Factoring By Grouping: Solve X³ - 9x² + 5x - 45

Hey guys! Today, we're diving deep into the awesome world of algebra, specifically tackling a problem that involves factoring polynomials by grouping. This is a super useful technique when you're faced with a polynomial that has four terms, and you're not quite sure where to start. Our mission, should we choose to accept it, is to figure out one way to determine the factors of $x^3 - 9x^2 + 5x - 45$ by grouping. We'll break down each step, explore the options, and make sure you understand exactly how this method works. So, grab your notebooks, and let's get this math party started! Understanding how to factor polynomials is a fundamental skill in mathematics, opening doors to solving more complex equations, simplifying expressions, and graphing functions. The method of factoring by grouping is particularly elegant because it allows us to break down a larger problem into smaller, more manageable pieces. When we see a four-term polynomial like the one presented, $x^3 - 9x^2 + 5x - 45$, our first thought should be, "Can I use grouping here?" The beauty of grouping is that it leverages the distributive property in reverse. We look for common factors within pairs of terms. This isn't just about finding any factors; it's about finding a common binomial factor that emerges after we've grouped the terms. The goal is to manipulate the expression so that a shared parenthesis appears, which we can then factor out. This process simplifies the polynomial significantly, making it easier to find its roots or further factor it if possible. We'll be examining a specific example to illustrate this, and by the end, you'll be a pro at spotting when and how to apply this handy algebraic trick.

Understanding the Problem: Factoring $x^3 - 9x^2 + 5x - 45$

So, we're staring down the barrel of the polynomial $x^3 - 9x^2 + 5x - 45$. Our main keyword here is factoring by grouping. This method is a go-to strategy when you have a polynomial with four terms, just like this one. The core idea behind factoring by grouping is to divide the polynomial into two pairs of terms and then factor out the greatest common factor (GCF) from each pair. If you've done it correctly, you'll notice that the remaining binomial factor in each pair is identical. This identical binomial is the key to completing the factorization. Let's look at our expression: $x^3 - 9x^2 + 5x - 45$. We need to see how we can group these terms to reveal a common factor. Typically, we group the first two terms and the last two terms. So, we'd consider $(x^3 - 9x^2)$ and $(5x - 45)$. The first step is to find the GCF of the first pair. The GCF of $x^3$ and $-9x^2$ is $x^2$. Factoring this out, we get $x^2(x - 9)$. Now, we move to the second pair: $(5x - 45)$. The GCF here is 5. Factoring this out, we get $5(x - 9)$. Notice something super cool? We ended up with the same binomial factor, $(x - 9)$, in both parts! This is exactly what we want when using the factoring by grouping method. If we hadn't gotten the same binomial, we might need to try grouping the terms differently or check our calculations. But in this case, we've hit the jackpot. The expression now looks like $x^2(x - 9) + 5(x - 9)$. Since $(x - 9)$ is a common factor to both terms, we can now factor it out. Think of $(x - 9)$ as a single 'chunk'. We have $x^2$ chunks and 5 chunks, making a total of $(x^2 + 5)$ chunks. Therefore, the factored form of the polynomial is $(x^2 + 5)(x - 9)$. This process demonstrates how factoring by grouping simplifies a complex-looking polynomial into a product of simpler expressions. It’s all about recognizing patterns and applying the distributive property in reverse. The success of this method hinges on creating that common binomial factor, which acts as a bridge to connect the two parts of the polynomial.

Exploring the Options: Which Grouping Works?

Alright, so we know the goal is to factor $x^3 - 9x^2 + 5x - 45$ using grouping. We've already done the heavy lifting in our heads (or on scratch paper!), and we know the correct factored form should lead to a common binomial factor. Let's scrutinize the given options to see which one shows one way to determine the factors by grouping correctly. Remember, the essence of factoring by grouping is to split the four terms into two pairs, factor out the GCF from each pair, and end up with the same binomial factor in both results.

Option A: $x^2(x-9)-5(x-9)$

Let's analyze this option. It presents two terms: $x^2(x-9)$ and $-5(x-9)$. The binomial factor here is clearly $(x-9)$. This looks promising! If we were to 'un-group' this, we'd distribute the $-5$ to the $(x-9)$, giving us $-5x + 45$. So, this option represents the expression $x^3 - 9x^2 - 5x + 45$. Wait a minute! Our original polynomial was $x^3 - 9x^2 + 5x - 45$. The sign of the $5x$ term and the constant term are different. This means Option A, while it looks like it might be a step in factoring, doesn't actually represent the original polynomial after factoring by grouping. However, the question asks which shows one way to determine the factors by grouping. This implies we might be looking at an intermediate step. Let's reconsider the original polynomial: $x^3 - 9x^2 + 5x - 45$. If we group the first two terms and the last two terms, we get $(x^3 - 9x^2) + (5x - 45)$. Factoring out the GCF from the first pair gives $x^2(x - 9)$. Factoring out the GCF from the second pair (which is 5) gives $5(x - 9)$. So we have $x^2(x - 9) + 5(x - 9)$. Now, let's look at Option A again: $x^2(x-9)-5(x-9)$. This option seems to be suggesting a scenario where the second GCF was factored out with a negative sign, leading to $(x^2 - 5)(x - 9)$. If we expand this, we get $x^2(x-9) - 5(x-9) = x^3 - 9x^2 - 5x + 45$. This is not our original polynomial. The question is a bit tricky here. It asks for one way to determine the factors. Let's assume the options represent potential first steps or intermediate stages. If we consider the original polynomial $x^3 - 9x^2 + 5x - 45$, and we group it as $(x^3 - 9x^2) + (5x - 45)$, we factor out $x^2$ from the first pair to get $x^2(x-9)$, and we factor out $+5$ from the second pair to get $+5(x-9)$. This leads to $x^2(x-9) + 5(x-9)$. Option A has $-5(x-9)$. This would only happen if the original polynomial was $x^3 - 9x^2 - 5x + 45$. Since the question is asking about determining factors by grouping, and Option A presents a structure that could arise from grouping (even if it doesn't match the exact original polynomial's signs), we need to be careful. Let's re-examine the prompt and options. The prompt is asking which shows one way to determine the factors of $x^3-9 x^2+5 x-45$ by grouping?. This implies the expression shown in the option should be a result of applying the grouping method to the given polynomial. Our correct intermediate step was $x^2(x-9) + 5(x-9)$. Option A is $x^2(x-9) - 5(x-9)$. This doesn't match. Let's hold off on concluding about A and look at others.

Option B: $x^2(x+9)-5(x+9)$

This option presents the binomial factor $(x+9)$. Let's see if this could arise from our original polynomial. If we had a term like $x^3 + 9x^2$, factoring out $x^2$ would give $x^2(x+9)$. However, our polynomial has $-9x^2$. So, this option immediately seems incorrect because the signs within the binomial don't align with the original polynomial's structure when factoring by grouping in the standard way. The binomials here are $(x+9)$, not $(x-9)$. This wouldn't be the result of factoring $x^3 - 9x^2$ or $5x - 45$ with the appropriate GCFs. So, Option B is definitely out.

Option C: $x ext{}(x^2+5 ext{)}-9 ext{}(x^2+5 ext{})$

This option has the common binomial factor $(x^2+5)$. Let's see if this could be derived from our polynomial $x^3 - 9x^2 + 5x - 45$. To get $x(x^2+5)$, we would need $x^3 + 5x$. This matches the first and third terms of our polynomial. To get $-9(x^2+5)$, we would need $-9x^2 - 45$. This matches the second and fourth terms of our polynomial. So, if we group the polynomial as $(x^3 + 5x) + (-9x^2 - 45)$, and factor out $x$ from the first pair, we get $x(x^2+5)$. If we factor out $-9$ from the second pair, we get $-9(x^2+5)$. Putting it together, we get $x(x^2+5) - 9(x^2+5)$. This perfectly matches Option C! This is a valid way to group and factor the polynomial, showing a different grouping strategy than the one we first considered (grouping first two and last two). This option represents the polynomial $x^3 + 5x - 9x^2 - 45$, which when reordered is $x^3 - 9x^2 + 5x - 45$. Bingo! This option correctly shows an intermediate step in factoring the polynomial by grouping.

Option D: $x ext{}(x^2-5 ext{)}-9 ext{}(x^2-5 ext{})$

This option also has a common binomial factor, $(x^2-5)$. Let's check if this can be formed from our original polynomial $x^3 - 9x^2 + 5x - 45$. If we factor $x$ from some terms, we'd expect $x^3$ and $5x$. If we factor $-9$ from some terms, we'd expect $-9x^2$ and $-45$. If we try to get $(x^2-5)$, it means we are looking for factors like $x(x^2-5) = x^3 - 5x$ and $-9(x^2-5) = -9x^2 + 45$. Neither of these pairs matches the terms in our original polynomial $x^3 - 9x^2 + 5x - 45$. Therefore, Option D is not a correct representation of factoring our polynomial by grouping.

The Solution: Identifying the Correct Grouping

After carefully examining each option, we found that Option C: $x(x^2+5)-9(x^2+5)$ is the one that correctly shows a way to determine the factors of $x^3 - 9x^2 + 5x - 45$ by grouping. Let's recap why this is the case. The method of factoring by grouping involves taking a polynomial with four terms and splitting it into two pairs. We then factor out the greatest common factor (GCF) from each pair. The crucial part is that the remaining binomial factor must be the same for both pairs. In Option C, we see the common binomial factor $(x^2+5)$. To arrive at this form, the original polynomial must have been grouped differently than the most common approach (first two terms and last two terms). Instead, the grouping implied by Option C is $(x^3 + 5x) + (-9x^2 - 45)$.

Let's verify this by working forwards:

1. Original Polynomial: $x^3 - 9x^2 + 5x - 45$
2. Rearrange terms (optional but helpful for this grouping): $x^3 + 5x - 9x^2 - 45$
3. Group the terms: $(x^3 + 5x) + (-9x^2 - 45)$
4. Factor out the GCF from the first group: The GCF of $x^3$ and $5x$ is $x$. So, $x(x^2 + 5)$.
5. Factor out the GCF from the second group: The GCF of $-9x^2$ and $-45$ is $-9$. So, $-9(x^2 + 5)$.
6. Combine the factored groups: $x(x^2 + 5) - 9(x^2 + 5)$

This resulting expression, $x(x^2+5)-9(x^2+5)$, is exactly what is presented in Option C. It clearly shows the common binomial factor $(x^2+5)$ emerging from the grouping process. From this point, we would factor out the common binomial $(x^2+5)$ to get the final factored form: $(x^2+5)(x-9)$. This step-by-step breakdown confirms that Option C is the correct intermediate representation derived from applying the factoring by grouping technique to the given polynomial. It's a fantastic example of how there can be different ways to group terms, but the underlying principle of finding a common binomial factor remains the same.

Conclusion: Mastering Factoring by Grouping

So, there you have it, guys! We've successfully navigated the process of factoring $x^3 - 9x^2 + 5x - 45$ by grouping. The key takeaway is that factoring by grouping is a powerful tool for polynomials with four terms. It involves strategic pairing of terms and factoring out GCFs to reveal a common binomial factor. We saw that while the most intuitive grouping might be the first two and last two terms, sometimes rearranging the terms or considering different pairings can also lead to a valid factorization, as demonstrated by Option C. Remember, the goal is always to find that identical binomial that allows you to consolidate the expression into a product of simpler polynomials. In this case, Option C, $x(x^2+5)-9(x^2+5)$, beautifully illustrates one such valid intermediate step. It shows how, by grouping $(x^3 + 5x)$ and $(-9x^2 - 45)$, we can extract common factors to arrive at the desired structure. This method not only helps in solving polynomial equations but also in simplifying complex algebraic expressions. Keep practicing, and you'll soon be spotting these patterns like a pro! The ability to factor polynomials efficiently is a cornerstone of algebraic manipulation, and understanding techniques like grouping provides you with a versatile set of tools. Keep an eye out for those four-term polynomials – they're often prime candidates for this elegant factoring method. Happy factoring!