Question

guided practice
divide: \\(\frac{x^2}{x^2 + 2x + 1} \div \frac{3x}{x^2 - 1}\\)
a. \\(\frac{(x^2)(3x)}{(x + 1)(x - 1)}
\\)b. \\(\frac{x(x - 1)}{3(x + 1)}
\\)c. \\(\frac{x - 1}{3(x + 1)}
\\)d. \\(\frac{x}{3}\\)

Explanation:

Step1: Reciprocal and Multiply

To divide fractions, multiply by the reciprocal of the divisor. So, $\frac{x^2}{x^2 + 2x + 1} \div \frac{3x}{x^2 - 1} = \frac{x^2}{x^2 + 2x + 1} \times \frac{x^2 - 1}{3x}$.

Step2: Factor the Polynomials

Factor $x^2 + 2x + 1$ as $(x + 1)^2$ (perfect square trinomial) and $x^2 - 1$ as $(x + 1)(x - 1)$ (difference of squares). Now the expression becomes $\frac{x^2}{(x + 1)^2} \times \frac{(x + 1)(x - 1)}{3x}$.

Step3: Simplify the Fractions

Multiply the numerators and denominators: $\frac{x^2 \cdot (x + 1)(x - 1)}{(x + 1)^2 \cdot 3x}$. Then cancel common factors. Cancel one $x$ from $x^2$ and $3x$ (leaving $x$ in the numerator), and cancel one $(x + 1)$ from $(x + 1)$ and $(x + 1)^2$ (leaving $(x + 1)$ in the denominator). This gives $\frac{x \cdot (x - 1)}{3(x + 1)}$, which is $\frac{x(x - 1)}{3(x + 1)}$.

Answer:

B. $\frac{x(x - 1)}{3(x + 1)}$