Easy Math: Multiply Binomials $(2x+y)(3x+3y)$

Hey math whizzes and curious minds! Today, we're diving into the super fun world of algebra to tackle a common task: multiplying binomials. Specifically, we're going to break down how to multiply $(2x+y)$ by $(3x+3y)$. Don't let the letters and numbers get you down, guys. It's all about following a few simple steps, and once you get the hang of it, you'll be multiplying these expressions like a pro! We'll cover the distributive property, which is your best friend in these situations, and walk through the calculation step-by-step. We'll also touch on simplifying the resulting expression, because nobody likes a messy equation. So, grab your calculators (or just your brains!), and let's get this algebra party started!

Understanding the Basics: What are Binomials and Why Multiply Them?

Alright guys, before we jump into the actual multiplication, let's quickly chat about what we're working with. You've got binomials, which are basically algebraic expressions with two terms. Think of $(2x+y)$ – that's one binomial with the terms $2x$ and $y$. Similarly, $(3x+3y)$ is another binomial with terms $3x$ and $3y$. Now, why do we even bother multiplying these things? Well, in math, especially in algebra, multiplying expressions is a fundamental skill. It's how we expand equations, simplify complex problems, and build the foundation for more advanced topics like factoring and solving quadratic equations. It's like learning to add before you can do calculus, you know? When you multiply two binomials, you're essentially distributing each term in the first binomial to every term in the second binomial. This ensures that every part of one expression interacts with every part of the other, giving you the complete expanded form. We'll be using a trusty method called the distributive property, often remembered by the acronym FOIL (First, Outer, Inner, Last), which is a neat trick to make sure you don't miss any terms. So, keep that in mind as we move forward, because it's the key to unlocking this problem.

The Distributive Property: Your Secret Weapon for Multiplication

So, let's talk about the distributive property, which is the absolute MVP when it comes to multiplying expressions like our binomials $(2x+y)$ and $(3x+3y)$. This property basically says that if you have a term outside a set of parentheses multiplied by terms inside, you distribute that outside term to each inside term. For example, $a(b+c) = ab + ac$. Now, when we're multiplying two binomials, say $(a+b)(c+d)$, we apply this property twice! Think of it like this: you take the first binomial $(a+b)$ and treat it as a single unit that needs to be distributed to both terms in the second binomial $(c+d)$. So, you'd have $(a+b) imes c$ and $(a+b) imes d$. See? We've distributed the $(a+b)$ unit. Now, for each of those parts, we apply the distributive property again. So, $(a+b) imes c$ becomes $ac + bc$, and $(a+b) imes d$ becomes $ad + bd$. Put it all together, and you get $ac + bc + ad + bd$.

This is where the handy FOIL acronym comes in! FOIL is just a mnemonic device to help you remember the order in which you multiply the terms:

• First: Multiply the first terms in each binomial.
• Outer: Multiply the outer terms of the expression.
• Inner: Multiply the inner terms of the expression.
• Last: Multiply the last terms in each binomial.

Using FOIL ensures you've covered all the combinations. We'll use this exact method to tackle our specific problem, $(2x+y)(3x+3y)$, in the next section. It's a systematic way to make sure no term gets left out, and that's crucial for getting the correct answer, guys. So, get ready to put this property to work!

Step-by-Step: Calculating $(2x+y)(3x+3y)$

Alright team, let's get our hands dirty and actually multiply $(2x+y)$ by $(3x+3y)$ using the distributive property, or our trusty FOIL method. This is where all the prep work pays off! We're going to go through it one step at a time, so follow along closely. Remember, FOIL stands for First, Outer, Inner, Last.

1. First: Multiply the first term of the first binomial by the first term of the second binomial. Our first terms are $2x$ and $3x$. So, $(2x) imes (3x) = 6x^2$. Remember that when you multiply variables with exponents, you add the exponents. Here, $x$ is $x^1$, so $x^1 imes x^1 = x^{1+1} = x^2$.

2. Outer: Multiply the outer term of the first binomial by the outer term of the second binomial. Our outer terms are $2x$ (from the first binomial) and $3y$ (from the second binomial). So, $(2x) imes (3y) = 6xy$. When multiplying variables that are different, you just write them next to each other.

3. Inner: Multiply the inner term of the first binomial by the inner term of the second binomial. Our inner terms are $y$ (from the first binomial) and $3x$ (from the second binomial). So, $(y) imes (3x) = 3xy$. Again, we just combine the variables.

4. Last: Multiply the last term of the first binomial by the last term of the second binomial. Our last terms are $y$ and $3y$. So, $(y) imes (3y) = 3y^2$. Just like with the $x$'s, $y imes y = y^2$.

Now, we take all these results and add them together: $6x^2 + 6xy + 3xy + 3y^2$

See? We've successfully multiplied the two binomials. But we're not quite done yet! The next crucial step is to simplify this expression. We've got a couple of terms that look pretty similar, and that means we can combine them. Keep reading to find out how!

Simplifying the Result: Combining Like Terms

Alright guys, we've done the hard part of multiplying $(2x+y)(3x+3y)$ and we ended up with $6x^2 + 6xy + 3xy + 3y^2$. Now, we need to make this expression as neat and tidy as possible by combining like terms. What are like terms, you ask? They are terms that have the exact same variables raised to the exact same powers. In our expression, $6x^2$ has an $x^2$, $3y^2$ has a $y^2$, and both $6xy$ and $3xy$ have an $xy$. That means the $6xy$ and the $3xy$ are our like terms because they both have the variables $x$ and $y$ raised to the power of 1 (which we don't usually write).

To combine like terms, you simply add or subtract their coefficients (the numbers in front of the variables). So, for our $xy$ terms: $6xy + 3xy = (6+3)xy = 9xy$

The terms $6x^2$ and $3y^2$ don't have any other like terms to combine with, so they stay as they are.

Now, we put our simplified terms back together in a standard order (usually descending powers of one variable, like $x$, then moving to other variables). Our final, simplified expression is:

$6x^2 + 9xy + 3y^2$

And there you have it! We've successfully multiplied and simplified the expression. This is the final answer, and it's in its simplest form. Pretty neat, right? It's all about breaking it down, using the distributive property or FOIL, and then tidying up. You guys are algebra rockstars!

Practice Makes Perfect: More Examples and Tips

So, you've conquered $(2x+y)(3x+3y)$, which is awesome! But like they say, practice makes perfect, especially in math. The more you practice multiplying binomials, the more natural it will feel, and the faster you'll get. Let's quickly look at another example to solidify your understanding.

Imagine we need to multiply $(x+2)(x+5)$. Using FOIL:

• First: $x imes x = x^2$
• Outer: $x imes 5 = 5x$
• Inner: $2 imes x = 2x$
• Last: $2 imes 5 = 10$

Adding them up: $x^2 + 5x + 2x + 10$.

Now, we combine like terms. The $5x$ and $2x$ are like terms: $5x + 2x = 7x$.

So, the simplified answer is: $x^2 + 7x + 10$.

See how that works? It's the same process. Remember these key tips, guys:

• Pay Attention to Signs: Be extra careful with negative signs. Multiplying a positive by a negative results in a negative, and multiplying two negatives results in a positive. This is a common place where mistakes happen.
• Double-Check Your Work: After you've done the multiplication and simplification, take a moment to quickly review your steps. Did you multiply everything correctly? Did you combine the right terms?
• Don't Rush: It's better to take your time and get the correct answer than to rush and make silly errors. Math is a marathon, not a sprint!
• Understand the 'Why': Knowing that you're using the distributive property helps you understand why you're doing each step, rather than just memorizing FOIL. This deeper understanding is crucial for tackling more complex problems down the line.

Keep practicing, and don't be afraid to try different combinations of terms. You've got this!

Conclusion: You've Mastered Binomial Multiplication!

And that, my friends, is how you multiply binomials! We broke down how to multiply $(2x+y)(3x+3y)$, starting with understanding what binomials are and why multiplication is so important in algebra. We explored the power of the distributive property and the handy FOIL acronym. Then, we walked through the step-by-step calculation, multiplying each term to get $6x^2 + 6xy + 3xy + 3y^2$, and finally simplified it by combining like terms to reach our answer: $6x^2 + 9xy + 3y^2$.

Remember, math is all about building skills step-by-step. The more you practice these types of problems, the more confident and capable you'll become. Whether you're tackling homework, preparing for a test, or just enjoy the mental workout, mastering binomial multiplication is a fantastic achievement. So give yourselves a pat on the back, guys! You've done a great job today. Keep exploring, keep practicing, and never stop learning. Happy calculating!