Questions: Perform the indicated operations. 1/(x^2-3x-18)+(1/(x-6)-1/(x+3)) Select the correct choice below and fill in the answer box(es) to complete your choice. A. 1/(x^2-3x-18)+(1/(x-6)-1/(x+3))=□, x ≠ □ B. (Simplify your answers. Use integers or fractions for any numbers in the expression. Use a comma to separate answers as needed.)

Solution Steps

To simplify the given expression, we first need to factor the quadratic expression in the denominator of the first term. Then, we find a common denominator for all the fractions involved. After that, we combine the fractions and simplify the resulting expression. Finally, we identify any values of \( x \) that must be excluded from the domain due to division by zero.

The expression \(\frac{1}{x^2 - 3x - 18}\) has a quadratic denominator. We factor it as follows: \[ x^2 - 3x - 18 = (x - 6)(x + 3) \]

Substitute the factored form into the expression: \[ \frac{1}{(x - 6)(x + 3)} + \left(\frac{1}{x - 6} - \frac{1}{x + 3}\right) \]

The common denominator for all terms is \((x - 6)(x + 3)\). Rewrite each term with this common denominator: \[ \frac{1}{(x - 6)(x + 3)} + \frac{x + 3 - (x - 6)}{(x - 6)(x + 3)} \]

Combine the numerators: \[ \frac{1 + (x + 3) - (x - 6)}{(x - 6)(x + 3)} = \frac{1 + x + 3 - x + 6}{(x - 6)(x + 3)} = \frac{10}{(x - 6)(x + 3)} \]

The values of \(x\) that make the denominator zero are \(x = 6\) and \(x = -3\). These values must be excluded from the domain.

Final Answer