Find The Xy^2 Term Coefficient In (x-2y)^3

Hey math whizzes and future mathematicians! Today, we're diving deep into the exciting world of binomial expansions. If you've ever wondered how to break down a seemingly complex expression like $(x-2y)^3$ and pinpoint specific terms, you're in the right place. We're going to tackle a classic problem: finding the coefficient of the $xy^2$ term. This isn't just about memorizing formulas; it's about understanding the patterns and logic behind algebraic expansion. So, grab your thinking caps, guys, because we're about to unravel this challenge step-by-step.

The Power of the Binomial Theorem

Alright, let's talk about the star of the show: the Binomial Theorem. This theorem is your ultimate best friend when dealing with expressions raised to a power, especially when those powers get larger. For an expression of the form $(a+b)^n$, the theorem gives us a systematic way to expand it. The general term in this expansion is given by the formula: $ T_{r+1} = inom{n}{r} a^{n-r} b^r $ Here, inom{n}{r} represents the binomial coefficient, which is calculated as rac{n!}{r!(n-r)!}. The term $a^{n-r}$ represents the power of the first element in the binomial, and $b^r$ represents the power of the second element. The beauty of this formula is that it allows us to find any term in the expansion without having to write out the entire thing. For our specific problem, we have $(x-2y)^3$. Here, $a = x$, $b = -2y$, and $n = 3$. Our mission is to find the term where the powers of $x$ and $y$ combine to give us $xy^2$. According to the general term formula, we need the power of $a$ (which is $x$) to be 1 and the power of $b$ (which is $-2y$) to be 2, so that when we multiply them, we get $x^1 imes (-2y)^2 = x imes 4y^2 = 4xy^2$. This means we are looking for the term where $n-r = 1$ and $r = 2$. Since $n=3$, both conditions are met: $3-2 = 1$ and $r=2$. So, we can plug these values into the general term formula to find our specific term.

Calculating the Specific Term

Now that we know which term we're after, let's plug the values into our general term formula: $ T_r+1} = inom{n}{r} a^{n-r} b^r $ We've established that $n=3$, $r=2$, $a=x$, and $b=-2y$. So, the term is $ T_{2+1 = T_3 = inom{3}{2} (x)^{3-2} (-2y)^2 $ Let's break this down. First, we calculate the binomial coefficient inom{3}{2}. This is given by rac{3!}{2!(3-2)!} = rac{3!}{2!1!} = rac{3 imes 2 imes 1}{(2 imes 1)(1)} = rac{6}{2} = 3. Next, we look at the powers of $a$ and $b$. We have $(x)^{3-2} = x^1 = x$. And for the second part, we have $(-2y)^2 = (-2)^2 imes y^2 = 4y^2$. Now, we multiply all these parts together: $ T_3 = 3 imes x imes 4y^2 $ $ T_3 = 12xy^2 $ And there you have it! The term containing $xy^2$ in the expansion of $(x-2y)^3$ is $12xy^2$. The question specifically asks for the coefficient of this term. The coefficient is the numerical factor that multiplies the variables. In this case, the coefficient of $xy^2$ is 12. So, out of the options provided (A. 3, B. 6, C. 12, D. 2), the correct answer is C. 12. Pretty cool, right? It's all about following the pattern!

Alternative Approach: Direct Expansion

While the Binomial Theorem is super efficient, sometimes it's helpful to see the expansion directly, especially for smaller powers like 3. This can reinforce your understanding and show you why the theorem works. Expanding $(x-2y)^3$ means multiplying $(x-2y)$ by itself three times: $(x-2y)(x-2y)(x-2y)$. Let's do this in stages. First, multiply the first two terms: $ (x-2y)(x-2y) = x(x-2y) - 2y(x-2y) $ $ = x^2 - 2xy - 2xy + 4y^2 $ $ = x^2 - 4xy + 4y^2 $ Now, we take this result and multiply it by the remaining $(x-2y)$: $ (x^2 - 4xy + 4y^2)(x-2y) $ We distribute each term in the first polynomial to each term in the second: $ x(x^2 - 4xy + 4y^2) - 2y(x^2 - 4xy + 4y^2) $ Let's expand the first part: $ x^3 - 4x^2y + 4xy^2 $ Now, expand the second part: $ -2yx^2 + 8xy^2 - 8y^3 $ Remember, $yx^2$ is the same as $x^2y$. So, the second part becomes: $ -2x^2y + 8xy^2 - 8y^3 $ Now, we combine all the terms: $ x^3 - 4x^2y + 4xy^2 - 2x^2y + 8xy^2 - 8y^3 $ Let's group like terms. We have $x^3$ (only one), $x^2y$ terms, $xy^2$ terms, and $y^3$ terms: $ x^3 + (-4x^2y - 2x^2y) + (4xy^2 + 8xy^2) - 8y^3 $ Combine the coefficients for each group: $ x^3 - 6x^2y + 12xy^2 - 8y^3 $ Look at that! The full expansion is $x^3 - 6x^2y + 12xy^2 - 8y^3$. When we look for the term with $xy^2$, we find it right there: $12xy^2$. And just like before, the coefficient is 12. This direct expansion method really shows you how all the terms come together. It's a bit more work for higher powers, but for $n=3$, it's a solid way to check your understanding and verify the Binomial Theorem's results. You can see how the terms $4xy^2$ and $8xy^2$ from the distribution steps combine to give the final $12xy^2$. It's all about careful multiplication and combining like terms, guys!

Understanding the Structure of Binomial Expansions

Let's really drill down into why these coefficients and powers work the way they do. When you expand $(a+b)^n$, you're essentially choosing either 'a' or 'b' from each of the 'n' binomial factors and multiplying them together. For $(x-2y)^3$, we have three factors of $(x-2y)$. To get a term with $x^1 y^2$, we need to pick 'x' from one of the factors and '-2y' from the other two factors. How many ways can we choose which factor we pick 'x' from? We have 3 factors, and we need to choose 1 of them to contribute the 'x'. This is a combination problem, denoted as inom{3}{1}, which equals 3. Alternatively, we could think about choosing the '-2y' terms. We need to pick '-2y' from two of the three factors. The number of ways to do this is inom{3}{2}, which also equals 3. This is why the binomial coefficient inom{n}{r} appears – it counts the number of distinct ways to obtain a specific combination of terms from the factors. Once we know there are 3 ways to get a term with one 'x' and two '-2y's, we need to figure out what that term actually is. We pick 'x' once, so we get $x^1$. We pick '-2y' twice, so we get $(-2y)^2 = 4y^2$. Multiplying these together gives $x imes 4y^2 = 4xy^2$. Since there are 3 such combinations, the total term is $3 imes (4xy^2) = 12xy^2$. The coefficient is indeed 12. This combinatorial perspective is fundamental to understanding the Binomial Theorem. It connects algebra with counting principles, showing that the coefficients aren't arbitrary numbers but arise from the very structure of the expansion process. Every term in a binomial expansion $(a+b)^n$ is formed by selecting 'a' from $n-r$ factors and 'b' from $r$ factors, and the binomial coefficient inom{n}{r} precisely counts how many ways this selection can occur. This concept is super powerful and applies to much more complex scenarios in combinatorics and probability, guys!

Final Thoughts and Review

So, to wrap things up, we've explored how to find the coefficient of the $xy^2$ term in the expansion of $(x-2y)^3$. We used the powerful Binomial Theorem, calculating the specific term using the formula T_{r+1} = inom{n}{r} a^{n-r} b^r, where $n=3$, $r=2$, $a=x$, and $b=-2y$. This gave us inom{3}{2} x^{3-2} (-2y)^2 = 3 imes x imes 4y^2 = 12xy^2. We also confirmed this result by directly expanding $(x-2y)^3$, which yielded $x^3 - 6x^2y + 12xy^2 - 8y^3$. Both methods clearly show that the coefficient of the $xy^2$ term is 12. This reinforces the idea that understanding fundamental algebraic principles like binomial expansion is crucial for problem-solving. Keep practicing these concepts, guys, because the more you work with them, the more intuitive they become. The options provided were A. 3, B. 6, C. 12, and D. 2. Our calculated coefficient is 12, which matches option C. Fantastic work, everyone!