Complete The Synthetic Division Problem Below 2 8 6

Synthetic division simplifies polynomial division, offering a streamlined approach, especially when dividing by a linear factor. Mastering this technique unlocks a deeper understanding of polynomial behavior and factorization. Let's dive into completing the synthetic division problem for "2 8 6".

Understanding Synthetic Division: A Foundation

Synthetic division is a shorthand method of dividing a polynomial by a linear expression of the form x - k. It offers a more efficient way to perform polynomial division compared to long division, particularly when dealing with simpler divisors. It hinges on extracting the coefficients of the polynomial and using a systematic process of multiplication and addition.

Before we jump into the specific problem "2 8 6," let's clarify the broader context. Polynomials are expressions consisting of variables and coefficients, combined using addition, subtraction, and non-negative integer exponents. Examples include x² + 3x - 2 or 5x⁴ - x + 7. The degree of a polynomial is the highest power of the variable.

The division algorithm for polynomials states that for any two polynomials, P(x) (the dividend) and D(x) (the divisor), where D(x) is not zero, there exist unique polynomials Q(x) (the quotient) and R(x) (the remainder) such that:

P(x) = D(x) * Q(x) + R(x)

Synthetic division provides a practical way to find Q(x) and R(x) when D(x) is linear.

Setting Up the Synthetic Division Problem "2 8 6"

The expression "2 8 6" represents the coefficients of a polynomial. Assuming the polynomial is in descending order of powers of x, we can interpret it as:

2x² + 8x + 6

Therefore, our dividend P(x) is 2x² + 8x + 6. To perform synthetic division, we also need a divisor of the form x - k. The problem "2 8 6" doesn't explicitly give us k. We'll need to choose a value for k. Let's start by exploring a few options, and then we'll consider how to find suitable values of k through factoring.

Scenario 1: Dividing by (x - 1), i.e., k = 1

• Write down the coefficients of the polynomial: 2 8 6
• Write the value of k (which is 1) to the left: 1 | 2 8 6
• Draw a horizontal line below the coefficients, leaving space for the results.

The setup looks like this:

1 | 2  8  6
    ---------

Scenario 2: Dividing by (x + 1), i.e., k = -1

• Write down the coefficients of the polynomial: 2 8 6
• Write the value of k (which is -1) to the left: -1 | 2 8 6
• Draw a horizontal line below the coefficients, leaving space for the results.

The setup looks like this:

-1 | 2  8  6
     ---------

Scenario 3: Dividing by (x + 3), i.e., k = -3

• Write down the coefficients of the polynomial: 2 8 6
• Write the value of k (which is -3) to the left: -3 | 2 8 6
• Draw a horizontal line below the coefficients, leaving space for the results.

The setup looks like this:

-3 | 2  8  6
     ---------

Choosing the right value for k often involves factoring or finding roots of the polynomial. Let's consider the polynomial 2x² + 8x + 6. We can factor out a 2:

2*(x² + 4x + 3)

Then, we can factor the quadratic:

2*(x + 1)(x + 3)

This tells us that x = -1 and x = -3 are roots of the polynomial. Therefore, dividing by (x + 1) or (x + 3) will result in a remainder of zero.

Let's proceed with the synthetic division using k = -3, as it will give us a clean result and demonstrate the process clearly.

Step-by-Step Synthetic Division with k = -3

Here's how we complete the synthetic division with k = -3:

1. Bring down the first coefficient: Bring down the '2' from the first position to below the line.

-3 | 2  8  6
     ---------
       2

2. Multiply and add:
   • Multiply the '2' (below the line) by k (-3): 2 * -3 = -6
   • Write the result (-6) under the next coefficient (8).
   • Add the numbers in that column: 8 + (-6) = 2

-3 | 2  8  6
     -6
    ---------
       2  2

3. Repeat the process:
   • Multiply the '2' (below the line) by k (-3): 2 * -3 = -6
   • Write the result (-6) under the next coefficient (6).
   • Add the numbers in that column: 6 + (-6) = 0

-3 | 2  8  6
     -6 -6
    ---------
       2  2  0

Interpreting the Results

The numbers below the line represent the coefficients of the quotient and the remainder. Reading from left to right:

• The last number (0) is the remainder.
• The preceding numbers (2 and 2) are the coefficients of the quotient.

Since the original polynomial was of degree 2 (quadratic), and we divided by a linear factor (degree 1), the quotient will be of degree 1 (linear). Therefore, the quotient is:

2*x + 2

The remainder is 0, which confirms that (x + 3) is a factor of 2x² + 8x + 6.

Therefore, we can write:

2x² + 8x + 6 = (x + 3)(2x + 2)

Synthetic Division with k = -1

Let's perform the synthetic division again, this time with k = -1:

-1 | 2  8  6
     ---------

1. Bring down the first coefficient:

-1 | 2  8  6
     ---------
       2

2. Multiply and add:
   • Multiply the '2' by k (-1): 2 * -1 = -2
   • Write the result (-2) under the next coefficient (8).
   • Add the numbers in that column: 8 + (-2) = 6

-1 | 2  8  6
     -2
    ---------
       2  6

3. Repeat the process:
   • Multiply the '6' by k (-1): 6 * -1 = -6
   • Write the result (-6) under the next coefficient (6).
   • Add the numbers in that column: 6 + (-6) = 0

-1 | 2  8  6
     -2 -6
    ---------
       2  6  0

The quotient is 2x + 6, and the remainder is 0.

Therefore:

2x² + 8x + 6 = (x + 1)(2x + 6)

Synthetic Division with k = 1

Now, let's try synthetic division with k = 1:

1 | 2  8  6
    ---------

1. Bring down the first coefficient:

1 | 2  8  6
    ---------
      2

2. Multiply and add:
   • Multiply the '2' by k (1): 2 * 1 = 2
   • Write the result (2) under the next coefficient (8).
   • Add the numbers in that column: 8 + 2 = 10

1 | 2  8  6
    2
    ---------
      2  10

3. Repeat the process:
   • Multiply the '10' by k (1): 10 * 1 = 10
   • Write the result (10) under the next coefficient (6).
   • Add the numbers in that column: 6 + 10 = 16

1 | 2  8  6
    2  10
    ---------
      2  10 16

In this case, the quotient is 2x + 10, and the remainder is 16. This means (x - 1) is not a factor of 2x² + 8x + 6.

Therefore:

2x² + 8x + 6 = (x - 1)(2x + 10) + 16

Choosing the Right 'k' and the Remainder Theorem

The examples above highlight the importance of choosing the correct value for k. When we factored the polynomial, we found the roots x = -1 and x = -3. Using these values in synthetic division resulted in a remainder of 0, meaning (x + 1) and (x + 3) are factors of the polynomial. When we used k = 1, we obtained a non-zero remainder.

This is directly related to the Remainder Theorem. The Remainder Theorem states that if a polynomial P(x) is divided by (x - k), then the remainder is P(k). In other words, to find the remainder when dividing by (x - k), you can simply substitute k into the polynomial.

In our example:

• P(x) = 2x² + 8x + 6
• If k = -1, then P(-1) = 2(-1)² + 8(-1) + 6 = 2 - 8 + 6 = 0 (remainder is 0)
• If k = -3, then P(-3) = 2(-3)² + 8(-3) + 6 = 18 - 24 + 6 = 0 (remainder is 0)
• If k = 1, then P(1) = 2(1)² + 8(1) + 6 = 2 + 8 + 6 = 16 (remainder is 16)

This reinforces the idea that finding the roots of the polynomial is crucial for efficient synthetic division, especially when you want to completely factor the polynomial.

Applications of Synthetic Division

Synthetic division has various applications in algebra and calculus:

• Finding roots of polynomials: As demonstrated, if the remainder is 0 after synthetic division, then k is a root of the polynomial.
• Factoring polynomials: When you find a root k, you can factor the polynomial as (x - k) times the quotient obtained from synthetic division.
• Evaluating polynomials: The Remainder Theorem provides a quick way to evaluate a polynomial at a specific value k. Simply perform synthetic division by (x - k), and the remainder is the value of the polynomial at k.
• Simplifying rational expressions: Synthetic division can be used to simplify rational expressions where the numerator is a polynomial and the denominator is a linear expression.
• Solving polynomial equations: By finding the roots of a polynomial using synthetic division (combined with other techniques), you can solve polynomial equations.

Potential Pitfalls and Considerations

• Missing Terms: When setting up the synthetic division, ensure you include a '0' as a placeholder for any missing terms in the polynomial. For example, if you have x³ + 2x - 1, you would write the coefficients as 1 0 2 -1. The '0' accounts for the missing x² term.
• Only Linear Divisors: Synthetic division is only directly applicable when dividing by linear expressions of the form (x - k). For divisors of higher degree, you'll need to use long division.
• Fractional Roots: While factoring can help find integer roots, synthetic division can also be used to test fractional roots. However, the arithmetic can be more cumbersome.
• Complex Roots: Synthetic division works with complex roots as well, but the calculations involve complex numbers.

Beyond the Basics: Connecting to Polynomial Long Division

It's helpful to understand the relationship between synthetic division and polynomial long division. Both methods achieve the same goal: dividing one polynomial by another. However, synthetic division is a condensed and streamlined version that works only for linear divisors.

If we were to perform long division on 2x² + 8x + 6 divided by (x + 3), we would arrive at the same quotient (2x + 2) and remainder (0) as we did with synthetic division. The advantage of synthetic division is its speed and simplicity, once you understand the process. However, long division is more general and can handle divisors of any degree.

Practice Problems

To solidify your understanding, try these practice problems:

1. Divide x³ - 4x² + x + 6 by (x - 2) using synthetic division.
2. Divide 3x³ + 7x² - 4x + 4 by (x + 2) using synthetic division.
3. Use synthetic division and the Remainder Theorem to find P(3) if P(x) = x⁴ - 5x² + 4x - 1.
4. Factor the polynomial x³ + 6x² + 11x + 6 completely, given that (x + 1) is a factor. (Hint: use synthetic division to find the quotient, then factor the quotient).

Conclusion

Synthetic division is a powerful tool for simplifying polynomial division. By understanding the underlying principles, practicing the steps, and recognizing its limitations, you can effectively use it to solve a variety of algebraic problems. Remember to connect it to the broader concepts of factoring, roots, and the Remainder Theorem for a deeper understanding of polynomial behavior. While the problem "2 8 6" initially seemed simple, it opened the door to a comprehensive exploration of this essential algebraic technique.