Problem 52 - Elementary Algebra

Question

Factor completely. $$ x 4-81 y 4 $$

Step-by-Step Solution

Answer

The factorization is \((x - 3y)(x + 3y)(x^2 + 9y^2)\).

Step 1: Identify the Expression

The given expression is \( x^4 - 81y^4 \). The first step is to recognize that this is a difference of two terms, each of which is a perfect square.

Step 2: Rewrite as a Difference of Squares

Express each term as a square: \( x^4 = (x^2)^2 \) and \( 81y^4 = (9y^2)^2 \). So, the expression becomes \((x^2)^2 - (9y^2)^2 \), representing a difference of squares.

Step 3: Apply the Difference of Squares Formula

The difference of squares formula is \(a^2 - b^2 = (a-b)(a+b)\). Apply this to \((x^2)^2 - (9y^2)^2 \), substituting \(a = x^2\) and \(b = 9y^2\), to factor the expression. This gives \((x^2 - 9y^2)(x^2 + 9y^2)\).

Step 4: Further Factor the First Term

Notice that \(x^2 - 9y^2\) is again a difference of squares since \(9y^2\) can be written as \((3y)^2\). Factor \(x^2 - 9y^2\) using the formula \(a^2 - b^2 = (a-b)(a+b)\). This gives \((x - 3y)(x + 3y)\).

Step 5: Combine All Factors

Substituting back, the complete factorization of the original expression \(x^4 - 81y^4\) is \((x - 3y)(x + 3y)(x^2 + 9y^2)\). The term \(x^2 + 9y^2\) cannot be factored further over the real numbers.

Key Concepts

Difference of Squares

The concept of the difference of squares is a powerful tool in algebraic factoring. It is useful in simplifying and solving various algebraic problems.

A difference of squares occurs when you have an expression in the form of \(a^2 - b^2\). This is a binomial (two-term expression) where both terms are perfect squares, and they are separated by a subtraction sign. The difference of squares formula is very helpful here:

• \(a^2 - b^2 = (a-b)(a+b)\).

This means you can factor a difference of squares into the product of two conjugate binomials, \((a-b)\) and \((a+b)\).

For example, consider \(x^2 - 16\). Both \(x^2\) and \(16\) are perfect squares, since \(16 = 4^2\). So using the difference of squares formula, we factor it as:

• \((x - 4)(x + 4)\)

Perfect Squares

Perfect squares are numbers or expressions that can be expressed as the square of another number or expression. Understanding perfect squares is crucial for identifying the difference of squares and other factorization techniques.

In algebra, perfect squares appear often in expressions like \(x^2\), \(9y^2\), or numbers like 16 and 81, which are squares of integers (4 and 9). Here is what makes an expression a perfect square:

• An expression \(a\) is a perfect square if \(a = b^2\) for some expression \(b\).

For instance, \(x^4\) is a perfect square because it can be rewritten as \((x^2)^2\), and similarly, \(81y^4\) is a perfect square since it can be expressed as \((9y^2)^2\). Recognizing perfect squares allows one to utilize various factorization strategies, such as the difference of squares technique, simplifying complex algebraic expressions.

Factorization Techniques

Factorization is the process of breaking down an expression into a product of simpler expressions, or factors. To master algebraic expressions, one needs a repertoire of factorization techniques.

• Difference of Squares: Use the formula \(a^2 - b^2 = (a-b)(a+b)\) when you recognize perfect squares in a difference of terms.
• Perfect Square Trinomials: Recognize and factor trinomials, like \(a^2 + 2ab + b^2\), into \((a+b)^2\).

Sometimes an expression might require multiple factorization processes. For instance, in the solved exercise, the expression \(x^4 - 81y^4\) was first identified as a difference of squares, factored into \((x^2 - 9y^2)(x^2 + 9y^2)\). The term \(x^2 - 9y^2\) is also a difference of squares, requiring further factorization into \((x-3y)(x+3y)\).

These techniques build the foundation for simplifying algebraic expressions and solving equations efficiently. Recognizing when each technique is applicable is key to mastering algebra.