Question

question 4 (1 point)

when fully simplified, excluding restrictions, \\(\frac{x^2 + x}{2x^2 + 7x + 3} \times \frac{x + 3}{x^2 - 1}\\) is equal to

a) \\(-\frac{1}{2x + 1}\\)

b) \\(\frac{x}{(2x - 1)(x + 1)}\\)

c) \\(\frac{x}{(2x + 1)(x - 1)}\\)

d) \\(\frac{(x^2 + x)(x + 3)}{(2x^2 + 7x + 3)(x^2 - 1)}\\)

Explanation:

Factor all numerators and denominators

$$
LATEXBLOCK0
$$

Substitute factored forms into the expression

$$ \frac{x(x + 1)}{(2x + 1)(x + 3)} \times \frac{x + 3}{(x - 1)(x + 1)} $$

Cancel common factors

$$ \frac{x \cdot \cancel{(x + 1)} \cdot \cancel{(x + 3)}}{(2x + 1)\cancel{(x + 3)}(x - 1)\cancel{(x + 1)}} = \frac{x}{(2x + 1)(x - 1)} $$

Answer:

• A) \(-\frac{1}{2x + 1}\)
• B) \(\frac{x}{(2x - 1)(x + 1)}\)
• C) \(\frac{x}{(2x + 1)(x - 1)}\) (Correct answer)
• D) \(\frac{(x^2 + x)(x + 3)}{(2x^2 + 7x + 3)(x^2 - 1)}\)