Synthetic Division Solve (x³ + 1) ÷ (x - 1) Quotient Explained
Hey guys! 👋 Today, we're diving into the fascinating world of synthetic division, a neat little shortcut for dividing polynomials. We'll be tackling the problem (x³ + 1) ÷ (x - 1) and figuring out the quotient. Don't worry, it's not as scary as it sounds! We'll break it down step-by-step, so you'll be a synthetic division pro in no time.
What is Synthetic Division?
Before we jump into the problem, let's quickly recap what synthetic division is all about. Basically, it's a streamlined method for dividing a polynomial by a linear expression (something like x - a). It's way faster and less cumbersome than long division, especially for more complex polynomials. Think of it as the fast lane for polynomial division!
The beauty of synthetic division lies in its simplicity. It focuses on the coefficients of the polynomial, making the process less prone to errors and easier to manage. Instead of dealing with variables and exponents all over the place, we work with numbers, making the whole process much cleaner and more efficient. This is especially handy when you're dealing with higher-degree polynomials, where long division can become a real headache.
Now, why should you care about synthetic division? Well, apart from being a time-saver, it's a crucial tool in algebra and calculus. It helps us find factors and roots of polynomials, which is essential for solving equations and analyzing functions. Understanding synthetic division opens doors to more advanced concepts in mathematics, so it's definitely a skill worth mastering. Plus, it's kinda cool to have a mathematical trick up your sleeve, right?
So, let's get started! We're going to take that seemingly intimidating polynomial division problem and turn it into a walk in the park. Get your pencils ready, and let's dive into the world of synthetic division!
Setting Up the Synthetic Division
Okay, let's get our hands dirty and set up the synthetic division for our problem: (x³ + 1) ÷ (x - 1). This is where the magic begins! The first step is crucial: we need to identify the coefficients of our polynomial and the divisor's root. This initial setup is the foundation of the entire process, so let's make sure we get it right.
First, let's look at our polynomial, x³ + 1. Notice anything missing? We have an x³ term and a constant term (+1), but where are the x² and x terms? This is a common trick in math problems, and we need to account for these missing terms by using placeholders. We'll represent them with coefficients of 0. So, our polynomial, in its full glory, is actually 1x³ + 0x² + 0x + 1. See how those zeros fill in the gaps? This is super important because it ensures we keep everything aligned correctly during the division process.
Now, let's identify the coefficients. They are simply the numbers in front of each term: 1 (for x³), 0 (for x²), 0 (for x), and 1 (the constant term). These coefficients are the stars of our synthetic division show! We'll be working with them directly, so make sure you have them clearly identified.
Next up, the divisor: (x - 1). We need to find the root of this linear expression. Remember, the root is the value of x that makes the expression equal to zero. So, we set x - 1 = 0 and solve for x. Adding 1 to both sides, we get x = 1. This, my friends, is the number we'll be using in the "outside corner" of our synthetic division setup. The root of the divisor is the key player in determining what we're actually dividing by.
Now that we have our coefficients (1, 0, 0, 1) and our divisor's root (1), we're ready to set up the synthetic division "scaffold." We'll write the root (1) in the top-left corner and the coefficients (1, 0, 0, 1) across the top row. Draw a horizontal line underneath the coefficients, leaving space below the line for our calculations. This setup might look a little strange at first, but trust me, it's the key to unlocking the synthetic division process. We've laid the groundwork; now, the fun part begins!
Performing the Synthetic Division
Alright, now for the main event: performing the synthetic division! This is where we put our setup into action and watch the magic happen. Don't worry, it's a rhythmic process, and once you get the hang of it, it's almost like a dance. We'll go through each step carefully, so you can see exactly how it works. The synthetic division process is a series of additions and multiplications, so a little focus is all you need.
The first step is super simple: bring down the first coefficient. In our case, the first coefficient is 1. We just copy it down below the horizontal line. This is our starting point, the foundation upon which we'll build our quotient. Think of it as the first piece of the puzzle falling into place. This initial step gets the ball rolling and sets the stage for the rest of the calculations.
Now comes the fun part: multiplication! We take the number we just brought down (1) and multiply it by the root of our divisor (which is also 1 in this case). So, 1 multiplied by 1 equals 1. We write this result (1) under the next coefficient, which is 0. This is the core of the synthetic division process: multiply and move, multiply and move. This multiplication step is what links the divisor and the polynomial together, allowing us to find the quotient.
Next, we add! We add the number we just wrote down (1) to the coefficient above it (0). So, 0 plus 1 equals 1. We write this sum (1) below the line. See the pattern? We multiply, then we add. This addition step is crucial for combining the terms and simplifying the polynomial. It's where the magic of division really happens.
We repeat the process: multiply and add. We take the new number we just got (1) and multiply it by the root of the divisor (1). 1 multiplied by 1 equals 1. We write this result (1) under the next coefficient, which is another 0. Then, we add: 0 plus 1 equals 1. We write this sum (1) below the line. We're on a roll! Each repetition of this process brings us closer to the final quotient and remainder.
One last time! We take the new number (1) and multiply it by the root (1). 1 multiplied by 1 equals 1. We write this result (1) under the last coefficient, which is 1. Then, we add: 1 plus 1 equals 2. We write this sum (2) below the line. We've reached the end of the line! This final calculation reveals the remainder of the division, which is a crucial piece of information.
Now, let's take a step back and look at the numbers we have below the line. These are the coefficients of our quotient and the remainder. We're almost there! In the next section, we'll interpret these numbers and write out the final answer. You've successfully navigated the synthetic division process – congratulations!
Interpreting the Results and Finding the Quotient
Fantastic! We've performed the synthetic division, and now we have a row of numbers below the line. But what do they mean? This is where we translate those numbers back into polynomial language and find our quotient and remainder. Understanding how to interpret the results is the final piece of the synthetic division puzzle.
Remember, our original polynomial was x³ + 1, which we rewrote as 1x³ + 0x² + 0x + 1 to account for the missing terms. We divided this by (x - 1). The numbers we have below the line (1, 1, 1, and 2) represent the coefficients of our quotient and the remainder. These numbers hold the key to unlocking the solution to our division problem.
The last number on the right (2) is our remainder. It's what's "left over" after the division. If the remainder is 0, it means the divisor divides the polynomial evenly. In our case, the remainder is 2, which means (x - 1) doesn't divide x³ + 1 perfectly, but we're still going to get a valid quotient with a remainder term. The remainder tells us how cleanly the division occurs.
The other numbers (1, 1, and 1) are the coefficients of our quotient. But what do they belong to? Well, we started with a cubic polynomial (x³), and we divided by a linear expression (x - 1). This means our quotient will be a polynomial one degree lower than the original, which is a quadratic (x²) polynomial. The degree of the quotient is always one less than the degree of the original polynomial when dividing by a linear expression.
So, the first 1 is the coefficient of x², the second 1 is the coefficient of x, and the third 1 is the constant term. This means our quotient is 1x² + 1x + 1, which we can simplify to x² + x + 1. Isn't that neat? The coefficients below the line directly translate into the coefficients of the quotient polynomial.
But we're not quite done yet! We have a remainder of 2, so we need to add the remainder term to our quotient. We write the remainder (2) over the divisor (x - 1), giving us the fraction 2/(x - 1). The remainder is always expressed as a fraction with the divisor as the denominator.
Finally, we can write out the complete answer: the quotient is x² + x + 1, and the remainder term is 2/(x - 1). So, the full result of the division (x³ + 1) ÷ (x - 1) is x² + x + 1 + 2/(x - 1). Congratulations! You've successfully used synthetic division to solve the problem. We did it, guys! 🎉
Identifying the Correct Answer
Now that we've gone through the entire synthetic division process and found our answer, let's take a look at the multiple-choice options and identify the correct one. This is a crucial step to ensure we've not only done the work correctly but also understand how to present the solution in the context of the question. Matching our solution to the given options reinforces our understanding and confirms our answer.
We found that (x³ + 1) ÷ (x - 1) gives us a quotient of x² + x + 1 and a remainder term of 2/(x - 1). So, our complete answer is x² + x + 1 + 2/(x - 1). Now, let's compare this to the options provided:
A. x² + x + 1 + 2/(x + 1) B. x² - x + 1 C. x² + x + 1 + 2/(x - 1) D. x³ - x² + x
Looking closely, we can see that option C, x² + x + 1 + 2/(x - 1), perfectly matches our solution. The quotient is x² + x + 1, and the remainder term is 2/(x - 1), just as we calculated. Careful comparison is key to avoiding careless errors in selecting the correct option.
Options A, B, and D are all incorrect. Option A has the wrong divisor in the remainder term (x + 1 instead of x - 1). Option B only gives the quotient but misses the remainder term altogether. Option D is a cubic polynomial, which doesn't make sense for a quotient when we divided a cubic by a linear expression. Analyzing why the incorrect options are wrong helps solidify our understanding of the correct process and answer.
Therefore, the correct answer is C. We've not only solved the problem using synthetic division but also confidently identified the correct option from the choices provided. Great job! This final step of verification is essential for ensuring accuracy and building confidence in our problem-solving abilities.
Conclusion: Mastering Synthetic Division
Woohoo! 🎉 You've made it to the end, and you've successfully conquered synthetic division! We took on the challenge of dividing (x³ + 1) by (x - 1), and we emerged victorious. You now have a solid understanding of how synthetic division works, from setting it up to interpreting the results and finding the quotient. Mastering synthetic division is a valuable asset in your mathematical toolkit, and you should feel proud of your accomplishment.
We started by understanding the purpose of synthetic division: a faster, more efficient way to divide polynomials by linear expressions. We then meticulously set up our problem, being careful to account for missing terms with zero coefficients. The initial setup is crucial for accurate synthetic division, and you've learned how to do it correctly.
Next, we performed the synthetic division process itself, a rhythmic dance of multiplication and addition. You saw how each step builds upon the previous one, leading us closer to the solution. The repetitive nature of synthetic division makes it easier to learn and remember, and you've practiced this process step-by-step.
We then interpreted the numbers below the line, understanding that they represent the coefficients of our quotient and the remainder. You learned how to translate these numbers back into polynomial language, giving us our final answer. Knowing how to interpret the results is just as important as performing the calculations, and you've mastered this skill.
Finally, we identified the correct answer from the multiple-choice options, reinforcing our understanding and ensuring we could confidently select the right solution. Verifying your answer against the given options is a crucial step in problem-solving, and you've practiced this skill as well.
Synthetic division might have seemed intimidating at first, but you've shown that with a clear understanding of the steps and a little practice, it's a powerful tool you can use to solve polynomial division problems. Keep practicing, and you'll become a synthetic division whiz in no time! Consistent practice is the key to solidifying your understanding and building confidence.
So, the next time you encounter a polynomial division problem, don't shy away – embrace the challenge and use your newfound synthetic division skills to conquer it! You've got this! 💪
Answer: C.