Question

ifying rational expressions

1. \\(\frac{x - 3}{x^2 - 6x - 9}\\)

Explanation:

Step1: Analyze the denominator

The denominator is \(x^2 - 6x - 9\). Let's check if it can be factored. The quadratic formula for \(ax^2+bx+c\) is \(x=\frac{-b\pm\sqrt{b^2 - 4ac}}{2a}\). Here, \(a = 1\), \(b=-6\), \(c=-9\). Then \(b^2-4ac=(-6)^2-4\times1\times(-9)=36 + 36=72\), which is not a perfect square, so the denominator doesn't factor into linear terms with integer coefficients.

Step2: Check for common factors

The numerator is \(x - 3\) and the denominator \(x^2-6x - 9\) has no common factors with the numerator (since we can't factor the denominator to get \(x - 3\) as a factor). So the rational expression \(\frac{x - 3}{x^2-6x - 9}\) is already in its simplified form (or we can note that the denominator can't be factored to cancel with the numerator).

Answer:

\(\frac{x - 3}{x^2-6x - 9}\) (or it is already simplified as the denominator does not factor to have a common factor with the numerator)