Problem 45 - Elementary and Intermediate Algebra

Question

Factor by first grouping the appropriate terms. \(c^{2}-d^{2}+c+d\)

Step-by-Step Solution

Answer

The factored form is \((c + d)(c - d)\).

Step 1: Understand the Expression

The expression given is \(c^{2} - d^{2} + c + d\). The exercise requires us to factor this expression by grouping. Grouping involves rearranging and combining terms to simplify the factorization.

Step 2: Group Terms Appropriately

Let's group the terms in a way that makes factoring easier. Consider the expression: \((c^{2} + c) + (-d^{2} + d)\). Notice that each group contains terms that can be factored.

Step 3: Factor Each Group

Factor out the common factor in each group. In the first group \(c^{2} + c\), \(c\) is common, so factor it as \(c(c + 1)\). In the second group \(-d^{2} + d\), \(-d\) is common, so factor it as \(-d(d - 1)\). This gives us: \(c(c + 1) - d(d - 1)\).

Step 4: Simplify the Expression

Notice that \(-d(d - 1)\) can be rewritten as \(- (d)(d - 1)\) or \(-d^2 + d\). Thus, the expression becomes \(c(c + 1) - d(d - 1)\).

Step 5: Rearrange the Factored Expression

Since \(-d(d - 1) = -d^2 + d\), this indicates there is a simpler way to reorder terms, resulting in \((c + 1)(c - d) + (d - 1)(c - d)\).

Step 6: Recognize the Common Binomial Factor

Recognize that both parts of the expression contain the common factor \((c - d)\). We can express the entire expression as \((c + 1 + d - 1)(c - d)\) or \((c + d)(c - d)\).

Step 7: Final Factored Form

Therefore, the factored form of the original expression is \((c + d)(c - d)\) using the grouping and factoring approach. This is the product of the sum and difference of two squares.

Key Concepts

Grouping Method

The grouping method is a handy technique when factoring polynomials, especially those that don't have a common factor across all terms. This involves rearranging the terms to form groups, each of which can be factored separately.
By creating smaller subsets or groups, you can often spot patterns or common factors more easily. The key is to rearrange and factor each group to simplify the entire expression.

For example, in the expression given,

• The terms are grouped as \( (c^2 + c) + (-d^2 + d) \) to allow for easier factorization.
• We then factor out the common factor in each group: \(c(c + 1)\) and \(-d(d - 1)\).

This technique focuses on simplifying complex expressions into more manageable pieces, making it easier to solve or further break down.

Difference of Squares

The difference of squares is a specific pattern that occurs when you subtract one squared term from another. This format follows the identity \( a^2 - b^2 = (a - b)(a + b) \), which is extremely useful in polynomial factorization.

In our exercise, the terms \( c^2 \) and \(-d^2\) suggest a potential difference of squares due to their squared nature. Although they aren't directly next to each other initially, grouping and rearranging allow us to recognize and utilize this identity.

• The exercise shows \(c^2 - d^2\) is a difference of squares and can be quickly split into \( (c - d)(c + d) \).

Understanding this concept is crucial as it frequently appears in algebra and can significantly simplify polynomial factoring tasks.

Common Factors

Identifying common factors is a fundamental step in simplifying polynomial expressions. A common factor is a number or term that divides each component of an expression evenly without leaving a remainder.

In the context of the original exercise, seeking common factors transforms seemingly complex polynomials into simpler, more approachable expressions.

• For instance, in the group \(c^2 + c\), the letter \(c\) is common, leading to the factor \(c(c + 1)\).
• Similarly, in \(-d^2 + d\), \(-d\) is a common factor, which helps us rewrite it as \(-d(d - 1)\).

Spotting and pulling out these common factors is a critical algebraic skill that simplifies the process of finding the final factored form and reveals the overall structure of the polynomial.