Simplifying $-3(x+3)^2-3+3x$ And Expressing In Standard Form

by ADMIN 61 views
Iklan Headers

Hey guys! Let's dive into simplifying this algebraic expression and get it into that neat standard form we all love. We're tackling the expression βˆ’3(x+3)2βˆ’3+3x-3(x+3)^2-3+3x. It might look a bit intimidating at first, but trust me, we'll break it down step-by-step, and you'll see it's totally manageable. Our goal? To rewrite this expression in the standard form of a quadratic equation, which is ax2+bx+cax^2 + bx + c, where a, b, and c are constants. So, let’s roll up our sleeves and get started!

Breaking Down the Expression

Our journey begins with the original expression: βˆ’3(x+3)2βˆ’3+3x-3(x+3)^2-3+3x. The first thing that catches our eye is the squared term, (x+3)2(x+3)^2. Remember, squaring a binomial means multiplying it by itself. So, we need to expand (x+3)(x+3)(x+3)(x+3) first. Let's use the FOIL method (First, Outer, Inner, Last) to make sure we get every term. When we expand (x+3)(x+3)(x+3)(x+3), we get:

  • First: xβˆ—x=x2x * x = x^2
  • Outer: xβˆ—3=3xx * 3 = 3x
  • Inner: 3βˆ—x=3x3 * x = 3x
  • Last: 3βˆ—3=93 * 3 = 9

Adding these together, we have x2+3x+3x+9x^2 + 3x + 3x + 9, which simplifies to x2+6x+9x^2 + 6x + 9. Great! We've expanded the binomial. Now, we need to substitute this back into our original expression. This gives us βˆ’3(x2+6x+9)βˆ’3+3x-3(x^2 + 6x + 9) - 3 + 3x. The next step is to distribute the βˆ’3-3 across the terms inside the parentheses. This means multiplying each term inside the parentheses by βˆ’3-3. When we do this, we get:

  • βˆ’3βˆ—x2=βˆ’3x2-3 * x^2 = -3x^2
  • βˆ’3βˆ—6x=βˆ’18x-3 * 6x = -18x
  • βˆ’3βˆ—9=βˆ’27-3 * 9 = -27

So, after distributing, our expression looks like this: βˆ’3x2βˆ’18xβˆ’27βˆ’3+3x-3x^2 - 18x - 27 - 3 + 3x. See? We're making progress! We've gotten rid of the parentheses and now have a series of terms that we can combine. This part is all about bringing together like terms – terms that have the same variable and exponent.

Combining Like Terms

Now comes the fun part – combining like terms! Looking at our expression βˆ’3x2βˆ’18xβˆ’27βˆ’3+3x-3x^2 - 18x - 27 - 3 + 3x, we can identify a few terms that can be combined. We have terms with x2x^2, terms with xx, and constant terms (numbers without any variables). Let’s start by identifying the like terms. We have:

  • One x2x^2 term: βˆ’3x2-3x^2
  • Two xx terms: βˆ’18x-18x and +3x+3x
  • Two constant terms: βˆ’27-27 and βˆ’3-3

Now, let's combine them. The βˆ’3x2-3x^2 term is the only term with x2x^2, so it stays as is. For the xx terms, we have βˆ’18x+3x-18x + 3x. Think of this as starting at -18 and moving 3 places to the right on a number line. This gives us βˆ’15x-15x. So, the combined xx term is βˆ’15x-15x. Next, let's combine the constant terms: βˆ’27βˆ’3-27 - 3. This is like starting at -27 and moving 3 places further to the left on the number line, resulting in βˆ’30-30. Putting it all together, we have βˆ’3x2βˆ’15xβˆ’30-3x^2 - 15x - 30. Awesome! We've combined all the like terms. But we're not quite done yet. We need to make sure our expression is in standard form.

Expressing in Standard Form

So, what exactly is standard form? Standard form for a quadratic expression (an expression with an x2x^2 term) is ax2+bx+cax^2 + bx + c, where aa, bb, and cc are constants. This means we want to arrange our terms in descending order of their exponents. In other words, the x2x^2 term comes first, followed by the xx term, and then the constant term. Looking at our simplified expression, βˆ’3x2βˆ’15xβˆ’30-3x^2 - 15x - 30, we can see that it's already in standard form! We have the x2x^2 term, βˆ’3x2-3x^2, followed by the xx term, βˆ’15x-15x, and finally the constant term, βˆ’30-30. This is perfect! So, the simplified expression in standard form is βˆ’3x2βˆ’15xβˆ’30-3x^2 - 15x - 30.

To recap, we started with a somewhat complex expression, βˆ’3(x+3)2βˆ’3+3x-3(x+3)^2-3+3x, and through careful expansion, distribution, and combining like terms, we transformed it into the standard form of a quadratic expression. This process not only simplifies the expression but also makes it easier to work with in further algebraic manipulations or when graphing. Mastering these simplification techniques is a fundamental skill in algebra, and it's something you'll use time and time again. So, take pride in your accomplishment – you've just conquered a potentially tricky problem!

Let's Summarize and Emphasize Key Steps

Alright, let's take a moment to summarize the steps we took to simplify the expression βˆ’3(x+3)2βˆ’3+3x-3(x+3)^2-3+3x and express it in standard form. This recap will help solidify the process in your mind and highlight the key techniques involved. Remember, practice makes perfect, so understanding these steps is crucial for tackling similar problems in the future.

  1. Expand the Squared Term: The first thing we did was tackle the squared binomial, (x+3)2(x+3)^2. This involved expanding it using the FOIL method or the distributive property. We multiplied (x+3)(x+3) by itself, which gave us x2+6x+9x^2 + 6x + 9. This step is essential because it removes the parentheses and sets us up for further simplification.
  2. Distribute the Constant: Next, we distributed the βˆ’3-3 across the terms in the expanded binomial. This means multiplying each term inside the parentheses by βˆ’3-3. Doing this, we transformed βˆ’3(x2+6x+9)-3(x^2 + 6x + 9) into βˆ’3x2βˆ’18xβˆ’27-3x^2 - 18x - 27. Distribution is a fundamental algebraic operation, and it’s vital for simplifying expressions that contain parentheses.
  3. Identify Like Terms: After distribution, we identified like terms in the expression βˆ’3x2βˆ’18xβˆ’27βˆ’3+3x-3x^2 - 18x - 27 - 3 + 3x. Like terms are terms that have the same variable raised to the same power. In our expression, we had terms with x2x^2, terms with xx, and constant terms. Recognizing these like terms is a key step toward simplification.
  4. Combine Like Terms: This is where we actually simplified the expression by adding or subtracting the coefficients of the like terms. We combined the xx terms (βˆ’18x-18x and +3x+3x) to get βˆ’15x-15x, and we combined the constant terms (βˆ’27-27 and βˆ’3-3) to get βˆ’30-30. The x2x^2 term, βˆ’3x2-3x^2, remained unchanged because it was the only term with x2x^2. Combining like terms is a cornerstone of algebraic simplification.
  5. Express in Standard Form: Finally, we made sure our simplified expression was in standard form, which for a quadratic expression is ax2+bx+cax^2 + bx + c. In our case, the simplified expression βˆ’3x2βˆ’15xβˆ’30-3x^2 - 15x - 30 was already in standard form. Standard form is important because it provides a consistent way to write expressions, which makes them easier to compare and analyze.

By following these steps, we successfully simplified the expression and expressed it in standard form. Remember, each step is important, and mastering these techniques will greatly enhance your algebraic skills. Keep practicing, and you'll become more confident and proficient in simplifying expressions. Great job, guys!

Common Pitfalls to Avoid

Alright guys, let's chat about some common mistakes that people often make when simplifying expressions like this. Knowing these pitfalls can help you avoid them and ensure you get the correct answer every time. We're aiming for accuracy and efficiency, so let's dive into these common errors.

  1. Incorrectly Expanding the Squared Term: One of the most frequent mistakes is messing up the expansion of the squared term, like (x+3)2(x+3)^2. It's tempting to think that (x+3)2(x+3)^2 is simply x2+32x^2 + 3^2, but that's not correct! Remember, squaring a binomial means multiplying it by itself: (x+3)(x+3)(x+3)(x+3). You need to use the FOIL method (First, Outer, Inner, Last) or the distributive property to expand it correctly. The correct expansion is x2+6x+9x^2 + 6x + 9. So, always take that extra step to expand the binomial fully and accurately.

  2. Forgetting to Distribute the Negative Sign: When you have a negative sign in front of parentheses, like in βˆ’3(x2+6x+9)-3(x^2 + 6x + 9), it's crucial to distribute the negative sign to every term inside the parentheses. This means multiplying each term by βˆ’3-3. A common mistake is to distribute the 3 but forget about the negative sign, which can lead to errors in the subsequent steps. So, always double-check that you've distributed the negative sign correctly.

  3. Mixing Up Like Terms: Another pitfall is combining terms that are not like terms. Remember, like terms have the same variable raised to the same power. For example, you can combine βˆ’18x-18x and +3x+3x because they both have xx to the power of 1. But you can't combine βˆ’18x-18x with βˆ’27-27 because βˆ’27-27 is a constant term (it doesn't have a variable). Make sure you're only combining terms that are truly like terms.

  4. Arithmetic Errors: Simple arithmetic mistakes can throw off the entire solution. This could be anything from adding or subtracting numbers incorrectly to miscalculating the product of coefficients. For example, when combining βˆ’27-27 and βˆ’3-3, make sure you get βˆ’30-30 and not βˆ’24-24. It's always a good idea to double-check your arithmetic, especially when dealing with negative numbers.

  5. Not Expressing the Final Answer in Standard Form: The question specifically asks for the simplified expression in standard form, which is ax2+bx+cax^2 + bx + c. Make sure your final answer is in this format. This means arranging the terms in descending order of their exponents: the x2x^2 term first, followed by the xx term, and then the constant term. If your terms are in the wrong order, you haven't fully answered the question.

By being aware of these common pitfalls, you can minimize your chances of making mistakes and improve your accuracy when simplifying algebraic expressions. Remember, practice and attention to detail are key! So, keep these tips in mind as you tackle similar problems, and you'll be well on your way to mastering algebraic simplification.

Practice Problems to Sharpen Your Skills

Alright, guys, now that we've walked through the solution and discussed common pitfalls, it's time to put your knowledge to the test! Practice is absolutely crucial for mastering algebraic simplification. The more you practice, the more comfortable and confident you'll become. So, let's dive into some practice problems that will help you sharpen your skills. I'll provide a few expressions similar to the one we just tackled, and I encourage you to work through them step-by-step, applying the techniques we've discussed.

Problem 1: Simplify the expression βˆ’2(x+2)2βˆ’4+4x-2(x+2)^2 - 4 + 4x and express it in standard form.

Problem 2: Simplify the expression βˆ’4(x+1)2βˆ’5+2x-4(x+1)^2 - 5 + 2x and express it in standard form.

Problem 3: Simplify the expression βˆ’3(xβˆ’2)2+6βˆ’2x-3(x-2)^2 + 6 - 2x and express it in standard form.

Problem 4: Simplify the expression βˆ’(x+4)2βˆ’3+5x-(x+4)^2 - 3 + 5x and express it in standard form.

Problem 5: Simplify the expression βˆ’2(xβˆ’3)2+5βˆ’3x-2(x-3)^2 + 5 - 3x and express it in standard form.

Remember, the key to success is to break each problem down into manageable steps. Start by expanding the squared term, then distribute any constants, combine like terms, and finally, express your answer in standard form. Don't rush, and pay close attention to the details, especially when dealing with negative signs. If you get stuck, revisit the steps we outlined earlier in this guide or review the common pitfalls to avoid. These practice problems are designed to reinforce your understanding of the simplification process. As you work through them, you'll not only improve your algebraic skills but also develop a deeper appreciation for the elegance and logic of mathematics. So, grab a pencil and paper, and let's get started! Happy simplifying, guys!

In conclusion, guys, simplifying algebraic expressions like βˆ’3(x+3)2βˆ’3+3x-3(x+3)^2-3+3x and expressing them in standard form is a fundamental skill in algebra. We've walked through the step-by-step process, from expanding the squared term to combining like terms and expressing the final answer in standard form. Remember, the key to success is practice, attention to detail, and a solid understanding of the underlying algebraic principles. By mastering these techniques, you'll not only be able to tackle similar problems with confidence but also build a strong foundation for more advanced mathematical concepts. So, keep practicing, keep exploring, and keep challenging yourself. You've got this! Happy algebra-ing!