Problem 51 - Introductory Algebra for College Students

Question

Divide as indicated. $$\frac{x^{2}-25}{2 x-2}+\frac{x^{2}+10 x+25}{x^{2}+4 x-5}$

Step-by-Step Solution

Answer

The result is \( \frac{3x+15}{2(x-1)(x+5)} \).

Step 1: Simplify the First Fraction

Factor the difference of two squares in the numerator and take out the common factor in the denominator: \( \frac{x^{2}-25}{2 x-2} = \frac{(x - 5)(x + 5)}{2(x - 1)} \)

Step 2: Simplify the Second Fraction

Factor the perfect square trinomial in the numerator and factor the denominator: \( \frac{x^{2}+10 x+25}{x^{2}+4 x-5} = \frac{(x+5)^2}{(x-1)(x+5)} \)

Step 3: Add the Fractions

The common denominator is \(2(x-1)(x+5)\). Multiply the first fraction by \( (x+5) / (x+5) \) and the second by \( 2 / 2 \) to get the common denominator: \(\frac{x(x+5)}{2(x-1)(x+5)} + \frac{2(x+5)}{2(x-1)(x+5)}\)

Step 4: Combine the Numerators over the Common Denominator

Add the fractions together, combining their numerators: \( \frac{3x+15}{2(x-1)(x+5)} \)

Step 5: Simplify the Result

The result is already in its simplest form so no further simplification is needed.

Key Concepts

Polynomial Division

Polynomial division is the process of dividing one polynomial by another. Think of it like the long division you learned in elementary school, but now with variables and exponents. In this exercise, we start by dividing two algebraic fractions. Each of these fractions has a polynomial in both the numerator and the denominator. Polynomials can be divided using either synthetic division or long division. For simpler polynomials, factoring (as performed in the given solution) is often more straightforward. By factoring the polynomials, you simplify each fraction so you can see where terms cancel out before the division even starts. Embedding polynomial division within the context of algebraic fractions helps us simplify equations and find a common denominator more efficiently in later steps.

Factoring Polynomials

Factoring polynomials is like turning a big problem into smaller, easier problems. When you factor a polynomial, you are writing it as the product of its simplest building blocks, called factors. In the solution given, the polynomial in the numerator of the first fraction, \[x^2 - 25\] is factored as \[(x - 5)(x + 5)\]. This is a difference of squares, a specific type of polynomial that splits neatly into two binomials.

• Difference of Squares: \(a^2 - b^2 = (a - b)(a + b)\)
• Perfect Square Trinomials: These look like \(a^2 \pm 2ab + b^2 = (a \pm b)^2\)

For the second fraction, the polynomial \[x^2 + 10x + 25\] is a perfect square trinomial, factoring into \[(x + 5)^2\].Factoring makes polynomials manageable by breaking them down into these simpler terms, thus simplifying the entire expression in the process.

Simplifying Expressions

Simplifying expressions is a crucial skill in algebra. It involves reducing expressions to their simplest form. We simplify expressions so that we can work with them more easily, spot patterns, and solve equations efficiently.In this particular exercise, each algebraic fraction goes through a simplification process:

• The expressions are factored to reveal what can be canceled or simplified.
• The common terms in the numerators and denominators are identified, and if possible, eliminated to simplify the fraction further.

The final expression in our example, \[\frac{3x + 15}{2(x - 1)(x + 5)}\], cannot be simplified further because there are no common factors in the numerator and the denominator that can be canceled. Simplicity in expressions allows more accessible computation and clarity when adding or subtracting fractions.

Finding Common Denominators

Finding common denominators is an essential skill for adding and subtracting algebraic fractions. A common denominator allows us to combine fractions because it gives them a shared backdrop, much like a unified language. In the given exercise:

• We need a denominator that is a multiple of each individual fraction's denominator.
• The least common denominator here is \[2(x - 1)(x + 5)\], a combination of the factors from both denominators.

To add these fractions, we adjust the numerators to reflect this new shared denominator, ensuring that the fractions themselves aren’t changed. This involves multiplying the numerators and denominators of each fraction by whatever terms are missing to reach the target common denominator.This step prepares the fractions for easy addition, where the numerators can be directly combined, resulting in a single simplified expression. Finding and using a common denominator streamlines calculations and ensures accuracy in solutions.