Question

Find the missing factor and state any domain restrictions necessary to make the two fractions equivalent.\(\frac{3 x}{x-3}=\frac{3 x(1)}{x^{2}-x-6}\)

Short answer

The missing factor is \(x+2\). The domain restrictions are \(x≠3\) and \(x≠-2\) in order to avoid the denominator being zero.

Step-by-step solution

Identifying the denominators

The denominators of our fractions are \(x-3\) and \(x^{2}-x-6\). These denominators must be equal

Factoring the second denominator

First, let's factor the second denominator. \(x^2-x-6\) can be factored into the form \((x-a)(x-b)\) such that \(a*b = -6\) and \(a+b = -1\). The numbers that satisfy these conditions are -3 and 2. So, \(x^2-x-6\) can be factored into \((x-2)(x+3)\)

Finding the missing factor

Now, if we compare our factored denominator with the first denominator, we find that we have \(x-3\) on the left side and \((x-3)(x+2)\) on the right side. So, \(x+2\) must be our missing factor. However, we must ensure that the denominator does not make it a zero. Therefore \(x≠3\) and \(x≠-2\), to avoid a zero denominator.

Key concepts

Factoring Quadratics

When we come across a quadratic equation in the form of ax^2 + bx + c, factoring is a technique used to break it down into a product of two binomials.

For the equation x^2 - x - 6, we search for two numbers that multiply to -6 (the constant term) and add to -1 (the coefficient of x). In this case, those numbers are -3 and 2. The quadratic then factors into (x - 3)(x + 2). Understanding how to factor quadratics is essential, as it is a tool used not just for simplifying algebraic expressions but also for solving quadratic equations, which are prevalent in various math and science problems.

• Identify a, b, and c in ax^2 + bx + c.
• Find two numbers that multiply to ac that also sum up to b.
• Rewrite the middle term bx as two terms using the two numbers found.
• Factor by grouping.

Finding Missing Factors

In the given exercise, we are to find the missing factor that will make two fractions equivalent. With one denominator factored into (x - 2)(x + 3) and the other simply as x - 3, it becomes evident that x + 2 is the missing factor, as x - 3 is common to both denominators.

If we treat the denominators as products and compare them, we can find what's missing in one of the products to make them identical. It's just like looking at two puzzles and spotting the missing piece that completes the picture. Here's how you would ideally proceed:

• Ensure that one of the denominators is fully factored.
• Compare both denominators to each other.
• Identify what factors are present in one denominator but absent in the other.
• Add the missing factor to the numerator of the fraction with the incomplete denominator.

This method enables us to work with equivalent expressions and solve for variables satisfying certain conditions.

Domain Restrictions

When we work with fractions in algebra, the domain of a function or expression is the set of all possible values of the variables that will not cause a division by zero or any other undefined operation.

In this example, since the denominators are x - 3 and (x - 3)(x + 2), we must exclude the values of x that would make either denominator zero — namely, 3 and -2. These are our domain restrictions. To properly state the domain of the expression, we write in interval notation, excluding 3 and -2, typically seen as: \(x \in \mathbb{R}, x eq 3, x eq -2\).

Understanding domain restrictions is critical to ensuring that mathematical expressions and functions are correctly evaluated and applied, as operations resulting in undefined terms can lead to incorrect or meaningless results. Through careful examination of denominators and roots, domain restrictions can be established to avoid these pitfalls.