the Complex Rational Expressions knowledge point

$$ \frac{3(a+1) \cdot 3(a+2)}{(a+2)(a^2+2a+3)} = \frac{9(a+1)}{a^2+2a+3} = \frac{9a+9}{a^2+2a+3} $$

Check for further factorization

Using the Factoring Trinomials knowledge point

$$ \text{The trinomial } a^2+2a+3 \text{ has discriminant } 2^2 - 4(1)(3) = -8 < 0 \text{, so it is prime.} $$

Answer:

$$\frac{9a+9}{a^2+2a+3}$$