Synthetic Division Is X+3 A Factor Of 2x^6+57x^3+81

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Determining if a binomial like x + 3 is a factor of a polynomial such as 2x⁶ + 57x³ + 81 is a common task in algebra. One efficient method for this is synthetic division. In this comprehensive guide, we will explore how to use synthetic division to determine if x + 3 is indeed a factor of the given polynomial. We will break down the process step-by-step, explain the underlying concepts, and discuss the significance of the remainder in determining factors. Let's dive in!

Understanding Synthetic Division

Synthetic division is a streamlined method for dividing a polynomial by a linear divisor of the form x - k. It's a shortcut that avoids the more cumbersome long division process. The key principle behind synthetic division is the Factor Theorem, which states that a polynomial f(x) has a factor (x - k) if and only if f(k) = 0. In other words, if when we divide f(x) by (x - k), the remainder is zero, then (x - k) is a factor of f(x). Conversely, if the remainder is not zero, then (x - k) is not a factor.

In our case, we want to determine if x + 3 is a factor of 2x⁶ + 57x³ + 81. To use synthetic division, we first need to identify the value of k. Since our divisor is in the form x + 3, which can be rewritten as x - (-3), we have k = -3. Now we can set up the synthetic division process.

Setting Up Synthetic Division

To set up synthetic division, we follow these steps:

  1. Write the coefficients: Write down the coefficients of the polynomial in descending order of powers of x. It's crucial to include a zero as a placeholder for any missing terms. Our polynomial is 2x⁶ + 57x³ + 81. Notice that we have terms for x⁶ and x³, but no terms for x⁵, x⁴, x², and x. So, we write the coefficients as: 2, 0, 0, 57, 0, 0, 81.
  2. Write the value of k: Write the value of k (which is -3 in our case) to the left of the coefficients.
  3. Draw a line: Draw a horizontal line below the coefficients, leaving space for a row of numbers below the line.

The setup will look like this:

-3 | 2  0  0  57  0  0  81
   |________________________

Performing Synthetic Division

Now, let's perform the synthetic division:

  1. Bring down the first coefficient: Bring down the first coefficient (2) below the line.
-3 | 2  0  0  57  0  0  81
   |________________________
     2
  1. Multiply and add: Multiply the value of k (-3) by the number you just brought down (2), and write the result (-6) below the next coefficient (0).
-3 | 2  0  0  57  0  0  81
   |    -6
   |________________________
     2
  1. Add the numbers in the column: Add the numbers in the second column (0 and -6) and write the sum (-6) below the line.
-3 | 2  0  0  57  0  0  81
   |    -6
   |________________________
     2 -6
  1. Repeat: Repeat steps 2 and 3 for the remaining coefficients. Multiply -3 by -6 to get 18, write 18 below the next coefficient (0), and add them to get 18.
-3 | 2  0  0  57  0  0  81
   |    -6 18
   |________________________
     2 -6 18

Multiply -3 by 18 to get -54, write -54 below 57, and add them to get 3.

-3 | 2  0  0  57  0  0  81
   |    -6 18 -54
   |________________________
     2 -6 18  3

Multiply -3 by 3 to get -9, write -9 below 0, and add them to get -9.

-3 | 2  0  0  57  0  0  81
   |    -6 18 -54 -9
   |________________________
     2 -6 18  3 -9

Multiply -3 by -9 to get 27, write 27 below 0, and add them to get 27.

-3 | 2  0  0  57  0  0  81
   |    -6 18 -54 -9 27
   |________________________
     2 -6 18  3 -9 27

Finally, multiply -3 by 27 to get -81, write -81 below 81, and add them to get 0.

-3 | 2  0  0  57  0  0  81
   |    -6 18 -54 -9 27 -81
   |________________________
     2 -6 18  3 -9 27  0

Interpreting the Result

The last number below the line (0) is the remainder. The other numbers (2, -6, 18, 3, -9, 27) are the coefficients of the quotient polynomial. Since the remainder is 0, according to the Factor Theorem, x + 3 is a factor of 2x⁶ + 57x³ + 81.

The quotient polynomial is one degree lower than the original polynomial. So, in this case, the quotient is a fifth-degree polynomial: 2x⁵ - 6x⁴ + 18x³ + 3x² - 9x + 27.

Conclusion

In conclusion, by performing synthetic division, we found that the remainder is 0 when dividing 2x⁶ + 57x³ + 81 by x + 3. Therefore, x + 3 is indeed a factor of the given polynomial. This method provides a quick and efficient way to determine factors of polynomials, especially when dealing with linear divisors. Understanding synthetic division and the Factor Theorem is crucial for polynomial factorization and solving algebraic equations.

To summarize, we have shown through synthetic division that because the remainder is 0, x + 3 is a factor of the polynomial 2x⁶ + 57x³ + 81. This demonstrates the power and efficiency of synthetic division in polynomial factorization.

When working with polynomials, a fundamental task is to determine if a given binomial is a factor. A powerful technique for this purpose is synthetic division. This article will delve into the application of synthetic division to assess whether x + 3 is a factor of the polynomial 2x⁶ + 57x³ + 81. We will explore the methodology, the underlying principles, and the interpretation of results. Let's embark on this exploration.

The Essence of Synthetic Division

Synthetic division is an efficient algorithm for dividing a polynomial by a linear expression of the form x - k. It simplifies the process compared to traditional long division, especially when dealing with higher-degree polynomials. The cornerstone of synthetic division is the Factor Theorem, which posits that a polynomial f(x) has a factor (x - k) if and only if f(k) = 0. In practical terms, if the division of f(x) by (x - k) yields a remainder of zero, then (x - k) is confirmed as a factor of f(x). Conversely, a non-zero remainder indicates that (x - k) is not a factor.

Our objective is to ascertain whether x + 3 is a factor of the polynomial 2x⁶ + 57x³ + 81. To proceed with synthetic division, we must first identify the value of k. Given the divisor x + 3, which can be expressed as x - (-3), we deduce that k = -3. With this value, we can now set up the synthetic division process.

Setting the Stage for Synthetic Division

The setup for synthetic division involves the following steps:

  1. Extract the Coefficients: Write down the coefficients of the polynomial in descending order of the powers of x. It is critical to use zero as a placeholder for any missing terms. Our polynomial is 2x⁶ + 57x³ + 81. Note that we have terms for x⁶ and x³, but no terms for x⁵, x⁴, x², and x. Thus, the coefficients are written as: 2, 0, 0, 57, 0, 0, 81.
  2. Identify k Value: The k value which is -3 in our problem, is written to the left of the coefficients.
  3. Horizontal Line: Draw a horizontal line beneath the coefficients, providing space for a row of numbers below.

The initial setup is as follows:

-3 | 2  0  0  57  0  0  81
   |________________________

Executing Synthetic Division: A Step-by-Step Guide

The synthetic division process unfolds as follows:

  1. Bring Down the Lead: The first coefficient (2) is brought down below the line.
-3 | 2  0  0  57  0  0  81
   |________________________
     2
  1. Multiply and Place: Multiply k (-3) by the number brought down (2), resulting in -6, which is placed below the next coefficient (0).
-3 | 2  0  0  57  0  0  81
   |    -6
   |________________________
     2
  1. Sum the Column: Add the numbers in the second column (0 and -6) to get -6, placing the sum below the line.
-3 | 2  0  0  57  0  0  81
   |    -6
   |________________________
     2 -6
  1. Iterate: Repeat steps 2 and 3 for all remaining coefficients. Multiply -3 by -6 to obtain 18, place 18 below the next coefficient (0), and add them to get 18.
-3 | 2  0  0  57  0  0  81
   |    -6 18
   |________________________
     2 -6 18

Multiply -3 by 18 to get -54, place -54 below 57, and add them to get 3.

-3 | 2  0  0  57  0  0  81
   |    -6 18 -54
   |________________________
     2 -6 18  3

Multiply -3 by 3 to get -9, place -9 below 0, and add them to get -9.

-3 | 2  0  0  57  0  0  81
   |    -6 18 -54 -9
   |________________________
     2 -6 18  3 -9

Multiply -3 by -9 to get 27, place 27 below 0, and add them to get 27.

-3 | 2  0  0  57  0  0  81
   |    -6 18 -54 -9 27
   |________________________
     2 -6 18  3 -9 27

Finally, multiply -3 by 27 to get -81, place -81 below 81, and add them to get 0.

-3 | 2  0  0  57  0  0  81
   |    -6 18 -54 -9 27 -81
   |________________________
     2 -6 18  3 -9 27  0

The Remainder Significance

The final number on the bottom row (0) is the remainder. The preceding numbers (2, -6, 18, 3, -9, 27) are the coefficients of the quotient polynomial. Given that the remainder is 0, the Factor Theorem confirms that x + 3 is indeed a factor of 2x⁶ + 57x³ + 81.

The resulting quotient polynomial is one degree lower than the original polynomial, yielding 2x⁵ - 6x⁴ + 18x³ + 3x² - 9x + 27.

Conclusion: The Factor Theorem and Synthetic Division

In summation, we have successfully employed synthetic division to determine that x + 3 is a factor of 2x⁶ + 57x³ + 81, as evidenced by the zero remainder. This method offers a swift and efficient approach for identifying polynomial factors, particularly those of the linear form. A strong grasp of synthetic division and the Factor Theorem is indispensable for polynomial manipulation and algebraic problem-solving.

To reiterate, synthetic division definitively shows that the remainder is 0, thereby establishing x + 3 as a factor of the polynomial 2x⁶ + 57x³ + 81. This underscores the practical value of synthetic division in the realm of polynomial factorization.

In polynomial algebra, identifying factors of a polynomial is a crucial task. Synthetic division provides an efficient way to determine if a binomial of the form x - k is a factor of a given polynomial. This article will focus on using synthetic division to check whether x + 3 is a factor of the polynomial 2x⁶ + 57x³ + 81. We'll walk through the method, explain the underlying principles, and interpret the results. Let's get started.

The Power of Synthetic Division

Synthetic division is a simplified method for dividing a polynomial by a linear expression, specifically of the form x - k. It’s more efficient than traditional long division, especially when dealing with higher-degree polynomials. The cornerstone of this method is the Factor Theorem, which states that a polynomial f(x) has a factor (x - k) if and only if f(k) = 0. In simpler terms, if the remainder after dividing f(x) by (x - k) is zero, then (x - k) is a factor of f(x). Conversely, if the remainder is not zero, then (x - k) is not a factor.

Our goal is to determine if x + 3 is a factor of the polynomial 2x⁶ + 57x³ + 81. To use synthetic division, we first need to identify the value of k. Since our divisor is x + 3, which can be rewritten as x - (-3), we find that k = -3. Now, we can set up the synthetic division process.

Setting Up the Synthetic Division Process

Setting up synthetic division involves these key steps:

  1. List the Coefficients: Write down the coefficients of the polynomial in descending order of the powers of x. It’s vital to include zeros as placeholders for any missing terms. Our polynomial is 2x⁶ + 57x³ + 81. Notice that we have terms for x⁶ and x³, but no terms for x⁵, x⁴, x², and x. Therefore, the coefficients are written as: 2, 0, 0, 57, 0, 0, 81.
  2. Identify the k Value: The value of k, which is -3 in our case, is written to the left of the coefficients.
  3. Draw the Line: Draw a horizontal line below the coefficients, leaving space for a row of numbers below.

The initial setup looks like this:

-3 | 2  0  0  57  0  0  81
   |________________________

Performing Synthetic Division: A Step-by-Step Guide

Let's perform the synthetic division, step by step:

  1. Bring Down: Bring down the first coefficient (2) below the line.
-3 | 2  0  0  57  0  0  81
   |________________________
     2
  1. Multiply and Add: Multiply the value of k (-3) by the number you just brought down (2), which gives -6. Write -6 below the next coefficient (0).
-3 | 2  0  0  57  0  0  81
   |    -6
   |________________________
     2
  1. Add Down: Add the numbers in the second column (0 and -6), which gives -6. Write -6 below the line.
-3 | 2  0  0  57  0  0  81
   |    -6
   |________________________
     2 -6
  1. Repeat the Process: Repeat steps 2 and 3 for the remaining coefficients. Multiply -3 by -6 to get 18, write 18 below the next coefficient (0), and add them to get 18.
-3 | 2  0  0  57  0  0  81
   |    -6 18
   |________________________
     2 -6 18

Multiply -3 by 18 to get -54, write -54 below 57, and add them to get 3.

-3 | 2  0  0  57  0  0  81
   |    -6 18 -54
   |________________________
     2 -6 18  3

Multiply -3 by 3 to get -9, write -9 below 0, and add them to get -9.

-3 | 2  0  0  57  0  0  81
   |    -6 18 -54 -9
   |________________________
     2 -6 18  3 -9

Multiply -3 by -9 to get 27, write 27 below 0, and add them to get 27.

-3 | 2  0  0  57  0  0  81
   |    -6 18 -54 -9 27
   |________________________
     2 -6 18  3 -9 27

Finally, multiply -3 by 27 to get -81, write -81 below 81, and add them to get 0.

-3 | 2  0  0  57  0  0  81
   |    -6 18 -54 -9 27 -81
   |________________________
     2 -6 18  3 -9 27  0

Interpreting the Results: The Remainder is Key

The last number below the line (0) is the remainder. The other numbers (2, -6, 18, 3, -9, 27) are the coefficients of the quotient polynomial. Since the remainder is 0, according to the Factor Theorem, x + 3 is a factor of 2x⁶ + 57x³ + 81.

The quotient polynomial is one degree lower than the original polynomial, making it a fifth-degree polynomial: 2x⁵ - 6x⁴ + 18x³ + 3x² - 9x + 27.

Conclusion: Synthetic Division and the Factor Theorem

In summary, we've used synthetic division to determine that the remainder is 0 when dividing 2x⁶ + 57x³ + 81 by x + 3. This confirms that x + 3 is a factor of the given polynomial. This technique is a valuable tool for polynomial factorization and solving algebraic problems.

Therefore, through synthetic division, we have demonstrated that x + 3 is indeed a factor of the polynomial 2x⁶ + 57x³ + 81, as the remainder is 0. This highlights the usefulness of synthetic division in factor identification.