Problem 63 - Elementary and Intermediate Algebra

Question

Factor. $$ b^{4}-256 $$

Step-by-Step Solution

Answer

The expression factors to \((b - 4)(b + 4)(b^2 + 16)\).

Step 1: Recognize the Expression as a Difference of Squares

The given expression is \(b^4 - 256\). This can be recognized as a difference of squares because \(b^4 = (b^2)^2\) and \(256 = 16^2\). Therefore, the expression can be rewritten as \((b^2)^2 - 16^2\).

Step 2: Apply the Difference of Squares Formula

Use the difference of squares formula, \(a^2 - c^2 = (a - c)(a + c)\), to factor \((b^2)^2 - 16^2\). Let \(a = b^2\) and \(c = 16\). Thus, it factors to \((b^2 - 16)(b^2 + 16)\).

Step 3: Recognize Another Difference of Squares in \(b^2 - 16\)

Notice that \(b^2 - 16\) is still a difference of squares. Rewrite it as \((b)^2 - 4^2\).

Step 4: Factor \(b^2 - 16\) Using Difference of Squares Again

Apply the difference of squares formula again to \(b^2 - 16 = (b - 4)(b + 4)\).

Step 5: Combine All Factors

Given that \(b^4 - 256 = (b^2 - 16)(b^2 + 16)\) and \(b^2 - 16 = (b - 4)(b + 4)\), substitute back to get the complete factorization: \((b - 4)(b + 4)(b^2 + 16)\). \(b^2 + 16\) cannot be factored further using real numbers.

Key Concepts

Difference of Squares

The difference of squares is a valuable technique used in algebraic factoring. It's perfect for expressions that follow the form of a squared term subtracted from another squared term. The general form for this is:

• \( a^2 - b^2 = (a - b)(a + b) \)

When you look at the expression \( b^4 - 256 \), it might seem complex at first. However, by recognizing it as a difference of squares, you can break it down.
Here, we notice \( b^4 \) is actually \((b^2)^2 \) and \( 256 \) is \( 16^2 \). So, it fits the difference of squares model as:

• \( (b^2)^2 - 16^2 = (b^2 - 16)(b^2 + 16) \)

This method is powerful because it simplifies complex polynomials into more easily manageable parts.

Polynomial Expressions

Polynomial expressions include variables raised to whole number powers and coefficients. They can be as simple as \(x + 1\) or as complex as \(b^4 - 256\). When you factor polynomial expressions, you break them down into the product of simpler polynomials.
Understanding how to work with polynomial expressions is fundamental in algebra. This exercise started with the polynomial \( b^4 - 256 \). By recognizing and applying the difference of squares multiple times, the expression was simplified into a product of smaller, digestible parts.

• Identify the entire polynomial and decide if a common factor exists.
• See if components fit into recognizable factoring techniques like the difference of squares.
• Simplify each part as far as possible to fully factor the expression.

Simplifying polynomials makes solving equations and understanding graph behaviors much easier.

Quadratic Expressions

Quadratic expressions are polynomial expressions where the highest exponent is 2. The standard form is \( ax^2 + bx + c \). In this exercise, although we began with a fourth-degree polynomial, a quadratic appeared during the factoring process.
A crucial point in the factorization came when \( b^2 - 16 \) was identified. This itself is a quadratic in the form of \( b^2 - c \), which is another difference of squares:

• \( (b - 4)(b + 4) \)

Notably, \( b^2 + 16 \) is a quadratic expression that cannot be factored further using real numbers. If attempting to factor such an expression, remember the limits of real number factorization.
Understanding quadratic expressions is key as they frequently appear in all forms of algebra.