Simplify 3a(-3a^2 - 9) Algebraically

by Andrew McMorgan 37 views

Hey guys! Today, we're diving into the fascinating world of algebra to tackle a cool expression: 3a(βˆ’3a2βˆ’9)3a\left(-3a^2-9\right). If you're looking to sharpen your algebraic skills, especially when it comes to simplifying expressions, you've come to the right place. We'll break down this problem step-by-step, making sure you understand every bit of the process. So, grab your calculators, your notebooks, and let's get this done!

Understanding the Expression: 3a(βˆ’3a2βˆ’9)3a\left(-3a^2-9\right)

Alright, let's start by looking at the expression we need to simplify: 3a(βˆ’3a2βˆ’9)3a\left(-3a^2-9\right). This expression involves a term outside the parentheses, 3a3a, and an expression inside the parentheses, βˆ’3a2βˆ’9-3a^2-9. The key operation here is multiplication, indicated by the parentheses. When you see a term right next to parentheses like this, it means you need to distribute that outside term to each term inside the parentheses. This is a fundamental concept in algebra, often referred to as the distributive property. It's like sharing a pizza – the person outside has to give a slice to everyone inside. In our case, 3a3a has to multiply with βˆ’3a2-3a^2 and also with βˆ’9-9.

Before we jump into the multiplication, let's quickly review some basic rules of exponents and multiplication with signed numbers, because these are crucial for getting the simplification right. When you multiply terms with the same base, you add their exponents. For example, a1Γ—a2=a1+2=a3a^1 \times a^2 = a^{1+2} = a^3. Also, remember the rules for multiplying signs: a positive number times a positive number is positive (+Γ—+=++\times+=+), a negative number times a negative number is positive (βˆ’Γ—βˆ’=+-\times- = +), a positive number times a negative number is negative (+Γ—βˆ’=βˆ’+\times- = -), and a negative number times a positive number is negative (βˆ’Γ—+=βˆ’-\times+ = -). Keeping these rules in mind will help us avoid silly mistakes as we move forward.

Now, let's focus on the terms themselves. We have 3a3a, which is 3Γ—a13 \times a^1. Inside the parentheses, we have βˆ’3a2-3a^2, which is βˆ’3Γ—a2-3 \times a^2, and βˆ’9-9, which is just a constant term. Our goal is to apply the distributive property to combine these terms and present the expression in its simplest polynomial form, which typically means having terms ordered by their exponents from highest to lowest. This form is often called the standard form of a polynomial. So, we're not just multiplying; we're also aiming for a neat, organized final answer. Let's get ready to perform that multiplication!

Applying the Distributive Property

Now for the fun part, guys: applying the distributive property to 3a(βˆ’3a2βˆ’9)3a\left(-3a^2-9\right). Remember, we need to multiply the term outside the parentheses, 3a3a, by each term inside the parentheses. So, we'll do two separate multiplications:

  1. Multiply 3a3a by βˆ’3a2-3a^2: This is our first step. Here, we're multiplying (3Γ—a1)(3 \times a^1) by (βˆ’3Γ—a2)(-3 \times a^2). Let's break this down:

    • Multiply the coefficients: The coefficients are the numbers in front of the variables. We have 33 and βˆ’3-3. So, 3Γ—(βˆ’3)=βˆ’93 \times (-3) = -9.
    • Multiply the variables: We have a1a^1 and a2a^2. Using the exponent rule for multiplication (add the exponents), we get a1Γ—a2=a1+2=a3a^1 \times a^2 = a^{1+2} = a^3.
    • Combine the results: Putting it all together, 3aΓ—(βˆ’3a2)=βˆ’9a33a \times (-3a^2) = -9a^3.
  2. Multiply 3a3a by βˆ’9-9: This is our second multiplication. Here, we're multiplying (3Γ—a1)(3 \times a^1) by (βˆ’9)(-9).

    • Multiply the coefficients: The coefficients are 33 and βˆ’9-9. So, 3Γ—(βˆ’9)=βˆ’273 \times (-9) = -27.
    • Multiply the variables: We only have a1a^1 in this term. So, the variable part remains a1a^1, or simply aa.
    • Combine the results: Putting it all together, 3aΓ—(βˆ’9)=βˆ’27a3a \times (-9) = -27a.

After performing both multiplications, we combine the results. The first multiplication gave us βˆ’9a3-9a^3, and the second gave us βˆ’27a-27a. Since these two terms have different variable parts (one is a3a^3 and the other is aa), they are unlike terms and cannot be combined further. Therefore, we simply add them together.

So, the simplified expression is βˆ’9a3βˆ’27a-9a^3 - 27a. We've successfully applied the distributive property and combined the terms. It's always a good idea to double-check your work, especially the signs and exponents, to ensure accuracy. We multiplied a positive term (3a3a) by a negative term (βˆ’3a2-3a^2), which correctly results in a negative term (βˆ’9a3-9a^3). We also multiplied a positive term (3a3a) by a negative term (βˆ’9-9), which correctly results in a negative term (βˆ’27a-27a). The exponents for the variable aa were handled correctly as well. This step is crucial for understanding how to manipulate algebraic expressions effectively.

Final Simplified Expression and Standard Form

We've reached the end of our simplification journey for 3a(βˆ’3a2βˆ’9)3a\left(-3a^2-9\right). After applying the distributive property, we found that 3aΓ—(βˆ’3a2)=βˆ’9a33a \times (-3a^2) = -9a^3 and 3aΓ—(βˆ’9)=βˆ’27a3a \times (-9) = -27a. Combining these results, we get βˆ’9a3βˆ’27a-9a^3 - 27a. This is our simplified expression.

Now, let's talk about standard form. For polynomials, the standard form means arranging the terms in descending order of their exponents. In our expression, βˆ’9a3βˆ’27a-9a^3 - 27a, the term βˆ’9a3-9a^3 has an exponent of 33, and the term βˆ’27a-27a has an exponent of 11 (since a=a1a = a^1). Since 33 is greater than 11, the terms are already in the correct descending order of exponents. So, βˆ’9a3βˆ’27a-9a^3 - 27a is indeed the expression in its standard form.

It's important to remember that simplifying expressions like this is a foundational skill in algebra. It allows us to work with more complex equations and functions. For example, if this simplified expression were part of a larger problem, having it in this neat form makes it much easier to substitute values, factor, or perform other operations. Think of it as tidying up your workspace before starting a big project – everything is organized and ready to go.

Let's do a quick recap of what we did:

  • We identified the expression to be simplified: 3a(βˆ’3a2βˆ’9)3a\left(-3a^2-9\right).
  • We recognized the need to use the distributive property.
  • We multiplied the outer term (3a3a) by each inner term (βˆ’3a2-3a^2 and βˆ’9-9).
  • We correctly handled the multiplication of coefficients and variables, paying close attention to exponents and signs.
  • We combined the resulting terms to get our final simplified expression: βˆ’9a3βˆ’27a-9a^3 - 27a.
  • We confirmed that this expression is in standard form, with terms arranged from highest exponent to lowest.

This process is super useful for all sorts of algebra problems, from solving equations to graphing functions. The more you practice, the more natural it becomes. Keep experimenting with different expressions, and don't be afraid to check your work. Mastering these basic algebraic manipulations is key to success in math, so keep up the great work, everyone!