Find Rectangle Width Using Synthetic Division: Step-by-Step

Hey guys! Let's dive into a fun math problem today that involves finding the width of a rectangle using synthetic division. It might sound a little intimidating at first, but trust me, we'll break it down step by step, and you'll see it's totally manageable. So, grab your pencils, and let’s get started!

Understanding the Problem

In this math problem, we're given the area of a rectangle, which is expressed as a polynomial: $5x^3 + 19x^2 + 6x - 18$. We also know the length of the rectangle, which is $x + 3$. Our mission, should we choose to accept it, is to find the width of the rectangle. Remember that the area of a rectangle is calculated by multiplying its length and width. So, if we know the area and the length, we can find the width by dividing the area by the length. This is where synthetic division comes into play, making our task much easier than traditional polynomial long division.

When dealing with polynomial division, synthetic division offers a streamlined approach, especially when dividing by a linear factor like $x + 3$. This method simplifies the division process by focusing on the coefficients of the polynomial and a specific value derived from the divisor. The beauty of synthetic division lies in its efficiency and reduced risk of errors compared to long division, making it an invaluable tool for solving problems like ours.

The relationship between area, length, and width is fundamental here. The area of a rectangle is given by the formula: Area = Length × Width. In our case, we have the area as a polynomial expression and the length as a binomial. Finding the width requires us to reverse this process, essentially dividing the area polynomial by the length binomial. This is where the magic of synthetic division helps us to simplify the process and arrive at the correct answer efficiently. So, let's roll up our sleeves and see how it works!

What is Synthetic Division?

Okay, so what exactly is synthetic division? Think of it as a shortcut for dividing a polynomial by a linear expression (something like x + a or x - a). It's way less messy than long division, and it's perfect for problems like this one. Synthetic division is particularly useful in algebra for dividing a polynomial by a linear factor. It's a streamlined process that simplifies the division, making it easier to find the quotient and remainder. This method is especially handy when dealing with higher-degree polynomials, where long division can become quite cumbersome.

To perform synthetic division, we focus on the coefficients of the polynomial and a specific value derived from the divisor. This value is the root of the linear expression we're dividing by. For example, if we're dividing by x + 3, the value we use in synthetic division is -3. The process involves a series of multiplications and additions, which efficiently lead us to the quotient and remainder. The quotient represents the result of the division, while the remainder tells us if the division is exact or if there's a leftover term.

The real advantage of synthetic division is its efficiency and reduced risk of errors. Compared to long division, it requires less writing and fewer steps, making it quicker and easier to execute. This is especially beneficial in problems where time is of the essence, or when dealing with complex polynomials. Plus, the structured format of synthetic division helps to minimize mistakes, as it keeps the numbers neatly organized. So, when faced with dividing a polynomial by a linear factor, synthetic division is definitely your best friend.

Setting Up Synthetic Division

Alright, let's get practical. First things first, we need to set up our synthetic division. Remember, our area is $5x^3 + 19x^2 + 6x - 18$, and our length is $x + 3$. The first step in setting up synthetic division involves identifying the coefficients of the polynomial we want to divide. In our case, the polynomial is $5x^3 + 19x^2 + 6x - 18$. So, the coefficients are 5, 19, 6, and -18. These numbers will form the top row of our synthetic division setup. Make sure to include the coefficients for every term, even if the coefficient is zero (which would indicate a missing term in the polynomial).

Next, we need to find the value we'll use as our divisor. Since we're dividing by $x + 3$, we take the opposite of the constant term, which is -3. This value goes on the outside of our division symbol. Think of this number as the root of the equation x + 3 = 0. This number is crucial because it determines the value we use in the synthetic division process, guiding us through the calculations that will reveal the quotient and remainder.

Now, we arrange these numbers in the synthetic division format. Draw a horizontal line and a vertical line to create a sort of upside-down L-shape. Write the coefficients (5, 19, 6, and -18) in a row at the top, and place the divisor (-3) to the left. Leave some space below the coefficients for the numbers we'll be calculating. This setup is the foundation of synthetic division, providing a clear visual structure for the steps that follow. With our setup complete, we're ready to dive into the actual division process. Let's do this!

Performing Synthetic Division

Okay, guys, the setup is done, so let's get to the nitty-gritty of actually performing synthetic division! This is where the magic happens, so pay close attention. The first step in performing synthetic division is to bring down the first coefficient. In our case, the first coefficient is 5. Simply copy this number down below the horizontal line. This number will be the first coefficient of our quotient, so it's an important starting point.

Next, we multiply the number we just brought down (5) by our divisor (-3). So, 5 times -3 equals -15. We write this result (-15) under the second coefficient, which is 19. Now, we add the two numbers in the second column: 19 plus -15. This gives us 4. This addition step is a key part of synthetic division, as it combines the effects of the divisor and the original coefficients.

We repeat this process for the remaining coefficients. Multiply the result we just obtained (4) by the divisor (-3), which gives us -12. Write -12 under the next coefficient, which is 6. Add these two numbers: 6 plus -12, which equals -6. We're on a roll! Finally, multiply -6 by the divisor (-3), which gives us 18. Write 18 under the last coefficient, which is -18. Add these two numbers: -18 plus 18, which equals 0. That zero at the end is super important – it tells us we have no remainder, meaning the division is exact.

These steps of multiplication and addition form the core of synthetic division. By systematically working through each coefficient, we transform the original polynomial into the quotient and remainder. The numbers we obtain below the line represent the coefficients of the quotient, and the last number is the remainder. With a remainder of 0, we know that the divisor divides the polynomial evenly, which is great news for our problem. Now, let's see what these numbers mean for finding the width of our rectangle!

Interpreting the Results

Woo-hoo! We've done the synthetic division, and now we need to figure out what it all means. The numbers we got at the bottom are the key to unlocking the width of our rectangle. The numbers along the bottom row (excluding the last one) represent the coefficients of the quotient. In our case, we have 5, 4, and -6. Since we started with a cubic polynomial (degree 3) and divided by a linear expression (degree 1), our quotient will be a quadratic polynomial (degree 2). So, these numbers correspond to the coefficients of our quadratic expression.

This means our quotient is $5x^2 + 4x - 6$. This quadratic expression represents the width of our rectangle! Remember, we divided the area by the length, and the result is the width. So, we've successfully found the width using synthetic division. The interpretation of these results is crucial because it connects the numerical output of synthetic division to the algebraic expression that represents the width. Understanding how the coefficients translate into the quotient allows us to solve the original problem and find the missing dimension of the rectangle.

The last number in the bottom row is the remainder. In our case, it's 0. A remainder of 0 tells us that the division is exact, meaning that $x + 3$ divides evenly into $5x^3 + 19x^2 + 6x - 18$. This is a good sign because it confirms that $x + 3$ is indeed a factor of the area polynomial, which we expected since it's the length of the rectangle. A non-zero remainder would indicate that the division is not exact, and we might need to double-check our work or consider other approaches.

Finding the Width

Okay, drumroll please! We've done the synthetic division, interpreted the results, and now we're ready to state the width of the rectangle. As we determined earlier, the quotient we obtained from the synthetic division, $5x^2 + 4x - 6$, represents the width of the rectangle. This quadratic expression is the answer we've been working towards, and it completes the solution to our problem. We've successfully found the width by dividing the area polynomial by the length binomial using synthetic division.

So, the width of the rectangle is $5x^2 + 4x - 6$. This is the final answer. Make sure to include the expression as your answer, not just the coefficients. The width is a polynomial expression, just like the area was, and it provides a complete description of the rectangle's dimension in terms of x. This result showcases the power of synthetic division in simplifying polynomial division and allowing us to solve practical problems involving geometric shapes.

Conclusion

And there you have it, guys! We successfully found the width of the rectangle using synthetic division. It might have seemed a bit tricky at first, but we broke it down step by step, and you nailed it. Remember, synthetic division is a super useful tool for dividing polynomials, so keep practicing, and you'll become a pro in no time! The ability to use synthetic division is a valuable skill in algebra, allowing you to efficiently solve problems involving polynomial division and factoring.

We started with a problem that seemed complex, but by understanding the principles of synthetic division and the relationship between area, length, and width, we were able to arrive at a clear and concise solution. This demonstrates the importance of breaking down problems into smaller, manageable steps and applying the appropriate techniques. So, keep up the great work, and don't be afraid to tackle challenging math problems. You've got this!