Problem 42 - Elementary and Intermediate Algebra

Question

Simplify each rational expression. See Example 4 $$\frac{x^{2}-2 x-15}{25-x^{2}}$$

Step-by-Step Solution

Answer

Simplified expression is \(-\frac{x + 3}{5 + x}\).

Step 1: Factor the Numerator

The numerator of the expression is a quadratic expression \(x^2 - 2x - 15\). To factor this, we need to find two numbers that multiply to \(-15\) and add to \(-2\). These numbers are \(3\) and \(-5\). So, the factorization of the numerator is \((x - 5)(x + 3)\).

Step 2: Factor the Denominator

The denominator \(25 - x^2\) is a difference of squares. It can be factored using the identity \(a^2 - b^2 = (a - b)(a + b)\). Here, \(a = 5\) and \(b = x\), so the factorization is \((5 - x)(5 + x)\).

Step 3: Rewrite the Expression with Factors

Once the numerator and denominator are factored, rewrite the expression:\[\frac{(x - 5)(x + 3)}{(5 - x)(5 + x)}\]

Step 4: Simplify the Expression with Cancellation

Now, observe that \(x - 5\) and \(5 - x\) are negatives of each other: \(x - 5 = -(5 - x)\). This means that \((x - 5) = -(5 - x)\), allowing us to cancel \((x - 5)\) with \((5 - x)\), leaving a \(-1\). Thus, the expression simplifies to: \[-\frac{(x + 3)}{(5 + x)}\]

Key Concepts

Factoring Quadratics

When simplifying rational expressions, factoring quadratics is a key skill to master. Quadratic expressions often take the form \(ax^2 + bx + c\). The goal is to express the quadratic as a product of two binomials, making it easier to simplify the expression.
In the example given, we have the quadratic \(x^2 - 2x - 15\). To factor it, you need two numbers that multiply to \(-15\) and add up to \(-2\). Think of it as solving a puzzle and finding the right pair of numbers—here, they are \(3\) and \(-5\).
You can then rewrite the quadratic as \((x - 5)(x + 3)\). This process not only simplifies the numerator but also sets the stage for canceling out any common factors in the rational expression. With practice, identifying these factor pairs becomes quicker and more intuitive.

Difference of Squares

The difference of squares is a special technique used in algebra to factor expressions of the form \(a^2 - b^2\). With this method, expressions are factored into the product of two binomials: \((a - b)(a + b)\). This technique is particularly useful in simplifying rational expressions and is applicable when dealing with certain quadratic denominators or numerators.
Consider the denominator in our exercise: \(25 - x^2\). Recognize that 25 is a perfect square \((5^2)\) and so is \(x^2\). According to the difference of squares rule, we can factor it as \((5 - x)(5 + x)\).
Understanding the structure of differences of squares saves time and reduces errors in simplification processes, allowing you to clearly see any cancelable factors that might simplify the expression further.

Algebraic Simplification

Algebraic simplification involves combining like terms and reducing expressions to their simplest forms. In the context of rational expressions, this often involves canceling common factors in the numerator and denominator.
In our problem, after factoring both the numerator \((x - 5)(x + 3)\) and the denominator \((5 - x)(5 + x)\), we notice that \((x - 5)\) and \((5 - x)\) are negatives of each other. Specifically, \(x - 5 = -(5 - x)\). Correspondingly, this allows these terms to be canceled out, while taking into account to multiply the remaining part by \(-1\).
Hence, the expression simplifies to \(-\frac{(x + 3)}{(5 + x)}\). Understanding this principle of simplification is crucial for working with rational expressions and ensuring they are in their simplest form. Moreover, it highlights the importance of factoring as a preliminary step for simplification.

• Factor all parts of the expression.
• Identify common factors in the numerator and denominator.
• Cancel out these factors carefully, keeping negative signs in mind.

By mastering these steps, you can effectively simplify any rational expression into its most concise form.