Question

question 2 of 2
select all the correct answers.
for which values is this expression undefined?
\\(\frac{3x}{x^2 - 9} - \frac{x + 4}{x^2 + 2x - 15}\\)
\\(\square\\) \\(x = 3\\)
\\(\square\\) \\(x = 5\\)
\\(\square\\) \\(x = -3\\)
\\(\square\\) \\(x = -5\\)
\\(\square\\) \\(x = -4\\)
\\(\square\\) \\(x = 0\\)

Explanation:

Step1: Factor denominators

First, factor the denominators of the two fractions. For the first denominator \(x^2 - 9\), we use the difference of squares formula \(a^2 - b^2=(a + b)(a - b)\), so \(x^2 - 9=(x + 3)(x - 3)\). For the second denominator \(x^2+2x - 15\), we factor it as \(x^2+2x - 15=(x + 5)(x - 3)\) (by finding two numbers that multiply to -15 and add to 2, which are 5 and -3).

Step2: Find values that make denominators zero

A rational expression is undefined when its denominator is zero. So we set each denominator equal to zero and solve for \(x\).

For the first denominator \((x + 3)(x - 3)=0\), we have \(x+3 = 0\) or \(x - 3=0\), which gives \(x=-3\) or \(x = 3\).

For the second denominator \((x + 5)(x - 3)=0\), we have \(x+5 = 0\) or \(x - 3=0\), which gives \(x=-5\) or \(x = 3\).

Combining all the values that make either denominator zero, we get \(x=-5\), \(x=-3\), and \(x = 3\).

Answer:

A. \(x = 3\), C. \(x=-3\), D. \(x=-5\)