Problem 144 - College Algebra

Question

Will help you prepare for the material covered in the next section.Exercises \(144-146\) 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 expression after factoring and dividing the common factors is \[\frac{x+1}{x-5}\]

Step 1: Factor the Quadratic Function

The numerator \(x^{2}+6 x+5\) can be factored into \((x+1)(x+5)\) as \(1\) and \(5\) are the roots which add up to \(6\) (the coefficient of x) and multiply to \(5\) (the constant term). Similarly, the denominator \(x^{2}-25\) is a difference of perfect squares and can be factored into \((x-5)(x+5)\). This changes the expression into: \[\frac{(x+1)(x+5)}{(x-5)(x+5)}\]

Step 2: Cancel Common Factors

We can see that \((x + 5)\) exists in both the numerator and the denominator, so we can cancel these factors out. This simplifies the expression to: \[\frac{x+1}{x-5}\]

Key Concepts

Factoring Quadratics

Factoring quadratics is an essential skill in algebra. It involves breaking down a quadratic equation into the product of two binomials. A quadratic equation typically looks like this: \( ax^2 + bx + c \). Here, the goal is to express it in the form \((x + p)(x + q)\). In this form, \( p \) and \( q \) are numbers that have two important properties:

• They add up to \( b \), the coefficient of the \( x \) term.
• They multiply together to give \( c \), the constant term in the equation.

In the given exercise, the quadratic expression in the numerator is \(x^2 + 6x + 5\). We need to find two numbers that add to 6 and multiply to 5. These numbers are 1 and 5, so we can rewrite the expression as \((x + 1)(x + 5)\). This step utilizes your knowledge of product-sum numbers, helping you to factor the quadratic quickly and efficiently.

Difference of Squares

The difference of squares is another factoring method where you deal with expressions that take the form \( a^2 - b^2 \). This type of expression can be quickly rewritten as \((a - b)(a + b)\). The reason this works is because of the properties of multiplication and subtraction:

• When you multiply the terms \((a - b)(a + b)\), you get \(a^2 - b^2\).

In the exercise, the denominator \(x^2 - 25\) is a difference of squares. Here, \(x^2\) is \((x)^2\) and 25 is \(5^2\). Thus, the expression \(x^2 - 25\) factors neatly into \((x - 5)(x + 5)\). Recognizing a difference of squares can save you time and simplify the process of factoring complex expressions.

Simplifying Fractions

Simplifying fractions is all about making expressions as straightforward as possible by canceling out common factors. Once you have factored both the numerator and the denominator, you should look for terms that appear in both. These are your common factors.In our example, after factoring, the expression becomes \(\frac{(x+1)(x+5)}{(x-5)(x+5)}\). Notice that \((x+5)\) appears in both the numerator and the denominator. This means you can cancel \((x+5)\) from both, reducing the expression to \(\frac{x+1}{x-5}\).

• Canceling common factors is like simplifying a basic fraction, such as reducing \(\frac{4}{8}\) to \(\frac{1}{2}\).
• This process leads to a simpler, more manageable expression, helping you resolve algebraic fractions easily.