Factored Completely? Polynomial Examples & Solutions

Hey guys! Ever get stuck trying to figure out if a polynomial is really factored all the way? It's a common head-scratcher in math, so let's tackle this with a clear explanation and examples. We'll walk through each option, making sure you understand the why behind the answer. So, grab your thinking caps, and let's dive into the world of complete factorization!

Understanding Complete Factorization

Before we jump into the options, it's super important to understand what "factored completely" actually means. A polynomial is factored completely when it's broken down into its simplest possible factors. This means you can't factor it any further. Think of it like prime factorization with numbers – you keep breaking it down until you only have prime numbers left. With polynomials, you're looking for irreducible factors, which are factors that can't be factored anymore using rational numbers.

Key things to remember:

• Look for common factors first (like a 'g' or a number that divides all terms).
• Check if any of the resulting factors can be further factored (like a difference of squares or a quadratic).
• Keep going until none of the factors can be broken down further.

Let's use an example: Consider the expression $x^2 - 4$. We can factor this as $(x - 2)(x + 2)$. Now, can we factor $(x - 2)$ or $(x + 2)$ any further? Nope! So, $(x - 2)(x + 2)$ is the complete factorization of $x^2 - 4$.

Now, let's apply this understanding to the problem at hand and dissect each option to see which polynomial is indeed factored completely. We will go through each option methodically, explaining why each one either is or is not completely factored. This way, you'll not only get the correct answer but also grasp the underlying concepts, making you a factorization pro in no time!

Analyzing the Options

Let's examine each polynomial to determine if it's factored completely.

A. $g^5 - g$

Okay, so we have $g^5 - g$. The first thing we should always look for is a common factor. In this case, both terms have 'g' in them, so we can factor out a 'g':

$g^5 - g = g(g^4 - 1)$

Now, take a closer look at the term inside the parentheses: $g^4 - 1$. Does this look familiar? It should! It's a difference of squares (since $g^4$ is $(g^2)^2$ and 1 is $1^2$). So, we can factor it further:

$g(g^4 - 1) = g(g^2 - 1)(g^2 + 1)$

But wait, there's more! The term $(g^2 - 1)$ is also a difference of squares! So, we can factor it again:

$g(g^2 - 1)(g^2 + 1) = g(g - 1)(g + 1)(g^2 + 1)$

Now, let's check each factor: 'g', '(g - 1)', and '(g + 1)' can't be factored any further. But what about $(g^2 + 1)$? Since it's a sum of squares, it cannot be factored using real numbers. Therefore, the complete factorization is:

$g(g - 1)(g + 1)(g^2 + 1)$

Since we could factor it multiple times, option A, $g^5 - g$, is not factored completely in its initial form.

B. $4g^3 + 18g^2 + 20g$

Alright, let's tackle option B: $4g^3 + 18g^2 + 20g$. Again, the first step is to look for common factors. Notice that all the terms are divisible by 2 and also have 'g' in them. So, we can factor out $2g$:

$4g^3 + 18g^2 + 20g = 2g(2g^2 + 9g + 10)$

Now, we need to see if the quadratic expression inside the parentheses, $2g^2 + 9g + 10$, can be factored further. To do this, we look for two numbers that multiply to $(2 * 10 = 20)$ and add up to 9. Those numbers are 4 and 5. So, we can rewrite the quadratic and factor by grouping:

$2g^2 + 9g + 10 = 2g^2 + 4g + 5g + 10 = 2g(g + 2) + 5(g + 2) = (2g + 5)(g + 2)$

Therefore, the complete factorization of the original polynomial is:

$2g(2g + 5)(g + 2)$

Since we could factor it further, option B, $4g^3 + 18g^2 + 20g$, is not factored completely in its initial form.

C. $24g^2 - 6g^4$

Now, let's analyze option C: $24g^2 - 6g^4$. Once again, we start by looking for common factors. Both terms are divisible by 6, and they both have $g^2$ in them. So, we can factor out $6g^2$:

$24g^2 - 6g^4 = 6g^2(4 - g^2)$

Notice that the term inside the parentheses, $(4 - g^2)$, is a difference of squares (since 4 is $2^2$). So, we can factor it further:

$6g^2(4 - g^2) = 6g^2(2 - g)(2 + g)$

Now, let's check each factor: $6g^2$, $(2 - g)$, and $(2 + g)$ cannot be factored any further. Thus, the complete factorization is:

$6g^2(2 - g)(2 + g)$

Since we could factor it multiple times, option C, $24g^2 - 6g^4$, is not factored completely in its initial form.

D. $2g^2 + 5g + 4$

Finally, let's examine option D: $2g^2 + 5g + 4$. This is a quadratic expression. To determine if it can be factored, we look for two numbers that multiply to $(2 * 4 = 8)$ and add up to 5. Can we find such numbers?

Let's list the factor pairs of 8: (1, 8) and (2, 4). Neither of these pairs adds up to 5. Therefore, the quadratic expression $2g^2 + 5g + 4$ cannot be factored using integers.

Thus, the polynomial $2g^2 + 5g + 4$ is already in its simplest form and cannot be factored further. Therefore, option D, $2g^2 + 5g + 4$, is factored completely.

The Answer!

So, after carefully analyzing each option, we can confidently say that the polynomial that is factored completely is:

D. $2g^2 + 5g + 4$

Key Takeaways

• Always look for common factors first. This simplifies the polynomial and makes it easier to factor further.
• Recognize special patterns like the difference of squares. This can significantly speed up the factoring process.
• If you have a quadratic, check if it can be factored. If you can't find integer roots, it might be irreducible.

Understanding these concepts will make you a factorization master in no time! Keep practicing, and you'll be able to spot completely factored polynomials with ease. Happy factoring!