Problem 98 - Beginning and Intermediate Algebra

Question

Factor completely. $$p^{6}-1$$

Step-by-Step Solution

Answer

The completely factored expression for \(p^6 - 1\) is \((p + 1)(p^2 - p + 1)(p - 1)(p^2 + p + 1)\).

Step 1: Factoring out the Greatest Common Factor (GCF)

In this case, there is no common factor to factor out, so we can move on to the next step.

Step 2: Factoring using Difference of Squares

Recall that a difference of squares can be factored as follows: \[a^2 - b^2 = (a + b)(a - b)\] We can use this to factor the given expression as follows: \[(p^6 - 1) = (p^3 + 1)(p^3 - 1)\]

Step 3: Factoring further using Difference of Squares

We can continue factoring by applying the difference of squares technique to the second factor, \(p^3 - 1\): \[(p^3 - 1) = (p - 1)(p^2 + p + 1)\]

Step 4: Factoring using Sum of Cubes on the other term

Recall that the sum of cubes can be factored as follows: \[a^3 + b^3 = (a + b)(a^2 - ab + b^2)\] We can use this to factor the first term, \(p^3 + 1\), as follows: \[(p^3 + 1) = (p + 1)(p^2 - p + 1)\]

Step 5: Putting it all together

Now that we have factored both terms, we can combine the factored terms to find the complete factorization of the given expression: \[(p^6 - 1) = (p^3 + 1)(p^3 - 1) = (p + 1)(p^2 - p + 1)(p - 1)(p^2 + p + 1)\] So the completely factored expression is \((p + 1)(p^2 - p + 1)(p - 1)(p^2 + p + 1)\).

Key Concepts

Difference of Squares

In the world of algebra, the term "difference of squares" refers to an expression where two perfect squares are subtracted from one another. The formula to factor a difference of squares is:

• \[a^2 - b^2 = (a + b)(a - b)\]

This formula is a useful tool for breaking down more complex polynomials into simpler factors. When you see an expression of the form \(p^6 - 1\), the first step is often to identify it as a difference of squares.
The expression \(p^6 - 1\) can be rewritten where each term is a square: \((p^3)^2 - 1^2\). By applying the difference of squares formula, we factor this to:

• \[p^3 + 1\]
• \[p^3 - 1\]

Recognizing expressions as differences of squares is crucial because it allows further factorization into linear or irreducible quadratic terms, making polynomial simplification much easier and leading to the solution's completion.

Sum of Cubes

The sum of cubes is another important concept in algebra that describes expressions combining cubic terms in addition. The factoring formula for the sum of cubes is:

• \[a^3 + b^3 = (a + b)(a^2 - ab + b^2)\]

When analyzing the problem \(p^6 - 1\), after factoring the difference of squares, you get the term \(p^3 + 1\) which fits the sum of cubes pattern.
Using the sum of cubes formula, we can factor \(p^3 + 1\) further:

• \[p^3 + 1 = (p + 1)(p^2 - p + 1)\]

This step is essential because it decomposes the cubic expression into a linear factor \(p + 1\) and a quadratic \(p^2 - p + 1\), simplifying the original expression. Recognizing the sum of cubes allows for streamlined polynomial manipulation.

Greatest Common Factor

The Greatest Common Factor (GCF) is a basic concept often used as an initial step in polynomial factorization. The GCF is the largest factor shared by all terms in the polynomial. Extracting it simplifies the polynomial by reducing the degrees of the terms.
In the example of \(p^6 - 1\), a quick inspection reveals that it does not have a GCF other than 1, as there is no common term or number divisible across all components of the expression.
Knowing when there's no GCF helps because it directs the approach to alternative factoring methods, such as the difference of squares or other identities like the sum of cubes. Although not always applicable, searching for a GCF is a standard initial step in polynomial factorization as it can significantly simplify further operations.