Factoring 125x^6 - 1 A Step By Step Guide

In the realm of mathematics, particularly algebra, factoring polynomials stands as a fundamental skill. It involves breaking down a polynomial expression into simpler expressions that, when multiplied together, yield the original polynomial. This process is crucial for solving equations, simplifying expressions, and gaining deeper insights into the behavior of polynomial functions. In this comprehensive guide, we delve into the intricacies of factoring the polynomial 125x⁶ - 1, exploring various techniques and strategies to identify its factors.

Understanding the Problem: Deconstructing 125x⁶ - 1

Before we embark on the factoring journey, let's dissect the polynomial 125x⁶ - 1. This expression is a difference of two terms: 125x⁶ and 1. Notably, both terms are perfect cubes. 125x⁶ can be expressed as (5x²)³, and 1 is simply 1³. This observation is pivotal because it allows us to leverage the difference of cubes factorization formula.

The difference of cubes formula states that:

a³ - b³ = (a - b)(a² + ab + b²)

In our case, a = 5x² and b = 1. Applying the formula, we get:

125x⁶ - 1 = (5x²)³ - 1³ = (5x² - 1)((5x²)² + (5x²)(1) + 1²) = (5x² - 1)(25x⁴ + 5x² + 1)

This factorization reveals two factors: (5x² - 1) and (25x⁴ + 5x² + 1). However, our quest for factors doesn't end here. We need to examine each factor further to determine if they can be factored further.

Diving Deeper: Factoring 5x² - 1

The factor 5x² - 1 presents an interesting scenario. It is a difference of squares, albeit with a slight twist. 5x² is not a perfect square in the traditional sense, but it can be expressed as (√5x)², and 1 is 1². This allows us to employ the difference of squares factorization formula:

a² - b² = (a - b)(a + b)

Here, a = √5x and b = 1. Applying the formula, we obtain:

5x² - 1 = (√5x)² - 1² = (√5x - 1)(√5x + 1)

While this factorization is mathematically correct, it introduces irrational coefficients (√5), which might not be desirable in some contexts. Furthermore, the original list of potential factors provided in the problem does not include these factors. Therefore, while (√5x - 1) and (√5x + 1) are indeed factors of 5x² - 1, we will not consider them in our final answer, as we are looking for factors with integer coefficients that match the provided options. We will consider 5x² - 1 as one of the factors for now, but we must remember that its own factors with irrational coefficients exist.

The Quartic Factor: Unraveling 25x⁴ + 5x² + 1

The second factor, 25x⁴ + 5x² + 1, is a quartic expression (a polynomial of degree 4). Factoring quartics can be more challenging than factoring quadratics or cubics. In this case, we can attempt to rewrite the expression to see if it fits a recognizable pattern. This particular quartic does not factor easily using traditional methods such as grouping or direct application of quadratic formulas. It also does not neatly fit the pattern for a perfect square trinomial or a difference of squares. Therefore, we consider 25x⁴ + 5x² + 1 as another irreducible factor over the integers.

Examining the Provided Options: Matching Factors

Now, let's revisit the list of potential factors provided in the original problem:

• 5x + 1
• 5x² - 1
• x² - 5
• 25x⁴ + 5x² + 1
• 5x² + 5x + 1
• 25x⁴ + 1
• 25x⁴ - 1

Comparing these options with our factorization, we can identify the following factors of 125x⁶ - 1:

• 5x² - 1: This factor directly matches one of our intermediate results from applying the difference of cubes formula.
• 25x⁴ + 5x² + 1: This quartic expression is the other factor we obtained from the difference of cubes factorization.

Let's consider the option 5x + 1. If 5x + 1 were a factor of 125x⁶ - 1, then the polynomial 125x⁶ - 1 should be divisible by 5x + 1. We can perform polynomial long division or synthetic division to check this. However, there's a more direct approach. If 5x + 1 is a factor, then setting 5x + 1 = 0 gives us x = -1/5. Substituting this value into the polynomial 125x⁶ - 1 should yield zero if it is indeed a root.

125(-1/5)⁶ - 1 = 125(1/15625) - 1 = 1/125 - 1 ≠ 0

Thus, 5x + 1 is not a factor of 125x⁶ - 1.

Now, let’s analyze the option 5x - 1. We know that:

5x² - 1 = (√5x - 1)(√5x + 1)

However, if we consider the possibility of a typo and the original intention was to ask if 5x - 1 (without the square) is a factor of some other related polynomial, then setting 5x - 1 = 0 gives us x = 1/5. Substituting this into 125x⁶ - 1:

125(1/5)⁶ - 1 = 125(1/15625) - 1 = 1/125 - 1 ≠ 0

So, 5x - 1 is also not a direct factor of 125x⁶ - 1.

The option x² - 5 can be quickly ruled out because substituting x² = 5 into 125x⁶ - 1 gives us:

125(5)³ - 1 = 125 * 125 - 1 = 15624 ≠ 0

Therefore, x² - 5 is not a factor either.

The options 5x² + 5x + 1, 25x⁴ + 1, and 25x⁴ - 1 do not appear in our factorization and are not readily discernible as factors through any simple algebraic manipulation. Hence, they are not factors of 125x⁶ - 1.

Final Answer: The Factors of 125x⁶ - 1

Based on our analysis, the factors of the polynomial 125x⁶ - 1 from the given list are:

• 5x² - 1
• 25x⁴ + 5x² + 1

These two factors, when multiplied together, reconstitute the original polynomial, confirming their validity as factors.

Factoring polynomials is a cornerstone of algebra, and mastering this skill is essential for success in higher-level mathematics. In this guide, we meticulously factored the polynomial 125x⁶ - 1, demonstrating the application of the difference of cubes and difference of squares formulas. We also highlighted the importance of scrutinizing each factor to determine if further factorization is possible. By systematically analyzing the given options and employing various factoring techniques, we successfully identified the factors of the polynomial. This exercise underscores the power of algebraic manipulation and the elegance of mathematical problem-solving.

What are the common techniques for factoring polynomials?

Common techniques include:

• Greatest Common Factor (GCF): Factoring out the largest common factor from all terms.
• Difference of Squares: Applying the formula a² - b² = (a - b)(a + b).
• Difference and Sum of Cubes: Using the formulas a³ - b³ = (a - b)(a² + ab + b²) and a³ + b³ = (a + b)(a² - ab + b²).
• Trinomial Factoring: Factoring quadratic trinomials of the form ax² + bx + c.
• Grouping: Grouping terms in the polynomial to identify common factors.

How do I know if a polynomial is completely factored?

A polynomial is completely factored when it is expressed as a product of irreducible factors. An irreducible factor is a polynomial that cannot be factored further using elementary factoring techniques over a specific set of numbers (e.g., integers, real numbers).

Are there any shortcuts for factoring polynomials?

Yes, recognizing patterns like the difference of squares or cubes can significantly speed up the factoring process. Practice and familiarity with different factoring techniques also help in identifying shortcuts.

Can all polynomials be factored?

No, not all polynomials can be factored using elementary techniques over integers or real numbers. Some polynomials may have irrational or complex roots, which lead to factors with non-integer coefficients. Additionally, some polynomials are irreducible, meaning they cannot be factored into simpler polynomials.

What is the importance of factoring polynomials?

Factoring polynomials is crucial for:

• Solving Equations: Factoring allows us to find the roots or solutions of polynomial equations.
• Simplifying Expressions: Factoring can simplify complex algebraic expressions, making them easier to work with.
• Graphing Functions: The factors of a polynomial reveal the x-intercepts of its graph.
• Calculus: Factoring is essential in various calculus operations, such as finding limits and derivatives.

By understanding and applying the principles of polynomial factorization, you can unlock a deeper understanding of algebra and its applications in various fields.