Tomas Factored The Polynomial Completely What Is True About His Final Product

by Luna Greco 78 views

Polynomial factorization, guys, is like unlocking a secret code in the world of mathematics! It's a fundamental skill, especially in algebra, and can help us simplify expressions, solve equations, and understand the behavior of functions. In this article, we're going to dive deep into a specific factorization problem and break down the steps involved. Let's explore what happens when Tomas completely factors the polynomial 3x4โˆ’18x3+9x2โˆ’54x{3x^4 - 18x^3 + 9x^2 - 54x} and expresses it in the form Ax(x2+B)(x+C){Ax(x^2 + B)(x + C)}. We'll uncover what's true about the final factored product and clarify some common misconceptions about polynomial factorization.

Understanding Polynomial Factorization

Before we jump into the problem, let's quickly recap what polynomial factorization is all about. In simple terms, factoring a polynomial means breaking it down into a product of simpler polynomials or expressions. Think of it like finding the prime factors of a number, but instead of numbers, we're dealing with algebraic expressions. For example, factoring x2โˆ’4{x^2 - 4} gives us (x+2)(xโˆ’2){(x + 2)(x - 2)}. This process is super useful for solving polynomial equations because if the product of several factors is zero, then at least one of those factors must be zero. This principle allows us to find the roots or solutions of the equation. Factoring polynomials can seem tricky at first, but with practice, you'll start to recognize patterns and develop strategies to tackle different types of polynomials. Whether it's finding the greatest common factor, using special factoring patterns, or employing more advanced techniques, mastering polynomial factorization is a significant step in your mathematical journey. So, let's get started and see how Tomas cracked the code!

Initial Factoring: Spotting the Greatest Common Factor

The journey of factoring the polynomial 3x4โˆ’18x3+9x2โˆ’54x{3x^4 - 18x^3 + 9x^2 - 54x} begins with a crucial first step: identifying the greatest common factor (GCF). What's the GCF? Well, it's the largest factor that divides evenly into all terms of the polynomial. In this case, let's carefully examine each term. We have 3x4{3x^4}, โˆ’18x3{-18x^3}, 9x2{9x^2}, and โˆ’54x{-54x}. Looking at the coefficients (3, -18, 9, and -54), we can see that the largest number that divides all of them is 3. Now, let's consider the variable part. Each term has at least one x{x}, so x{x} is also a common factor. Combining these, we find that the GCF is 3x{3x}. Factoring out 3x{3x} from the polynomial is like extracting a common ingredient from a recipe. We divide each term by 3x{3x} and write the result in parentheses:

3x4โˆ’18x3+9x2โˆ’54x=3x(x3โˆ’6x2+3xโˆ’18){ 3x^4 - 18x^3 + 9x^2 - 54x = 3x(x^3 - 6x^2 + 3x - 18) }

This step simplifies the polynomial and makes further factoring easier. It's like reducing a fraction to its simplest form before performing other operations. Factoring out the GCF is a fundamental technique, and it's always a good idea to start your factoring adventure this way. By doing so, you often reveal the underlying structure of the polynomial, making it easier to spot further patterns and apply other factoring methods. So, what's next? We've pulled out the 3x{3x}, but we're not done yet. We still have a cubic polynomial inside the parentheses, and we need to see if we can factor it further. This is where the fun really begins!

Factoring by Grouping: A Powerful Technique

Now that we've extracted the GCF, we're left with the cubic polynomial x3โˆ’6x2+3xโˆ’18{x^3 - 6x^2 + 3x - 18}. At this stage, a technique called factoring by grouping comes into play. Factoring by grouping is a nifty method particularly useful for polynomials with four terms. The idea behind it is to pair up terms, factor out common factors from each pair, and then see if a common binomial factor emerges. Let's apply this technique to our cubic polynomial. We can group the first two terms and the last two terms:

(x3โˆ’6x2)+(3xโˆ’18){ (x^3 - 6x^2) + (3x - 18) }

Now, we factor out the GCF from each group. From the first group, x3โˆ’6x2{x^3 - 6x^2}, the GCF is x2{x^2}. Factoring this out, we get x2(xโˆ’6){x^2(x - 6)}. From the second group, 3xโˆ’18{3x - 18}, the GCF is 3. Factoring this out, we get 3(xโˆ’6){3(x - 6)}. So, our expression now looks like this:

x2(xโˆ’6)+3(xโˆ’6){ x^2(x - 6) + 3(x - 6) }

Do you see the magic happening? We now have a common binomial factor: (xโˆ’6){(x - 6)}. This is the key to factoring by grouping! We factor out this common binomial, treating it as a single term:

(xโˆ’6)(x2+3){ (x - 6)(x^2 + 3) }

And there you have it! We've successfully factored the cubic polynomial into (xโˆ’6)(x2+3){(x - 6)(x^2 + 3)}. Factoring by grouping is a powerful tool in your factoring arsenal. It's especially effective when dealing with polynomials that don't fit the standard factoring patterns. By strategically grouping terms and factoring out common factors, you can often break down complex polynomials into simpler, manageable factors. But our journey doesn't end here. Remember, we initially factored out 3x{3x}, so we need to bring that back into the picture to get the complete factorization of the original polynomial. Let's see what the final factored form looks like.

The Grand Finale: The Completely Factored Polynomial

We've come a long way, guys! We started with the polynomial 3x4โˆ’18x3+9x2โˆ’54x{3x^4 - 18x^3 + 9x^2 - 54x}, factored out the GCF 3x{3x}, and then used factoring by grouping to break down the resulting cubic polynomial. Now, it's time to put all the pieces together and see the completely factored form. Remember, we factored out 3x{3x} in the first step, leaving us with 3x(x3โˆ’6x2+3xโˆ’18){3x(x^3 - 6x^2 + 3x - 18)}. Then, we factored the cubic polynomial x3โˆ’6x2+3xโˆ’18{x^3 - 6x^2 + 3x - 18} into (xโˆ’6)(x2+3){(x - 6)(x^2 + 3)}. So, to get the complete factorization, we simply combine these results:

3x4โˆ’18x3+9x2โˆ’54x=3x(xโˆ’6)(x2+3){ 3x^4 - 18x^3 + 9x^2 - 54x = 3x(x - 6)(x^2 + 3) }

This is the fully factored form of the polynomial. We've broken it down into a product of irreducible factors โ€“ factors that cannot be factored further using real numbers. The factor 3x{3x} is a monomial, (xโˆ’6){(x - 6)} is a linear binomial, and (x2+3){(x^2 + 3)} is an irreducible quadratic because it has no real roots (you can't find a real number that, when squared and added to 3, equals zero). Now, let's connect this back to the original question. We were asked what's true about Tomas's final product when he expresses the polynomial in the form Ax(x2+B)(x+C){Ax(x^2 + B)(x + C)}. By comparing our factored form 3x(xโˆ’6)(x2+3){3x(x - 6)(x^2 + 3)} with the given form Ax(x2+B)(x+C){Ax(x^2 + B)(x + C)}, we can identify the values of A{A}, B{B}, and C{C}. This comparison is the key to answering the question and understanding the relationship between the factors of the polynomial. So, let's make the comparison and see what we find!

Decoding the Final Product: Identifying A, B, and C

Alright, let's put on our detective hats and decode the final product! We have the completely factored form of the polynomial:

3x(xโˆ’6)(x2+3){ 3x(x - 6)(x^2 + 3) }

And we're trying to match it with the form Tomas obtained:

Ax(x2+B)(x+C){ Ax(x^2 + B)(x + C) }

It's like comparing two different languages โ€“ we need to translate between them to understand the meaning. Let's start by matching the terms. The factor 3x{3x} in our factored form corresponds to Ax{Ax} in Tomas's form. This tells us immediately that A=3{A = 3}. Next, we have the quadratic factor (x2+3){(x^2 + 3)} in our form, which corresponds to (x2+B){(x^2 + B)} in Tomas's form. This means that B=3{B = 3}. Finally, we have the linear factor (xโˆ’6){(x - 6)} in our form. Notice that Tomas's form has (x+C){(x + C)}. To match these, we need to recognize that (xโˆ’6){(x - 6)} is the same as (x+(โˆ’6)){(x + (-6))}. Therefore, C=โˆ’6{C = -6}. So, we've successfully identified the values: A=3{A = 3}, B=3{B = 3}, and C=โˆ’6{C = -6}. This is like finding the missing pieces of a puzzle. Now that we know these values, we can evaluate the given options and determine which statement about A{A}, B{B}, and C{C} is true. This is the final step in our journey, and it will bring us to the answer we've been searching for. So, let's examine the options and see which one fits our findings!

Evaluating the Options: What's the Truth?

We've cracked the code and found that A=3{A = 3}, B=3{B = 3}, and C=โˆ’6{C = -6}. Now, it's time to put these values to the test. We're given a set of statements about A{A}, B{B}, and C{C}, and our mission is to determine which one is the truth. Let's take a look at the statements and evaluate them one by one.

Option A states that A{A} and B{B} are both 6. But we found that A=3{A = 3} and B=3{B = 3}, so this statement is false. It's like trying to fit a square peg in a round hole โ€“ it just doesn't work.

Option B states that A{A} and C{C} are both 6. We know that A=3{A = 3} and C=โˆ’6{C = -6}, so this statement is also false. It's important to be precise in math, and this option misses the mark.

Now, let's consider Option C carefully. It might state something about the relationship between A{A}, B{B}, and C{C}, or perhaps give a different set of values. We need to compare this option with our findings and see if it matches. If Option C states, for example, that A{A} is 3 and B{B} is 3, then it would be a true statement. Similarly, if it states something like B{B} is the additive inverse of C{C}, that would also be true because 3 is the additive inverse of -6. The key is to carefully analyze the statement and see if it aligns with the values we've determined.

By systematically evaluating each option, we can pinpoint the statement that accurately reflects the values of A{A}, B{B}, and C{C}. This is the final step in solving the problem, and it's the moment where all our hard work pays off. So, make sure to pay close attention to the wording of each option and compare it with your results. The truth is out there, and we're about to find it!

Key Takeaways and the Importance of Factoring

We've successfully navigated the world of polynomial factorization, guys! We started with a seemingly complex polynomial, 3x4โˆ’18x3+9x2โˆ’54x{3x^4 - 18x^3 + 9x^2 - 54x}, and broke it down into its fundamental factors: 3x(xโˆ’6)(x2+3){3x(x - 6)(x^2 + 3)}. Along the way, we used key techniques like factoring out the greatest common factor and factoring by grouping. These are powerful tools in your mathematical arsenal, and mastering them can open doors to solving a wide range of problems. But more than just the mechanics of factoring, we've also seen the importance of careful analysis and attention to detail. We had to match the factored form with the given expression Ax(x2+B)(x+C){Ax(x^2 + B)(x + C)}, and that required a precise comparison of terms. A small error in identifying A{A}, B{B}, or C{C} could have led us to the wrong conclusion. This highlights the importance of double-checking your work and being meticulous in your calculations.

Factoring polynomials is not just an abstract mathematical exercise; it has real-world applications in various fields. For example, in engineering, factoring can be used to simplify equations that describe the behavior of structures and systems. In computer science, it can be used to optimize algorithms and solve problems in cryptography. And in economics, it can be used to model and analyze financial markets. So, the skills you develop in factoring polynomials are valuable not only in the classroom but also in the broader world. As you continue your mathematical journey, remember that factoring is a fundamental skill that will serve you well in many contexts. Keep practicing, keep exploring, and keep unlocking the secrets of mathematics! This problem is a testament to the power and elegance of factoring. By breaking down a complex expression into simpler components, we gain a deeper understanding of its structure and behavior. And that, guys, is what mathematics is all about!