Problem 118 - Introductory Algebra for College Students

Question

will help you prepare for the material covered in the first section of the next chapter. 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

In the numerator, \(x^{2}+6x+5\), factors as \((x + 1)(x + 5)\). This is done by finding two numbers which multiply to 5 and add to 6, which are 1 and 5.

Step 2: Factor the denominator

In the denominator, \(x^{2}-25\), factors as \((x + 5)(x - 5)\). This is done by applying the difference of squares formula, \(a^2 - b^2 = (a - b)(a + b)\), with \(a=x\) and \(b=5\).

Step 3: Simplify the fraction

After factoring both the numerator and denominator, the fraction becomes \(\frac{(x + 1)(x + 5)}{(x + 5)(x - 5)}\). Now, divide out the common factor of \((x + 5)\) from the numerator and denominator to simplify the fraction, resulting in \(\frac{x + 1}{x - 5}\).

Key Concepts

Factoring Polynomials

Factoring polynomials involves expressing a polynomial as a product of its simplest polynomials that cannot be further factored. This process is essential for solving polynomial equations, simplifying expressions, or integrating functions. To factor a polynomial like \(x^2 + 6x + 5\):

• Identify two numbers that multiply to the constant term (5 in this case) and add to the coefficient of the linear term (6 in this case). These numbers are 1 and 5.
• The polynomial then splits into factors as \((x+1)(x+5)\).

This simplifies the equations and makes finding solutions easier. Factoring is like uncovering the skeleton upon which an equation is built.

Difference of Squares

Difference of squares is a specific technique used to factor polynomials where two terms are squared and subtracted. This is represented by the formula \(a^2 - b^2 = (a-b)(a+b)\). This formula is trigonometric because it provides a straightforward method of breaking down significant expressions.
In the exercise, the expression \(x^2-25\) is a difference of squares where:

• \(a=x\)
• \(b=5\)

Plug these values into the difference of squares formula to factor \(x^2-25\) as \((x+5)(x-5)\). This method swiftly reveals valuable insights into simplifying or solving equations.

Simplifying Fractions

Simplifying fractions helps in reducing expressions to their simplest form, making them easier to work with. This process involves canceling out common factors from the numerator and the denominator. Start simplifying by:

• Factoring both the numerator and denominator as was done in the exercise.
• Identifying common factors; in this exercise, both contain the factor \((x+5)\).
• Cancel the common factor from the numerator and the denominator.

Thus, the fraction \(\frac{(x+1)(x+5)}{(x+5)(x-5)}\) simplifies to \(\frac{x+1}{x-5}\) after canceling out \((x+5)\). Simplified fractions are pivotal in math to ensure expressions are concise and manageable.