Problem 39 - Elementary and Intermediate Algebra

Question

Perform the operations. Simplify, if possible. $$ \frac{4}{s^{2}+5 s+4}+\frac{s}{s^{2}+2 s+1} $$

Step-by-Step Solution

Answer

The result is \(\frac{s^2 + 8s + 4}{(s + 1)^2(s + 4)}\).

Step 1: Factor the Denominators

First, factor the denominators of each fraction. The first denominator is \(s^2 + 5s + 4\). Factoring this, we get \((s + 1)(s + 4)\). The second denominator is \(s^2 + 2s + 1\), which factors to \((s + 1)^2\).

Step 2: Find a Common Denominator

The common denominator of \((s + 1)(s + 4)\) and \((s + 1)^2\) is \((s + 1)^2(s + 4)\). This common denominator will allow us to add the fractions.

Step 3: Rewrite Each Fraction

Rewrite each fraction with the common denominator of \((s + 1)^2(s + 4)\). The first fraction becomes \(\frac{4(s + 1)}{(s + 1)^2(s + 4)}\), and the second fraction is \(\frac{s(s + 4)}{(s + 1)^2(s + 4)}\).

Step 4: Add the Numerators

Add the numerators of the fractions: \(4(s + 1) + s(s + 4)\). This simplifies to \(4s + 4 + s^2 + 4s\), which further reduces to \(s^2 + 8s + 4\).

Step 5: Write the Combined Fraction

The new fraction is \(\frac{s^2 + 8s + 4}{(s + 1)^2(s + 4)}\). This is the sum of the original fractions.

Step 6: Simplify, If Possible

Check if \(s^2 + 8s + 4\) can be factored further. Factorization does not yield simple factors, so no further simplification is possible. Leave the fraction as is.

Key Concepts

Factoring Polynomials

Factoring is a crucial skill in algebra, especially when dealing with expressions that include polynomials. Think of factoring as the process of breaking down a complex expression into simpler components. For instance, in our exercise, the polynomial in the denominator of the first fraction, \(s^2 + 5s + 4\), can be factored into \((s + 1)(s + 4)\). How do we do this? Recognize that we are looking for two numbers that multiply to give the constant term, 4, and add to give the coefficient of the middle term, 5. These are 1 and 4, respectively.

A similar approach is used for the second denominator, \(s^2 + 2s + 1\). Here, we find numbers that multiply to 1 and add to 2, which are both 1. Thus, we factor it as \((s + 1)^2\). Factoring simplifies many steps in algebra and helps us find common denominators for adding fractions.

Common Denominators

When adding algebraic fractions, finding a common denominator is essential. Just like with regular fractions, a common denominator allows fractions to be combined. The denominators in our problem are \((s + 1)(s + 4)\) and \((s + 1)^2\). To perform the addition, both fractions must share the same denominator. The best result is usually the least common denominator, which in this case is \((s + 1)^2(s + 4)\).

This common denominator is chosen because it includes each factor the highest number of times it appears in any original denominator. This step ensures that when we adjust each fraction to have this new denominator, we can combine the two seamlessly without losing any information from the original fractions. Finding a common denominator might seem tricky at first, but with practice, it becomes a straightforward task that significantly simplifies the process of adding fractions.

Simplifying Expressions

Simplifying algebraic expressions involves reducing them to their simplest form. This often involves combining like terms or factoring where possible. In our exercise, after combining the fractions under a common denominator, we are left with a new numerator: \(4s + 4 + s^2 + 4s\). This expression simplifies by combining the \(s\) terms: we add \(4s + 4s\) to get \(8s\), resulting in \(s^2 + 8s + 4\).

Next, we attempt to factor the numerator \(s^2 + 8s + 4\) as much as possible. Often, simplification involves recognizing when an expression can't be broken down further. In this example, after trying to factor, we find there are no simple integers to neatly reduce \(s^2 + 8s + 4\) into further factors. Thus, it remains as it is, and our fully simplified expression becomes: \(\frac{s^2 + 8s + 4}{(s + 1)^2(s + 4)}\). Remember, a simplified expression is one where all parts are in their most reduced form, ensuring clarity and ease of use in further calculations.