Rectangle Width Calculation Using Synthetic Division

The area of a rectangle is $5 x^3+19 x^2+6 x-18$ and the length is $x+3$. Using synthetic division, find the width of the rectangle.

In the realm of geometry and algebra, the relationship between a rectangle's area, length, and width is fundamental. The area of a rectangle is simply the product of its length and width. When presented with the area and one dimension (either length or width), we can determine the other dimension through division. This article delves into a specific problem where the area of a rectangle is given as a polynomial expression, and the length is another polynomial. Our mission is to find the width, and we'll employ the powerful tool of synthetic division to achieve this.

The Problem: Area, Length, and the Quest for Width

The problem at hand presents us with a rectangle whose area is expressed as the polynomial $5x^3 + 19x^2 + 6x - 18$. We are also given that the length of this rectangle is $x + 3$. Our objective is to find the width of the rectangle. Recalling the basic formula for the area of a rectangle,

Area = Length × Width

we can rearrange it to solve for the width:

Width = Area / Length

In our case, this translates to dividing the polynomial representing the area by the polynomial representing the length. Polynomial division can be performed using several methods, but synthetic division is particularly efficient when dividing by a linear expression of the form $x - a$. Since our length is $x + 3$, which can be rewritten as $x - (-3)$, synthetic division is the ideal technique for this problem.

Synthetic Division: A Step-by-Step Guide

Synthetic division is a streamlined method for dividing a polynomial by a linear expression. It focuses on the coefficients of the polynomials, making the process less cumbersome than long division. Let's walk through the steps of synthetic division as applied to our problem:

1. Identify the divisor's root: Our divisor is $x + 3$, so we set it equal to zero and solve for $x$: $x + 3 = 0$, which gives us $x = -3$. This value, -3, is the root we'll use in our synthetic division.
2. Write down the coefficients of the dividend: The dividend is the polynomial representing the area, $5x^3 + 19x^2 + 6x - 18$. We extract the coefficients: 5, 19, 6, and -18. It's crucial to ensure that the polynomial is written in descending order of powers of $x$, and if any powers are missing, we include a coefficient of 0 as a placeholder.
3. Set up the synthetic division table: We draw a horizontal line and a vertical line to create a table. We place the root (-3) to the left of the vertical line and the coefficients (5, 19, 6, -18) to the right of the vertical line, above the horizontal line.
4. Perform the division:
   • Bring down the first coefficient (5) below the horizontal line.
   • Multiply the root (-3) by the number we just brought down (5), which gives us -15. Write this product under the next coefficient (19).
   • Add the numbers in the second column (19 and -15), which gives us 4. Write this sum below the horizontal line.
   • Multiply the root (-3) by the result (4), which gives us -12. Write this product under the next coefficient (6).
   • Add the numbers in the third column (6 and -12), which gives us -6. Write this sum below the horizontal line.
   • Multiply the root (-3) by the result (-6), which gives us 18. Write this product under the last coefficient (-18).
   • Add the numbers in the last column (-18 and 18), which gives us 0. Write this sum below the horizontal line.
5. Interpret the results: The numbers below the horizontal line (except the last one) are the coefficients of the quotient polynomial, and the last number is the remainder. In our case, the numbers are 5, 4, -6, and 0. Since the original dividend was a cubic polynomial (degree 3) and we divided by a linear expression (degree 1), the quotient will be a quadratic polynomial (degree 2). Therefore, the quotient is $5x^2 + 4x - 6$, and the remainder is 0.

The Answer: Unveiling the Rectangle's Width

The result of our synthetic division tells us that when we divide the area polynomial ($5x^3 + 19x^2 + 6x - 18$) by the length ($x + 3$), we obtain the quotient $5x^2 + 4x - 6$ with a remainder of 0. This means that:

Width = (5x^3 + 19x^2 + 6x - 18) / (x + 3) = 5x^2 + 4x - 6

Therefore, the width of the rectangle is $5x^2 + 4x - 6$. Looking back at the given options, we find that option A, $5x^2 + 4x - 6$, is the correct answer.

Why Synthetic Division Works: A Deeper Dive

At its heart, synthetic division is a condensed form of polynomial long division. It leverages the structure of polynomial division to simplify the process. When we divide a polynomial by a linear expression $x - a$, we are essentially finding a quotient polynomial and a remainder such that:

Dividend = (Divisor × Quotient) + Remainder

Synthetic division efficiently calculates the coefficients of the quotient and the remainder by focusing on the numerical relationships between the coefficients. The root of the divisor ($a$) plays a crucial role in this process, as it allows us to systematically eliminate terms and determine the coefficients of the quotient. The remainder represents the portion of the dividend that is not evenly divisible by the divisor.

In our example, the remainder of 0 indicates that the length ($x + 3$) divides the area polynomial ($5x^3 + 19x^2 + 6x - 18$) perfectly, which is consistent with the fact that the width is also a polynomial. If the remainder were non-zero, it would imply that the length does not divide the area evenly, and the width would include a fractional term.

Beyond the Problem: Applications of Polynomial Division

The ability to divide polynomials, whether through synthetic division or long division, is a valuable skill in algebra and calculus. It has numerous applications, including:

• Factoring Polynomials: If dividing a polynomial by $x - a$ results in a remainder of 0, then $x - a$ is a factor of the polynomial. This can be used to factor higher-degree polynomials into simpler expressions.
• Solving Polynomial Equations: Polynomial division can help find the roots of polynomial equations. If we know one root, we can divide the polynomial by the corresponding linear factor to obtain a lower-degree polynomial, which may be easier to solve.
• Simplifying Rational Expressions: Rational expressions are fractions where the numerator and denominator are polynomials. Polynomial division can be used to simplify these expressions by dividing out common factors.
• Calculus: Polynomial division is used in calculus for various purposes, such as finding limits, integrating rational functions, and analyzing the behavior of functions.

Conclusion: Mastering Polynomial Division

In this article, we successfully determined the width of a rectangle given its area and length by employing the technique of synthetic division. This problem highlights the importance of understanding the relationship between a rectangle's dimensions and its area, as well as the power of algebraic tools like polynomial division. By mastering synthetic division, we can efficiently divide polynomials and solve a variety of problems in mathematics and related fields. The key takeaway is that synthetic division provides a streamlined approach to polynomial division, especially when dealing with linear divisors. Its applications extend beyond geometry and are fundamental in various areas of mathematics and beyond.