Complete The Synthetic Division Problem Below 2 1 7

Mastering Synthetic Division: A Comprehensive Guide to Solving 2x² + x + 7

Synthetic division is a shortcut method used to divide polynomials by binomials of the form (x - c), where 'c' is a constant. It's a powerful tool that simplifies polynomial division, making it faster and less prone to errors than long division. This comprehensive guide will walk you through the process of synthetic division, explaining the underlying principles and providing step-by-step instructions, specifically tackling the problem of dividing 2x² + x + 7 by a binomial. We will also delve into the theoretical underpinnings and address frequently asked questions.

Understanding the Fundamentals of Synthetic Division

Before diving into the problem, let's establish the foundation. Synthetic division is based on the Remainder Theorem, which states that when a polynomial P(x) is divided by (x - c), the remainder is P(c). This means that substituting 'c' into the polynomial will give you the remainder. This theorem is crucial because synthetic division cleverly manipulates the coefficients of the polynomial to directly calculate the quotient and remainder without the cumbersome process of long division. The method works only when dividing by a linear binomial (x - c). If the divisor is anything more complex (e.g., 2x - 1, x² + 1), you must resort to long division.

Step-by-Step Solution: Dividing 2x² + x + 7 by (x - c)

The problem statement "complete the synthetic division problem below 2 1 7" is incomplete. Synthetic division requires knowing the binomial divisor, (x - c). We need the value of 'c' to proceed. Let's assume different values of 'c' to illustrate the process:

Example 1: Dividing 2x² + x + 7 by (x - 2)

Here, c = 2. Let's perform synthetic division:

Set up the synthetic division table:
```
2 | 2   1   7
  |
  |---------
  |
```
Bring down the leading coefficient:
```
2 | 2   1   7
  |
  |---------
  | 2
```
Multiply the divisor (2) by the leading coefficient (2) and write the result under the next coefficient (1):
```
2 | 2   1   7
  |    4
  |---------
  | 2
```

Add the numbers in the second column:

2 | 2   1   7
  |    4
  |---------
  | 2   5

Multiply the divisor (2) by the result (5) and write it under the next coefficient (7):
```
2 | 2   1   7
  |    4  10
  |---------
  | 2   5
```

Add the numbers in the third column:

2 | 2   1   7
  |    4  10
  |---------
  | 2   5  17

Interpreting the Results:

The last number (17) is the remainder.
The other numbers (2 and 5) are the coefficients of the quotient.

Therefore, when 2x² + x + 7 is divided by (x - 2), the quotient is 2x + 5 and the remainder is 17. We can express this as:

2x² + x + 7 = (x - 2)(2x + 5) + 17

Example 2: Dividing 2x² + x + 7 by (x + 3)

Here, c = -3. Remember that (x + 3) is equivalent to (x - (-3)).

Set up the synthetic division table:
```
-3 | 2   1   7
   |
   |---------
   |
```

Bring down the leading coefficient:

-3 | 2   1   7
   |
   |---------
   | 2

Multiply the divisor (-3) by the leading coefficient (2) and write the result under the next coefficient (1):
```
-3 | 2   1   7
   |   -6
   |---------
   | 2
```

Add the numbers in the second column:

-3 | 2   1   7
   |   -6
   |---------
   | 2  -5

Multiply the divisor (-3) by the result (-5) and write it under the next coefficient (7):
```
-3 | 2   1   7
   |   -6  15
   |---------
   | 2  -5
```

Add the numbers in the third column:

-3 | 2   1   7
   |   -6  15
   |---------
   | 2  -5  22

Interpreting the Results:

The remainder is 22.
The quotient is 2x - 5.

Therefore, 2x² + x + 7 = (x + 3)(2x - 5) + 22

The Mathematical Proof Behind Synthetic Division

Synthetic division is a streamlined version of polynomial long division. Let's consider the general case of dividing a polynomial P(x) = aₙxⁿ + aₙ₋₁xⁿ⁻¹ + ... + a₁x + a₀ by (x - c). Long division would yield a quotient Q(x) and a remainder R, such that:

P(x) = (x - c)Q(x) + R

By the Remainder Theorem, R = P(c). Synthetic division cleverly extracts the coefficients of Q(x) and the remainder R without explicitly performing the long division steps. The process involves a series of multiplications and additions that directly compute these values. This efficiency is why synthetic division is preferred for polynomial division by linear binomials.

Frequently Asked Questions (FAQ)

Q1: What if the polynomial has a missing term?

A: If a term with a specific power of x is missing (e.g., 3x³ + 7, where the x² term is missing), you must insert a zero as a placeholder for that missing coefficient in the synthetic division table. For instance, for the polynomial 3x³ + 7, you'd set up the table as: c | 3 0 0 7.

Q2: Can synthetic division be used for dividing by binomials other than (x - c)?

A: No. Synthetic division only works for divisors of the form (x - c). For other divisors, you must use polynomial long division.

Q3: How do I check my answer?

A: You can verify your answer by expanding the product of the quotient and the divisor and adding the remainder. This should give you the original polynomial. For example, in Example 1: (x - 2)(2x + 5) + 17 = 2x² + 5x - 4x - 10 + 17 = 2x² + x + 7. This matches the original polynomial.

Q4: What are the real-world applications of synthetic division?

A: Synthetic division has various applications in mathematics and related fields, such as:

Finding roots of polynomials: The Remainder Theorem, which underlies synthetic division, is crucial in determining if a given value is a root of a polynomial.
Polynomial factoring: Synthetic division helps factor polynomials, making it easier to solve equations and analyze functions.
Curve sketching: Understanding the roots and behavior of a polynomial is critical in sketching its graph accurately.
Calculus: Synthetic division simplifies calculations involving derivatives and integrals of polynomials.

Conclusion: Mastering a Powerful Tool

Synthetic division is a valuable tool for anyone working with polynomials. Understanding its principles, steps, and limitations empowers you to solve polynomial division problems efficiently and accurately. While seemingly a simple technique, its underlying mathematical foundation is robust and has significant implications across various mathematical applications. Practice is key to mastering synthetic division; try different problems with various values of 'c' and polynomials to build your proficiency. By understanding the steps and the theoretical justification, you can confidently tackle more complex polynomial division problems and develop a deeper appreciation for the elegance and power of this method. Remember to always check your work to ensure accuracy. Through consistent practice and careful application, you'll find synthetic division becomes an intuitive and invaluable tool in your mathematical arsenal.