Synthetic Division Explained Solving For The Quotient Of Polynomials
In the realm of polynomial algebra, synthetic division emerges as a streamlined technique for dividing polynomials by linear divisors. It offers a more concise and efficient alternative to long division, particularly when dealing with divisors of the form x - k, where k is a constant. This article delves into the intricacies of synthetic division, providing a step-by-step guide to solving a problem and identifying the quotient in polynomial form. Specifically, we will tackle the problem:
-3 | 1 2 -2 3
Our goal is to dissect this synthetic division setup, perform the necessary calculations, and express the quotient as a polynomial. Understanding synthetic division is crucial not only for simplifying polynomial division but also for grasping concepts like the Remainder Theorem and the Factor Theorem. It serves as a foundational tool in various mathematical contexts, including finding polynomial roots, factoring polynomials, and solving algebraic equations. This article will provide a comprehensive explanation to help you master this essential technique.
Understanding Synthetic Division
Before we dive into solving the given problem, let’s first establish a solid understanding of synthetic division. Synthetic division is a shorthand method for dividing a polynomial by a linear divisor of the form x - k. It streamlines the division process by focusing solely on the coefficients of the polynomial and the constant k. This method is particularly useful because it reduces the amount of writing and the complexity involved in long division. To fully grasp the mechanics of synthetic division, it's essential to understand its underlying principles and how each step contributes to finding the quotient and remainder.
The setup for synthetic division involves writing the constant k (from the divisor x - k) to the left of a vertical line. Then, the coefficients of the polynomial to be divided (the dividend) are written to the right of the vertical line. It's crucial to include a zero for any missing terms in the polynomial to maintain the correct place values. For instance, if you are dividing x³ + 2x - 1 by x - 2, you would write the coefficients as 1, 0, 2, and -1, acknowledging the missing x² term. This meticulous preparation is vital for the accuracy of the synthetic division process.
The actual process of synthetic division involves a series of steps that are repeated for each coefficient. First, the leading coefficient of the dividend is brought down below the horizontal line. Then, this number is multiplied by k, and the result is written under the next coefficient. The numbers in this column are then added together, and the sum is written below the horizontal line. This process of multiplication and addition is repeated for each coefficient until the last one is reached. The numbers below the horizontal line, excluding the last one, represent the coefficients of the quotient, and the last number represents the remainder. Understanding this iterative process is key to performing synthetic division accurately and efficiently.
Step-by-Step Solution to the Problem
Now, let's apply our understanding of synthetic division to solve the given problem:
-3 | 1 2 -2 3
This notation indicates that we are dividing a polynomial with coefficients 1, 2, -2, and 3 by the linear divisor x + 3 (since k is -3). Let’s break down the solution step-by-step to make it clear and easy to follow.
Step 1: Set up the synthetic division.
Write down the coefficients of the polynomial (1, 2, -2, 3) and the value of k (-3) as shown:
-3 | 1 2 -2 3
------------
Step 2: Bring down the first coefficient.
Bring down the first coefficient (1) below the line:
-3 | 1 2 -2 3
------------
1
Step 3: Multiply and add.
Multiply the number you brought down (1) by k (-3) and write the result (-3) under the next coefficient (2):
-3 | 1 2 -2 3
---- -3
1
Add the numbers in the second column (2 and -3) and write the sum (-1) below the line:
-3 | 1 2 -2 3
---- -3
1 -1
Step 4: Repeat the process.
Multiply the latest number below the line (-1) by k (-3) and write the result (3) under the next coefficient (-2):
-3 | 1 2 -2 3
---- -3 3
1 -1
Add the numbers in the third column (-2 and 3) and write the sum (1) below the line:
-3 | 1 2 -2 3
---- -3 3
1 -1 1
Step 5: Final multiplication and addition.
Multiply the latest number below the line (1) by k (-3) and write the result (-3) under the last coefficient (3):
-3 | 1 2 -2 3
---- -3 3 -3
1 -1 1
Add the numbers in the last column (3 and -3) and write the sum (0) below the line:
-3 | 1 2 -2 3
---- -3 3 -3
1 -1 1 0
Step 6: Interpret the result.
The numbers below the line (1, -1, 1, 0) represent the coefficients of the quotient and the remainder. The last number (0) is the remainder, and the other numbers (1, -1, 1) are the coefficients of the quotient. Since we started with a cubic polynomial (degree 3) and divided by a linear term (degree 1), the quotient will be a quadratic polynomial (degree 2). Thus, the quotient is:
1x² - 1x + 1
This simplifies to:
x² - x + 1
Therefore, the quotient in polynomial form is x² - x + 1.
Identifying the Quotient in Polynomial Form
After performing synthetic division, the crucial step is to correctly interpret the result to identify the quotient in polynomial form. The numbers obtained in the bottom row of the synthetic division tableau, excluding the last number, represent the coefficients of the quotient. The last number represents the remainder. To express the quotient as a polynomial, it’s essential to understand how the degree of the quotient relates to the degree of the original polynomial (dividend) and the divisor.
When dividing a polynomial of degree n by a linear divisor (degree 1), the quotient will have a degree of n - 1. For example, if you divide a cubic polynomial (degree 3) by a linear term, the resulting quotient will be a quadratic polynomial (degree 2). This is a fundamental principle to keep in mind when constructing the quotient polynomial. The coefficients obtained from the synthetic division are then attached to the appropriate powers of x, starting with the power n - 1 for the first coefficient and decreasing by one for each subsequent coefficient.
In our example, we started with a cubic polynomial (represented by the coefficients 1, 2, -2, and 3) and divided by a linear divisor. The resulting coefficients for the quotient were 1, -1, and 1. This means the quotient is a quadratic polynomial. The first coefficient (1) corresponds to the x² term, the second coefficient (-1) corresponds to the x term, and the third coefficient (1) is the constant term. Therefore, the quotient in polynomial form is x² - x + 1. This systematic approach ensures accurate interpretation of the results of synthetic division and proper construction of the quotient polynomial.
Choosing the Correct Answer
Based on our step-by-step solution, we found that the quotient in polynomial form is:
x² - x + 1
Now, let's match this result with the given options:
- A. x² + x - 1
- B. x² - x + 3
- C. x² + x - 3
- D. x² - x + 1
Comparing our solution with the options, we can see that option D, x² - x + 1, matches our result exactly.
Therefore, the correct answer is D. This demonstrates the importance of not only performing the synthetic division correctly but also accurately interpreting the result and matching it with the given choices. Synthetic division, when executed meticulously, provides a reliable method for determining polynomial quotients and remainders.
Conclusion
In summary, we have successfully navigated through a synthetic division problem, systematically breaking down each step to arrive at the correct quotient. We began by understanding the fundamental principles of synthetic division, its setup, and the iterative process of multiplication and addition. We then applied these principles to solve the given problem:
-3 | 1 2 -2 3
By meticulously following the steps, we determined that the quotient in polynomial form is x² - x + 1, which corresponds to option D. This exercise underscores the efficiency and accuracy of synthetic division as a method for dividing polynomials by linear divisors.
Furthermore, we highlighted the significance of correctly interpreting the results of synthetic division. Understanding how the coefficients in the bottom row relate to the quotient and remainder is crucial for expressing the quotient in polynomial form. This involves recognizing the degree of the quotient and assigning the coefficients to the appropriate powers of x. The ability to accurately perform and interpret synthetic division is a valuable skill in algebra and calculus, facilitating the simplification of polynomial expressions and the solution of algebraic equations.
Mastering synthetic division not only enhances your ability to solve polynomial division problems but also deepens your understanding of polynomial algebra as a whole. It provides a foundation for tackling more complex algebraic concepts and applications, making it an indispensable tool in your mathematical toolkit.
