Solving (3√5 - 2)(4√5 - 3) Step-by-Step

What's up, mathletes! Today, we're diving deep into the wonderful world of algebra to tackle a problem that might look a little intimidating at first glance: $(3 \sqrt{5}-2)(4 \sqrt{5}-3)=$. But don't you worry, by the end of this article, you'll be a radical expression-solving pro! We're going to break this down step-by-step, making sure every single one of you, no matter your math level, can follow along and feel super confident. So, grab your notebooks, maybe a snack, and let's get this math party started! We'll be using the trusty FOIL method, which is your best friend when multiplying binomials. Remember, FOIL stands for First, Outer, Inner, Last. It's a systematic way to ensure you multiply every term in the first set of parentheses by every term in the second set. It's like making sure everyone gets a handshake! We'll also be focusing on simplifying square roots, specifically $\sqrt{5}$, which, thankfully, is already in its simplest form. This means we won't need to do any complex simplification there, but it's always good to keep an eye out for perfect square factors within radicals. So, let's get ready to wield our mathematical tools and conquer this expression!

Unpacking the Problem: What Are We Actually Doing Here?

Alright guys, let's really understand what we're up against with this expression: $(3 \sqrt{5}-2)(4 \sqrt{5}-3)$. This is a multiplication of two binomials, and each binomial contains a term with a square root and a constant term. The core of this problem lies in understanding how to multiply these terms together, especially when a square root is involved. The key player here is $\sqrt{5}$. It's an irrational number, meaning it cannot be expressed as a simple fraction, and its decimal representation goes on forever without repeating. When we multiply terms involving square roots, we often apply the property that $\sqrt{a} \times \sqrt{b} = \sqrt{ab}$. Also, remember that $\sqrt{a} \times \sqrt{a} = a$. This is super important! For example, $\sqrt{5} \times \sqrt{5}$ will simply be $5$. This is because $\sqrt{5}$ is the number that, when multiplied by itself, gives you $5$. So, any time you see a term like $(\sqrt{x})^2$ or $\sqrt{x} \times \sqrt{x}$, it simplifies to just $x$. This is a fundamental rule in simplifying radical expressions. We also need to be comfortable multiplying a number by a radical, like $3 \times 4\sqrt{5}$. In this case, you just multiply the numbers outside the radical: $(3 \times 4)\sqrt{5} = 12\sqrt{5}$. The number under the square root, the radicand, stays the same unless further simplification is possible. Combining like terms is another crucial skill we'll be using. Like terms are terms that have the same variable raised to the same power, or in this case, terms that have the same radical part. So, you can add or subtract terms like $2\sqrt{5}$ and $7\sqrt{5}$ (which would give you $9\sqrt{5}$), but you can't directly combine $2\sqrt{5}$ and $3\sqrt{7}$ because the radical parts are different. Keep these fundamental rules in mind as we move forward. They are the building blocks for successfully navigating this type of algebraic problem.

The FOIL Method: Your Secret Weapon for Binomial Multiplication

Now, let's talk about the FOIL method, our go-to strategy for multiplying these binomials. FOIL is an acronym that stands for First, Outer, Inner, and Last. It’s a systematic way to ensure you multiply every single term in the first binomial by every single term in the second binomial. Missing even one of these multiplications can lead to an incorrect answer, so we gotta be diligent, guys!

Let's break down our expression $(3 \sqrt{5}-2)(4 \sqrt{5}-3)$ using FOIL:

• F (First): Multiply the first terms in each binomial. That's $(3 \sqrt{5})$ and $(4 \sqrt{5})$.

  • Calculation: $(3 \sqrt{5}) \times (4 \sqrt{5}) = (3 \times 4) \times (\sqrt{5} \times \sqrt{5})$.
  • We know $3 \times 4 = 12$, and as we discussed, $\sqrt{5} \times \sqrt{5} = 5$.
  • So, the first term simplifies to $12 \times 5 = 60$.

• O (Outer): Multiply the outer terms of the entire expression. That's $(3 \sqrt{5})$ and $(-3)$.

  • Calculation: $(3 \sqrt{5}) \times (-3) = (3 \times -3) \sqrt{5}$.
  • This gives us $-9\sqrt{5}$. Remember to include the negative sign!

• I (Inner): Multiply the inner terms of the expression. That's $(-2)$ and $(4 \sqrt{5})$.

  • Calculation: $(-2) \times (4 \sqrt{5}) = (-2 \times 4) \sqrt{5}$.
  • This results in $-8\sqrt{5}$. Again, watch that negative sign.

• L (Last): Multiply the last terms in each binomial. That's $(-2)$ and $(-3)$.

  • Calculation: $(-2) \times (-3) = 6$.
  • A negative times a negative equals a positive, so we get $+6$.

So, after applying FOIL, we have the sum of these four results: $60 - 9\sqrt{5} - 8\sqrt{5} + 6$. This is where the next crucial step comes in: combining like terms. You can see we have two terms with $\sqrt{5}$ and two constant terms. We'll tackle that next!

Combining Like Terms: Simplifying the Expression

Okay, team, we've successfully used the FOIL method and arrived at this expression: $60 - 9\sqrt{5} - 8\sqrt{5} + 6$. Now, it's time to simplify by combining like terms. Think of it like sorting your LEGO bricks – you group the red ones together, the blue ones together, and so on. In our case, the 'like terms' are the constant numbers ($60$ and $6$) and the terms with the $\sqrt{5}$ part ($-9\sqrt{5}$ and $-8\sqrt{5}$).

Let's group them:

• Constant Terms: We have $60$ and $+6$. Adding these together gives us $60 + 6 = 66$.

• Radical Terms: We have $-9\sqrt{5}$ and $-8\sqrt{5}$. To combine these, we just add or subtract the coefficients (the numbers in front of the $\sqrt{5}$). So, we do $-9 - 8$. Since both numbers are negative, we add their absolute values and keep the negative sign: $-9 - 8 = -17$. Therefore, $-9\sqrt{5} - 8\sqrt{5} = -17\sqrt{5}$.

Now, we put our combined terms back together. The simplified expression is the sum of our combined constant terms and our combined radical terms:

$66 - 17\sqrt{5}$

And there you have it! We've taken a seemingly complex expression and simplified it into a much more manageable form. This process highlights the power of systematic approaches like FOIL and the importance of understanding how to combine like terms, especially when dealing with radicals. The final answer, $66 - 17\sqrt{5}$, is in its simplest form because $66$ is a rational number and $-17\sqrt{5}$ is an irrational number, and you can't combine them further. Pretty neat, right?

Final Answer and Key Takeaways

So, after all that hard work, we’ve arrived at the final answer for $(3 \sqrt{5}-2)(4 \sqrt{5}-3)=$. Drumroll, please... it's $66 - 17\sqrt{5}$!

Let's quickly recap what we did, guys:

1. Identified the Operation: We recognized this as the multiplication of two binomials.
2. Applied the FOIL Method: We systematically multiplied the First, Outer, Inner, and Last terms.
   • $(3 \sqrt{5} \times 4 \sqrt{5}) = 60$
   • $(3 \sqrt{5} \times -3) = -9\sqrt{5}$
   • $(-2 \times 4 \sqrt{5}) = -8\sqrt{5}$
   • $(-2 \times -3) = 6$
3. Combined Like Terms: We grouped and added the constant terms ($60 + 6 = 66$) and the radical terms ($-9\sqrt{5} - 8\sqrt{5} = -17\sqrt{5}$).
4. Stated the Final Simplified Answer: Putting it all together, we got $66 - 17\sqrt{5}$.

Key Takeaways for Your Math Toolkit:

• FOIL is Your Friend: Always use the FOIL method (or a similar distributive property approach) when multiplying binomials. It ensures you don't miss any terms.
• Radical Rules: Remember that $\sqrt{a} \times \sqrt{a} = a$. This is crucial for simplifying terms involving square roots.
• Combine Like Terms: Only terms with the exact same variable part (or radical part) can be combined. You can combine constants with constants, and terms like $a\sqrt{b}$ with other terms like $c\sqrt{b}$.
• Simplify Radicals: While $\sqrt{5}$ was already simple here, always check if the number under the square root has any perfect square factors that can be pulled out (e.g., $\sqrt{12} = \sqrt{4 \times 3} = 2\sqrt{3}$).

Mastering these steps will not only help you solve this specific problem but will build a strong foundation for tackling more complex algebraic expressions involving radicals. Keep practicing, keep questioning, and you'll become a math whiz in no time! You guys got this!