Problem 31 - Algebra A Combined Function

Question

Find the \(L C D\) for each list of rational expressions. $$ \frac{2 x}{3 x^{2}+4 x+1}, \frac{7}{2 x^{2}-x-1} $$

Step-by-Step Solution

Answer

The LCD is \((3x + 1)(x + 1)(2x + 1)(x - 1)\).

Step 1: Identify Denominators

For the two expressions, the denominators are \(3x^2 + 4x + 1\) and \(2x^2 - x - 1\). We need to find the least common denominator (LCD) of these two polynomials.

Step 2: Factor First Denominator

Factor the first polynomial \(3x^2 + 4x + 1\). Looking for two numbers that multiply to \(3 \times 1 = 3\) and add to \(4\), we find \((3x + 1)(x + 1)\) as the factors.

Step 3: Factor Second Denominator

Factor the second polynomial \(2x^2 - x - 1\). Looking for two numbers that multiply to \(2 \times -1 = -2\) and add to \(-1\), we find \((2x + 1)(x - 1)\) as the factors.

Step 4: Combine All Factors

The LCD must include all unique factors from both factorizations. The factors are \((3x + 1)\), \((x + 1)\), \((2x + 1)\), and \((x - 1)\).

Step 5: Construct LCD

By combining the unique factors, the LCD is \((3x + 1)(x + 1)(2x + 1)(x - 1)\).

Key Concepts

Factoring Polynomials

Factoring polynomials is an important skill when working with algebraic expressions. It involves breaking down a polynomial into simpler, multiplicative components, known as factors. This process is crucial for finding the least common denominator (LCD) in problems involving rational expressions. To factor a quadratic polynomial such as \(ax^2 + bx + c\), one looks for two numbers that multiply to \(a \, c\) (the product of the leading coefficient and the constant term) and add to \(b\) (the linear coefficient). These numbers help break down the middle term and thus factor the polynomial into two binomial expressions. In the example given, we factored:

• \(3x^2 + 4x + 1\) into \((3x + 1)(x + 1)\)
• \(2x^2 - x - 1\) into \((2x + 1)(x - 1)\)

Practice these steps by trying different values that satisfy the conditions, as numerous polynomial forms can often stump learners. It's like solving a puzzle: find the perfect numbers to split the linear term! Each polynomial form tells a story of its own, so getting familiar with spotting factorable patterns is key.

Rational Expressions

Rational expressions are fractions that have polynomials in both the numerator and the denominator. They operate under the same rules as numerical fractions, requiring common denominators to add or subtract them. The process of finding a least common denominator (LCD) for rational expressions is essential because, without a common denominator, the expressions cannot be easily combined or compared. Taking the provided rational expressions as an example, we start with:

• \(\frac{2x}{3x^2+4x+1}\)
• \(\frac{7}{2x^2-x-1}\)

The key step in simplifying such expressions is first to factor the polynomials, then determine the LCD that represents the smallest expression which each rational can be expressed with the same denominator. This step simplifies comparison, addition, and subtraction, letting the algebra problems flow as smoothly as those you've encountered with simple numbers.

Algebraic Fractions

Algebraic fractions are expressions that contain variables in their numerators or denominators or both. Similar to numerical fractions, these must be manipulated carefully, especially when involved in processes like addition, subtraction, multiplication, and division.When working with algebraic fractions, it's necessary first to verify if they can be simplified. Simplification often involves factoring, as seen in the step-by-step solution. Also, finding and using the LCD is central in operations that involve combining algebraic fractions.In our example, to solve for the least common denominator of:

• \(\frac{2x}{(3x + 1)(x + 1)}\)
• \(\frac{7}{(2x + 1)(x - 1)}\)

We constructed an LCD: \((3x + 1)(x + 1)(2x + 1)(x - 1)\). This new denominator is a product of unique factors from each original expression's denominator, ensuring that any combination of these algebraic fractions can be efficiently managed under this common ground. The mastery of algebraic fractions opens up a more comprehensive understanding of algebra, letting you handle even more complex equations confidently.