Problem 101 - Algebra and Trigonometry with Analytic Geometry

Question

Factor the polynomial. $$ x^{16}-1 $$

Step-by-Step Solution

Answer

The polynomial \(x^{16} - 1\) factors as \((x - 1)(x + 1)(x^2 + 1)(x^4 + 1)(x^8 + 1)\).

Step 1: Recognize it as a difference of squares

The polynomial is in the form of a difference of squares, since you can rewrite it as \[x^{16} - 1^2.\] Use the identity for difference of squares: \[a^2 - b^2 = (a-b)(a+b).\] In this case, set \(a = x^8\) and \(b = 1\).

Step 2: Apply the difference of squares formula

Apply the formula to the expression \[x^{16} - 1 = (x^8 - 1)(x^8 + 1).\] Now factor each term further, if possible.

Step 3: Factor as sum and difference of squares again

Focus on the expression \(x^8 - 1\) and notice it is again a difference of squares: \[x^8 - 1 = [(x^4)^2 - 1^2] = (x^4 - 1)(x^4 + 1).\] Similarly, \(x^8 + 1\) cannot be factored with real coefficients using the difference of squares method.

Step 4: Factor \(x^4 - 1\) using difference of squares

Now look at \(x^4 - 1\), and see it as a difference of squares:\[x^4 - 1 = [(x^2)^2 - 1^2] = (x^2 - 1)(x^2 + 1).\]

Step 5: Recognize the simplest form for \(x^2 - 1\)

This can again be factored as a difference of squares:\[x^2 - 1 = (x - 1)(x + 1).\] However, \(x^2 + 1\) cannot be factored using real coefficients.

Step 6: Combine all the factors

We have fully factored the polynomial into:\[(x - 1)(x + 1)(x^2 + 1)(x^4 + 1)(x^8 + 1).\] This expression includes all possible factored terms using real coefficients.

Key Concepts

Difference of Squares

The difference of squares is a powerful technique used in algebra to simplify expressions. It is based on the identity \(a^2 - b^2 = (a-b)(a+b)\), which expresses that a squared term minus another squared term can be factored into two binomials. In this context:

• The expression \((x^{16} - 1)\) is initially recognized as a difference of squares because it can be rearranged to \(x^{16} - 1^2\).
• Applying the identity, we set \(a = x^8\) and \(b = 1\), giving \((x^8 - 1)(x^8 + 1)\).

This pattern of continuously breaking down even powers into simpler polynomial forms using the difference of squares highlights the elegance and repetitive nature of algebraic manipulations. Every subsequent expression that matches the difference of squares format can further be factored until no more such pairs exist.

Polynomial Division

Polynomial division is not directly involved in this problem, but it underpins some deeper algebraic manipulations. The process helps in understanding how larger polynomial expressions can be broken down into smaller, more manageable components. When engaging in polynomial division:

• We often search for common patterns or strategies such as factoring based on recognizable identities like the difference of squares.
• This division process echoes when factoring functions, allowing us to simplify expressions by cutting them down step-by-step.

Though not explicitly used here, polynomial division encourages the same systematic breaking down that we do with difference of squares, both leading to reduced forms of the polynomial that are easier to work with or analyze for specific properties.

Algebraic Manipulations

Algebraic manipulations refer to the various techniques used to rearrange, simplify, or solve polynomial equations and expressions. This exercise primarily relies on recognizing special patterns and identities to factor the polynomial effectively. Here are some fundamental approaches involved:

• Observe initial expressions for known identities such as difference of squares, often revealing factorable forms quickly.
• Take advantage of iterative factoring, repeatedly applying identities until reaching irreducible components, like \(x^2 + 1\), which can't be broken down using real numbers.

These manipulations require a keen eye for detail and sometimes demand creative thinking to see the simplest representations of complex equations. Each manipulation transforms the polynomial to a new form until one arrives at the most reduced and useful version.