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A rational function is a ratio of two polynomials. Think of it as a fraction with polynomials in both the numerator and the denominator. For example, in the given function:\[ R(x) = \frac{3(x^2 - x - 6)}{5(x^2 - 4)} \]The polynomial in the numerator is \(3(x^2 - x - 6)\) and the polynomial in the denominator is \(5(x^2 - 4)\).Rational functions can have restrictions on their domains due to the denominator. If the denominator equals zero, the function is undefined, because division by zero is not allowed. This is a crucial point to remember when dealing with rational functions.

The domain of a function includes all possible input values (x-values) that the function can accept. For rational functions, the domain is specifically restricted by the values that make the denominator zero. For our function:\[ R(x) = \frac{3(x^2 - x - 6)}{5(x^2 - 4)} \]We need to exclude the x-values that make \(5(x^2 - 4)\) equal to zero, because these values would make the function undefined. To find these restrictions, set the denominator to zero and solve the equation:\[ 5(x^2 - 4) = 0 \]Then simplify and solve for x.

Interval notation is a way to express the set of all possible values that belong to a function's domain. Once we find the restrictions due to the denominator, we can use interval notation to describe the allowed x-values. For our function, solving:\[ 5(x^2 - 4) = 0 \]Gives us the solutions \(x = 2\) and \(x = -2\). Therefore, these values must be excluded from the domain. The domain in interval notation is written as:\[ (-\infty, -2) \cup (-2, 2) \cup (2, +\infty) \]This notation tells us that the domain includes all real numbers except \(x = 2\) and \(x = -2\). The \(\cup\) symbol represents the union of the intervals, meaning that the function includes each piece of the domain separately.

Factoring polynomials is a technique used to simplify expressions and solve equations. In the context of rational functions, it helps us find the roots of the numerator and the denominator. For our function:\[ R(x) = \frac{3(x^2 - x - 6)}{5(x^2 - 4)} \]We need to factor both the numerator and the denominator. Starting with the denominator \(x^2 - 4\), we recognize it as a difference of squares, which factors into:\[ (x - 2)(x + 2) \]Next, factor the numerator \(x^2 - x - 6\). We look for two numbers that multiply to -6 and add to -1, giving:\[ (x - 3)(x + 2) \]So the function in its factored form is:\[ R(x) = \frac{3(x - 3)(x + 2)}{5(x - 2)(x + 2)} \]Factoring makes it easier to identify the restrictions on the domain and can help simplify the rational function further.