Problem 55 - Algebra A Combined Function

Question

Factor each completely. $$ m^{4}-1 $$

Step-by-Step Solution

Answer

The expression factors to \((m^2 + 1)(m + 1)(m - 1)\).

Step 1: Recognize the Difference of Squares

The expression \(m^4 - 1\) is a difference of squares because it can be written as \((m^2)^2 - 1^2\). A difference of squares has the form \(a^2 - b^2 = (a+b)(a-b)\).

Step 2: Factor the Difference of Squares

Apply the difference of squares formula to factor \((m^2)^2 - 1^2\). Set \(a = m^2\) and \(b = 1\). Substituting these gives: \((m^2 + 1)(m^2 - 1)\).

Step 3: Factor Further Using Difference of Squares Again

The term \(m^2 - 1\) is also a difference of squares, as it can be considered as \((m)^2 - (1)^2\). Factor it using \((m + 1)(m - 1)\).

Step 4: Write the Complete Factorization

Combine all factors. The original expression \(m^4 - 1\) factors to \((m^2 + 1)(m + 1)(m - 1)\). This is the complete factorization since \(m^2 + 1\) cannot be factored further over the real numbers.

Key Concepts

Difference of Squares

A difference of squares is a special algebraic expression that features two squared terms separated by a minus sign, like this: \(a^2 - b^2\). This type of expression can be factored into a product of two binomials as \((a + b)(a - b)\). This pattern is very handy in algebra because it allows us to simplify complex expressions quickly.

To identify a difference of squares, check for these characteristics:

• The expression has exactly two terms.
• Both terms are perfect squares.
• The terms are separated by a subtraction sign.

The process is straightforward but powerful. In our original exercise, we noticed that \(m^4 - 1\) fits this pattern because \(m^4 = (m^2)^2\) and \(1 = 1^2\). By applying the pattern, we found part of the solution: \((m^2 + 1)(m^2 - 1)\).

Polynomial Factorization

Polynomial factorization is the process of breaking down a polynomial expression into a product of simpler polynomials. Each factor is a polynomial of a lower degree than the original. This method helps simplify complex problems and solve equations effectively.

When factoring, we look for patterns such as common factors, grouping, or recognizable forms like the difference of squares. It can often be a multistep process, especially when a factor can be further decomposed, as seen in our exercise.

• Identify factors that can be broken down further.
• Apply known factoring techniques, such as the difference of squares.
• Continue factoring until every factor is in its simplest form.

In our step-by-step solution, after applying the difference of squares to \(m^4 - 1\), we were left with \(m^2 - 1\), and we noticed it was again a difference of squares. Factoring it resulted in \((m + 1)(m - 1)\). Thus, the complete factorization of \(m^4 - 1\) is \((m^2 + 1)(m + 1)(m - 1)\).

Algebraic Expressions

Algebraic expressions are a combination of numbers, variables, and operators that represent a mathematical phrase. They are fundamental in algebra as they express calculations simply and concisely. Algebraic expressions can range from simple monomials like \(7x\) to complex polynomials like \(m^4 - 1\).

Understanding how to manipulate these expressions is crucial for solving algebraic equations. Here are some key points:

• Expressions can be simplified or rearranged using laws of arithmetic.
• Polynomials are a type of algebraic expression that include terms with variables raised to whole number exponents.
• Separate algebraic expressions with like terms can sometimes be combined.

In our exercise, the expression \(m^4 - 1\) was treated as a polynomial, which we successfully factored using properties of algebraic expressions. By recognizing patterns like the difference of squares, we were able to manipulate the expression into its simplest form.