Synthetic Division Test Potential Root Step-by-Step Guide

In the realm of polynomial equations, finding the roots is a fundamental task. These roots, also known as zeros, are the values of the variable that make the polynomial equal to zero. Among the various methods available for finding these roots, synthetic division stands out as a particularly efficient and elegant technique. This article will delve into the intricacies of synthetic division, guiding you through the process with a detailed example and explaining its underlying principles.

Understanding Synthetic Division

Synthetic division is a streamlined method for dividing a polynomial by a linear factor of the form x - c. It offers a more compact and efficient alternative to long division, especially when dealing with higher-degree polynomials. The process involves using only the coefficients of the polynomial and the constant term c from the linear factor. By following a specific set of steps, we can quickly determine the quotient and remainder of the division. Synthetic division is not just a computational shortcut; it's also a powerful tool for testing potential roots of a polynomial.

The Power of the Remainder Theorem

The Remainder Theorem plays a crucial role in understanding the connection between synthetic division and finding roots. This theorem states that if a polynomial f(x) is divided by x - c, the remainder is equal to f(c). In other words, if we substitute c into the polynomial, the resulting value is the same as the remainder obtained from synthetic division. This theorem provides a direct way to test if a given value c is a root of the polynomial. If the remainder is zero, then c is indeed a root, and x - c is a factor of the polynomial.

Setting Up the Synthetic Division Problem

To illustrate the process, let's consider the polynomial equation:

f(x) = x^3 + 6x^2 - 7x - 60

We want to test if -5 is a root of this polynomial using synthetic division. Here's how we set up the problem:

1. Write the coefficients: List the coefficients of the polynomial in descending order of powers of x. In this case, the coefficients are 1, 6, -7, and -60. Make sure to include a 0 for any missing terms. For example, if the polynomial was x^3 + 2x - 5, we would write the coefficients as 1, 0, 2, -5.
2. Write the potential root: Write the value of c (the potential root) to the left. In our example, c is -5.
3. Draw the division symbol: Draw a horizontal line and a vertical line to create a space for the calculations. This setup visually organizes the synthetic division process.

Here's what the setup looks like for our example:

-5 | 1 6 -7 -60
   |__________

Performing the Synthetic Division

Now, let's perform the synthetic division step by step:

1. Bring down the first coefficient: Bring down the first coefficient (1 in our example) below the horizontal line. This is the first step in building the quotient.

-5 | 1 6 -7 -60
   |__________
   1

2. Multiply and add: Multiply the value we brought down (1) by the potential root (-5) and write the result (-5) under the next coefficient (6). Then, add the two numbers (6 and -5) and write the sum (1) below the line.

-5 | 1 6 -7 -60
   | -5
   |__________
 1 1

3. Repeat the process: Repeat the multiply-and-add process for the remaining coefficients. Multiply the last number below the line (1) by the potential root (-5) and write the result (-5) under the next coefficient (-7). Add the two numbers (-7 and -5) and write the sum (-12) below the line.

-5 | 1 6 -7 -60
   | -5 -5
   |__________
 1 1 -12

4. Final step: Repeat the process one last time. Multiply the last number below the line (-12) by the potential root (-5) and write the result (60) under the last coefficient (-60). Add the two numbers (-60 and 60) and write the sum (0) below the line.

-5 | 1 6 -7 -60
   | -5 -5 60
   |__________
 1 1 -12 0

Interpreting the Results

The numbers below the line represent the coefficients of the quotient and the remainder. The last number (0 in our example) is the remainder. The other numbers (1, 1, and -12) are the coefficients of the quotient, which is a polynomial one degree lower than the original polynomial.

In our example:

• Remainder: 0
• Quotient: x^2 + x - 12

Since the remainder is 0, this confirms that -5 is a root of the polynomial f(x) = x^3 + 6x^2 - 7x - 60. This also means that (x + 5) is a factor of the polynomial. We can write the original polynomial as:

f(x) = (x + 5)(x^2 + x - 12)

Factoring the Quotient

We can further factor the quadratic quotient x^2 + x - 12:

x^2 + x - 12 = (x + 4)(x - 3)

Therefore, the completely factored form of the original polynomial is:

f(x) = (x + 5)(x + 4)(x - 3)

And the roots of the polynomial are -5, -4, and 3.

Completing the Division Problem

Now, let's return to the original problem presented in the prompt:

-5 | 1 6 -7 -60
   | a c 60
   |__________
 1 b d 0

a = X => -5

We need to find the values of a, b, c, and d that complete the synthetic division problem. Based on our understanding of the process, we can fill in the missing numbers:

• a: This is the first multiplication step. We multiply -5 (the potential root) by 1 (the first coefficient), so a = -5 * 1 = -5.
• b: This is the sum of the second coefficient (6) and a (-5), so b = 6 + (-5) = 1.
• c: This is the second multiplication step. We multiply -5 (the potential root) by b (1), so c = -5 * 1 = -5.
• d: This is the sum of the third coefficient (-7) and c (-5), so d = -7 + (-5) = -12.

Therefore, the completed synthetic division problem is:

-5 | 1 6 -7 -60
   | -5 -5 60
   |__________
 1 1 -12 0

And the values are:

• a = -5
• b = 1
• c = -5
• d = -12

Key Takeaways

• Synthetic division is a powerful tool for dividing polynomials by linear factors.
• The Remainder Theorem connects synthetic division to finding roots of polynomials.
• A remainder of 0 indicates that the potential root is indeed a root of the polynomial.
• The numbers below the line in synthetic division represent the coefficients of the quotient and the remainder.

By mastering synthetic division, you gain a valuable technique for simplifying polynomial expressions, finding roots, and ultimately solving polynomial equations. This method not only streamlines the division process but also provides a deeper understanding of the relationship between roots, factors, and remainders in polynomial algebra.

In conclusion, synthetic division is an indispensable tool in algebra for simplifying polynomial division and identifying potential roots. Its efficiency and elegance make it a cornerstone technique for anyone delving into the world of polynomials. By understanding the steps involved and the underlying principles, you can confidently apply synthetic division to solve a wide range of algebraic problems.