Question

algebra 2: 3.4 solve rational equations
the algebros

\\(\frac{6}{x - 3} = \frac{8x^2}{x^2 - 9} - \frac{4x}{x + 3}\\)

pregunta de opción múltiple
factor the \\(x^2 - 9\\)
\\((x+3)(x+3)\\)
\\((x+3)(x-3)\\)
\\((x-3)(x-3)\\)

Explanation:

To factor \( x^2 - 9 \), we recognize it as a difference of squares. The formula for factoring a difference of squares is \( a^2 - b^2 = (a + b)(a - b) \). Here, \( a = x \) and \( b = 3 \) (since \( 9 = 3^2 \)). Applying the formula, we get \( x^2 - 3^2 = (x + 3)(x - 3) \).

Answer:

B. (x+3)(x-3) (assuming the options are labeled A, B, C with A: (x+3)(x+3), B: (x+3)(x-3), C: (x-3)(x-3))