Question

write as equivalent rational expressions with denominator \\((x + 3)(x - 4)(x + 4)\\): \\(\frac{2}{x^2 - x - 12}\\), \\(\frac{1}{x^2 - 16}\\).

Explanation:

Factor the denominators

We begin by factoring the denominators of both rational expressions to identify their current factors.

For the first expression, we factor the quadratic trinomial using Factoring Trinomials:

$$ x^2 - x - 12 = (x - 4)(x + 3) $$

For the second expression, we factor the binomial using Difference of Squares:

$$ x^2 - 16 = (x - 4)(x + 4) $$

Identify the missing factors

The target denominator is the Least Common Denominator:

$$ (x + 3)(x - 4)(x + 4) $$

Comparing each factored denominator to this target:

• For the first expression, \((x - 4)(x + 3)\) is missing the factor \((x + 4)\).
• For the second expression, \((x - 4)(x + 4)\) is missing the factor \((x + 3)\).

Convert the first expression

Multiply the numerator and denominator of the first expression by the missing factor \((x + 4)\):

$$ \frac{2}{x^2 - x - 12} = \frac{2}{(x - 4)(x + 3)} \cdot \frac{x + 4}{x + 4} = \frac{2(x + 4)}{(x + 3)(x - 4)(x + 4)} $$

Convert the second expression

Multiply the numerator and denominator of the second expression by the missing factor \((x + 3)\):

$$ \frac{1}{x^2 - 16} = \frac{1}{(x - 4)(x + 4)} \cdot \frac{x + 3}{x + 3} = \frac{x + 3}{(x + 3)(x - 4)(x + 4)} $$

Answer:

\(\frac{2(x + 4)}{(x + 3)(x - 4)(x + 4)}\), \(\frac{x + 3}{(x + 3)(x - 4)(x + 4)}\)