Problem 144 - Algebra and Trigonometry

Question

Will help you prepare for the material covered in the next section. Factor the numerator and the denominator. Then simplify by dividing out the common factor in the numerator and the denominator. $$\frac{x^{2}+6 x+5}{x^{2}-25}$$

Step-by-Step Solution

Answer

The simplified form of the given expression is \(\frac{x+1}{x-5}\).

Step 1: Factor the Numerator

Let's factorize the numerator \(x^{2}+6x+5\) using the formula for factorizing quadratic expression to get \((x+5)(x+1)\).

Step 2: Factor the Denominator

The denominator \(x^{2}-25\) can be factored using the rules for difference of squares. This rule states \(a^2 - b^2 = (a + b)(a - b)\). Applying this rule, we get \((x+5)(x-5)\) as the factored denominator.

Step 3: Simplify the Expression

Both the factored numerator and denominator have \((x+5)\) as a common factor. Divide both by this common factor to simplify the expression. We are left with \(\frac{x+1}{x-5}\) as the simplified expression.

Key Concepts

Factoring Quadratics

When working with quadratic expressions like the numerator \(x^2 + 6x + 5\), the goal is to express it as a product of two binomials. This process is important because it simplifies complex algebraic expressions, making them easier to work with.

Here's a simple way to factor quadratics:

• First, identify the terms: coefficient of \(x^2\), coefficient of \(x\), and the constant term.
• For \(x^2 + 6x + 5\), the coefficient of \(x^2\) is 1, for \(x\) is 6, and the constant is 5.
• Next, find two numbers that multiply to the constant (5) and add up to the coefficient of \(x\) (6).
• Here, the numbers are 5 and 1, because \(5 \times 1 = 5\) and \(5 + 1 = 6\).
• Thus, the expression factors into \((x + 5)(x + 1)\).

Understanding how to factor quadratics is crucial in algebra as it helps in reducing and simplifying expressions.

Difference of Squares

The denominator \(x^2 - 25\) in our expression is a classic case of a difference of squares. This special form occurs when you subtract one squared term from another. Recognizing this pattern allows you to factor quickly.

• The formula for factoring a difference of squares is \(a^2 - b^2 = (a + b)(a - b)\).
• In this case, \(x^2 - 25\) can be seen as \(x^2 - 5^2\).
• Applying the formula, it factors into \((x + 5)(x - 5)\).

Remembering this pattern will make it quick and easy to factor expressions with squared terms subtracted from one another.

Simplifying Expressions

The final goal of simplifying algebraic fractions involves cancelling out common factors in the numerator and the denominator. This makes the expression less complex and easier to understand or use.

• Start by fully factoring both the numerator and denominator, as we did earlier.
• Here, the numerator factors to \((x + 5)(x + 1)\), and the denominator to \((x + 5)(x - 5)\).
• Observe that both parts have a common factor, \((x + 5)\).
• These common factors can "cancel" one another, simplifying the fraction to \(\frac{x+1}{x-5}\).

Simplifying expressions is a valuable skill in algebra, allowing you to break down and organize information neatly, eliminating unnecessary complexity.