Problem 102 - College Algebra Essentials

Question

Factor and simplify each algebraic expression. $$-8(4 x+3)^{-2}+10(5 x+1)(4 x+3)^{-1}$$

Step-by-Step Solution

Answer

The simplified form of the given expression is \((90x +25)/ (4 x+3)^2\).

Step 1: Analyze the given expression

The given expression is \(-8(4 x+3)^{-2}+10(5 x+1)(4 x+3)^{-1}\). Here two terms are there. In the first term, \(-8(4 x+3)^{-2}\), the power of the binomial is -2. In the second term, \(10(5 x+1)(4 x+3)^{-1}\), we see the power is -1.

Step 2: Rewrite the terms using positive exponents

Rewrite the expression so that the exponents are positive. This will make factoring simpler. The given expression can be rewritten as \(-8/(4 x+3)^2 + 10(5 x+1)/(4 x+3)\)

Step 3: Find a common denominator

To factor or simplify further, we need to have the same denominator in both terms so that we can add or subtract them. So we multiply the second term by \((4 x+3)/ (4 x+3)\) to get the same denominator in both terms. Then, our expression becomes \(-8/(4 x+3)^2 + 10(5 x+1)(4 x+3)/ (4 x+3)^2\).

Step 4: Simplify the expression

Now, both terms have the same denominator. So, combine them into a single fraction. Therefore, the expression becomes \((-8 +10(5 x+1)(4 x+3))/ (4 x+3)^2\). Then distribute the 10 in the second term to get the final simplified expression: \((-8 +50x +30 +40x +3)/ (4 x+3)^2\). This further simplifies to \((90x +25)/ (4 x+3)^2\). This is the simplified form of the given expression.

Key Concepts

Simplifying Expressions

Simplifying expressions is the process of transforming an expression into its simplest form. This often involves reducing it by performing mathematical operations like addition, subtraction, multiplication, and division, as well as by applying rules and properties of algebra.

The goal in simplification is to make the expression as concise as possible. In algebra, this often means removing parentheses and combining like terms. By simplifying, one can make complex expressions more understandable and easier to work with.

Step 4 in our exercise involved simplifying the expression after obtaining a common denominator. This included distributing a multiplication over addition and combining all like terms together into a single, simplified expression. Examining each component carefully can make the bigger picture much clearer, allowing for true simplification.

Common Denominator

A common denominator is essential when working with algebraic fractions because it allows us to combine or compare fractions efficiently. When two fractions have the same denominator, they can be easily added or subtracted because we only need to operate on their numerators.

In our original problem, the expression initially had different denominators for two terms. By multiplying the second term by \((4 x+3)/(4 x+3)\) it became possible to rewrite both terms with the same denominator, \((4x+3)^2\). This was a crucial move as it allowed the terms to be combined into a single fraction, simplifying further operations.

This step shows how fundamental common denominators are when dealing with algebraic fractions, enabling ease in operations like subtraction and simplification.

Exponents

Exponents are a way to represent repeated multiplication. Negative exponents in particular denote reciprocal values because \(a^{-n} = \frac{1}{a^n}\), meaning the base is divided by one, "\(n\)-times".

In our expression, we saw terms with negative exponents, signaling the need to turn these into fractions with positive exponents to simplify the process. Acknowledge that rewriting \((4x+3)^{-2}\) as \(\frac{1}{(4x+3)^2}\) helps visualize the expression more clearly.

Understanding how to manipulate exponents, especially turning negative exponents into fractions, is key to solving and simplifying algebraic expressions.

Algebraic Fractions

Algebraic fractions, much like regular fractions, consist of a numerator and a denominator, but include algebraic expressions. They follow the same fundamental rules of arithmetic used with number fractions, with some added complexity due to the presence of variables.

In tackling problems with algebraic fractions, it is vital to become comfortable with simplification processes, such as finding a common denominator, factoring, and, if needed, canceling terms across the numerator and denominator.

Our example demonstrated these concepts clearly: converting each term to have the same denominator was essential before combining them into one. The outcome was a simplified algebraic fraction in its most reduced form, illustrating the power of methodical problem-solving in algebra.