Completely Factored Polynomial: Find The Solution

Dec 4, 2025 by Alex Johnson 50 views

Which Polynomial is Factored Completely? A Detailed Guide

Choosing the correctly factored polynomial from a set of options can sometimes feel like navigating a maze. In this guide, we'll break down the process step by step, ensuring you understand not only the solution but also the why behind it. We’ll explore what it means for a polynomial to be completely factored, and then we'll apply that understanding to the given options. This will equip you with the skills to tackle similar problems with confidence. Let's dive in!

Understanding Complete Factorization

When addressing the question of which polynomial is factored completely, it’s essential to first grasp the concept of complete factorization. A polynomial is said to be completely factored when it is expressed as a product of irreducible factors. In simpler terms, this means you've broken down the polynomial into its most basic building blocks – factors that cannot be factored any further. These factors can be monomials (single-term expressions), binomials (two-term expressions), or trinomials (three-term expressions), but the key is that none of these factors can be simplified or factored again.

To illustrate, consider the number 12. It can be factored as 2 x 6, but this isn't complete factorization because 6 can be further factored as 2 x 3. The complete factorization of 12 is 2 x 2 x 3, where each factor is a prime number and cannot be broken down any further. Similarly, with polynomials, we aim to break them down into factors that are analogous to prime numbers – expressions that resist further factorization.

Complete factorization often involves several techniques. The most common one is factoring out the greatest common factor (GCF). This is the largest term that divides each term of the polynomial. For example, in the expression 4x^2 + 8x, the GCF is 4x, which can be factored out to give 4x(x + 2). Another important technique is recognizing special patterns like the difference of squares (a^2 - b^2 = (a + b)(a - b)) or perfect square trinomials (a^2 + 2ab + b^2 = (a + b)^2). Applying these patterns helps to simplify the polynomial into its completely factored form.

Another aspect of complete factorization is ensuring that the factors themselves cannot be factored further. This often means checking if any of the resulting binomials or trinomials can be factored using methods such as grouping, trial and error, or more advanced techniques like the quadratic formula. For instance, if you end up with a quadratic trinomial, you should check if it can be factored into two binomials. If it can't, then that factor is irreducible.

The final step in complete factorization is ensuring that each factor is indeed in its simplest form. This means checking for any remaining common factors within the factors themselves and making sure that no further algebraic manipulations can simplify the expression. By following these steps, you can confidently determine when a polynomial is completely factored and ensure that you have reached the most simplified representation of the polynomial.

Analyzing the Given Options

Now, let's apply our understanding of complete factorization to the given options and determine which polynomial is factored completely. We have four options to consider, each presenting a different polynomial expression. Our task is to analyze each one, identify potential factors, and determine if the polynomial is indeed expressed in its most completely factored form.

Option A: g^5 - g

Starting with Option A, which is g^5 - g, the first step in factoring is typically to look for the greatest common factor (GCF). In this case, the GCF is 'g'. Factoring 'g' out, we get: g(g^4 - 1). Now, we have two factors: 'g' and (g^4 - 1). The factor 'g' is a monomial and cannot be factored further. However, the factor (g^4 - 1) is a difference of squares, which can be factored further. Recognizing this pattern is crucial in complete factorization. The expression (g^4 - 1) can be rewritten as (g²⁾2 - 1^2, which fits the form a^2 - b^2, and can be factored as (a + b)(a - b). Applying this to our expression, we get: (g^2 + 1)(g^2 - 1).

Now, we have three factors: 'g', (g^2 + 1), and (g^2 - 1). The factor (g^2 + 1) is a sum of squares, and over the real numbers, it cannot be factored further. However, the factor (g^2 - 1) is another difference of squares, which can be factored as (g + 1)(g - 1). Therefore, the completely factored form of g^5 - g is g(g^2 + 1)(g + 1)(g - 1). Since Option A in its original form can be factored further, it is not completely factored.

Option B: 4g^3 + 18g^2 + 20g

Moving on to Option B, which is 4g^3 + 18g^2 + 20g, we again start by looking for the GCF. In this case, the GCF is 2g. Factoring 2g out, we get: 2g(2g^2 + 9g + 10). Now, we need to determine if the quadratic trinomial (2g^2 + 9g + 10) can be factored further. To do this, we look for two numbers that multiply to give the product of the leading coefficient (2) and the constant term (10), which is 20, and add up to the middle coefficient (9). The numbers 4 and 5 satisfy these conditions (4 x 5 = 20 and 4 + 5 = 9). We can use these numbers to split the middle term and factor by grouping:

2g^2 + 9g + 10 = 2g^2 + 4g + 5g + 10

Now, we group the terms: (2g^2 + 4g) + (5g + 10)

Factor out the GCF from each group: 2g(g + 2) + 5(g + 2)

Factor out the common binomial factor (g + 2): (2g + 5)(g + 2)

Therefore, the completely factored form of 4g^3 + 18g^2 + 20g is 2g(2g + 5)(g + 2). Since Option B in its original form can be factored further, it is not completely factored.

Option C: 24g^2 - 6g^4

For Option C, 24g^2 - 6g^4, we once again begin by identifying the GCF. In this case, the GCF is 6g^2. Factoring out 6g^2, we get: 6g^2(4 - g^2). Now, we have two factors: 6g^2 and (4 - g^2). The factor 6g^2 is a monomial and cannot be factored further. However, the factor (4 - g^2) is a difference of squares, which can be factored as (2 + g)(2 - g). Therefore, the completely factored form of 24g^2 - 6g^4 is 6g^2(2 + g)(2 - g). Since Option C in its original form can be factored further, it is not completely factored.

Option D: 2g^2 + 5g + 4

Finally, we consider Option D, which is 2g^2 + 5g + 4. This is a quadratic trinomial, and we need to determine if it can be factored. We look for two numbers that multiply to give the product of the leading coefficient (2) and the constant term (4), which is 8, and add up to the middle coefficient (5). The pairs of factors of 8 are (1, 8) and (2, 4). Neither of these pairs adds up to 5. Therefore, the quadratic trinomial 2g^2 + 5g + 4 cannot be factored further using integer coefficients. This means that Option D is already in its simplest form and is considered completely factored.

Determining the Completely Factored Polynomial

After a thorough analysis of each option, we've arrived at a clear conclusion about which polynomial is factored completely. Let's recap our findings and pinpoint the correct answer.

We examined four polynomials:

Option A: g^5 - g, which factors to g(g^2 + 1)(g + 1)(g - 1)
Option B: 4g^3 + 18g^2 + 20g, which factors to 2g(2g + 5)(g + 2)
Option C: 24g^2 - 6g^4, which factors to 6g^2(2 + g)(2 - g)
Option D: 2g^2 + 5g + 4, which cannot be factored further

Our analysis revealed that Options A, B, and C could all be factored further, meaning they were not in their completely factored forms. Option A involved factoring out a common factor and recognizing the difference of squares pattern multiple times. Option B required factoring out the greatest common factor and then factoring a quadratic trinomial by splitting the middle term. Option C also involved factoring out a common factor and applying the difference of squares pattern.

However, Option D, the quadratic trinomial 2g^2 + 5g + 4, presented a different scenario. We attempted to find two numbers that multiply to 8 (the product of the leading coefficient and the constant term) and add up to 5 (the middle coefficient). Since no such integers exist, we concluded that this trinomial is irreducible over the integers. This means it cannot be factored further using basic factoring techniques.

Therefore, based on our comprehensive analysis, the polynomial that is factored completely is Option D: 2g^2 + 5g + 4. This polynomial stands alone as the only one that resists further factorization, meeting the criteria for complete factorization.

Conclusion

In conclusion, identifying the completely factored polynomial requires a systematic approach. It involves recognizing and applying various factoring techniques, such as factoring out the GCF, recognizing special patterns like the difference of squares, and determining whether quadratic trinomials can be factored further. By methodically applying these techniques, we can confidently determine when a polynomial is in its most simplified, factored form.

Option D, 2g^2 + 5g + 4, emerged as the correct answer because it could not be factored further using integer coefficients. This underscores the importance of not only factoring but also confirming that the resulting factors are irreducible. Understanding the nuances of polynomial factorization is a fundamental skill in algebra, essential for solving equations, simplifying expressions, and tackling more advanced mathematical concepts.

For further exploration of polynomial factorization and related topics, consider visiting trusted educational resources such as Khan Academy's Algebra I section.