Complete The Synthetic Division Problem Below 2 1 5

Let's dive into synthetic division, a streamlined method for dividing polynomials, particularly useful when the divisor is in the form of x - k. It's a powerful shortcut that simplifies the often tedious process of long division. We'll break down the process step-by-step, illustrating how to complete the synthetic division problem: 2 | 1 5.

What is Synthetic Division?

Synthetic division is a simplified way to divide a polynomial by a linear expression of the form x - k. It avoids the use of variables and exponents, focusing instead on the coefficients of the polynomial. This makes the division process faster and more efficient, especially for higher-degree polynomials. It's essentially a shortcut derived from polynomial long division. While it can only be used with linear divisors, it's an incredibly useful tool in algebra and calculus.

Setting Up the Problem: 2 | 1 5

Our problem is to understand what the expression "2 | 1 5" represents in the context of synthetic division and how to properly execute it. This notation tells us several things:

• Divisor: The number to the left of the vertical bar, which is 2, represents the value of k in the linear divisor x - k. In this case, k = 2, so the divisor is x - 2.
• Dividend: The numbers to the right of the vertical bar, which are 1 and 5, represent the coefficients of the polynomial being divided. We need to understand how these coefficients translate to a polynomial expression. Since we have two coefficients, the polynomial is of degree one (linear).

Interpreting the Dividend

The coefficients 1 and 5 represent the polynomial 1x + 5, or simply x + 5. Therefore, the problem "2 | 1 5" is equivalent to dividing the polynomial x + 5 by x - 2.

The Complete Synthetic Division Setup

Now that we understand what each part of the notation represents, we can set up the synthetic division problem:

2 | 1   5
  |
  |_________

Here, 2 is the value of k (from x - k), 1 is the coefficient of x in the dividend, and 5 is the constant term of the dividend.

The Steps of Synthetic Division

Now that we have the setup, let's walk through the steps of synthetic division:

1. Bring Down the First Coefficient: Bring the first coefficient of the dividend (which is 1 in our case) down below the horizontal line.

   2 | 1   5
     |
     |_________
       1
   

2. Multiply and Carry: Multiply the value of k (which is 2) by the number you just brought down (which is 1). Write the result (2 * 1 = 2) under the next coefficient of the dividend (which is 5).

   2 | 1   5
     |   2
     |_________
       1
   

3. Add Down: Add the numbers in the second column (5 + 2 = 7) and write the result below the horizontal line.

   2 | 1   5
     |   2
     |_________
       1   7
   

4. Interpret the Result: The numbers below the horizontal line represent the coefficients of the quotient and the remainder. In this case, 1 is the coefficient of the quotient, and 7 is the remainder.

Interpreting the Resulting Polynomial

The last row of the synthetic division gives us the coefficients of the quotient and the remainder. Reading from left to right:

• The '1' represents the coefficient of the quotient. Since we started with a linear polynomial (x + 5) and divided by a linear expression (x - 2), the quotient will be a constant. Therefore, the quotient is simply 1.
• The '7' is the remainder.

Therefore, the result of the synthetic division is:

• Quotient: 1
• Remainder: 7

Writing the Solution

The solution to the division problem can be expressed as:

x + 5 = (x - 2)(1) + 7

Or, dividing both sides by (x - 2):

(x + 5) / (x - 2) = 1 + 7/(x - 2)

This means that when you divide x + 5 by x - 2, you get a quotient of 1 and a remainder of 7.

Example Problems and Detailed Explanation

Let's explore a variety of examples to solidify your understanding of synthetic division. We'll work through each problem step-by-step, highlighting key concepts and potential pitfalls.

Example 1: Dividing a Quadratic by a Linear Expression

Divide x² + 3x - 10 by x - 2 using synthetic division.

1. Identify k and the coefficients:

   • The divisor is x - 2, so k = 2.
   • The coefficients of the dividend x² + 3x - 10 are 1, 3, and -10.

2. Set up the synthetic division:

   2 | 1   3  -10
     |
     |_________
   

3. Bring down the first coefficient:

   2 | 1   3  -10
     |
     |_________
       1
   

4. Multiply and carry:

   2 | 1   3  -10
     |   2
     |_________
       1
   

5. Add down:

   2 | 1   3  -10
     |   2
     |_________
       1   5
   

6. Multiply and carry:

   2 | 1   3  -10
     |   2   10
     |_________
       1   5
   

7. Add down:

   2 | 1   3  -10
     |   2   10
     |_________
       1   5   0
   

8. Interpret the result:

   • The coefficients of the quotient are 1 and 5. Since we started with a quadratic and divided by a linear expression, the quotient is linear. Thus, the quotient is 1x + 5, or x + 5.
   • The remainder is 0.

9. Write the solution:

   x² + 3x - 10 = (x - 2)(x + 5) + 0

   (x² + 3x - 10) / (x - 2) = x + 5

   Therefore, x² + 3x - 10 divided by x - 2 is x + 5.

Example 2: Dealing with Missing Terms

Divide 2x³ - 5x + 6 by x + 3 using synthetic division.

1. Identify k and the coefficients:

   • The divisor is x + 3, which can be written as x - (-3), so k = -3.
   • The coefficients of the dividend 2x³ - 5x + 6 are 2, 0, -5, and 6. Notice that we must include a 0 for the missing x² term. This is crucial for proper synthetic division.

2. Set up the synthetic division:

   -3 | 2   0  -5   6
      |
      |_________
   

3. Bring down the first coefficient:

   -3 | 2   0  -5   6
      |
      |_________
        2
   

4. Multiply and carry:

   -3 | 2   0  -5   6
      |  -6
      |_________
        2
   

5. Add down:

   -3 | 2   0  -5   6
      |  -6
      |_________
        2  -6
   

6. Multiply and carry:

   -3 | 2   0  -5   6
      |  -6  18
      |_________
        2  -6
   

7. Add down:

   -3 | 2   0  -5   6
      |  -6  18
      |_________
        2  -6  13
   

8. Multiply and carry:

   -3 | 2   0  -5   6
      |  -6  18 -39
      |_________
        2  -6  13
   

9. Add down:

   -3 | 2   0  -5   6
      |  -6  18 -39
      |_________
        2  -6  13 -33
   

10. Interpret the result:

    • The coefficients of the quotient are 2, -6, and 13. Since we started with a cubic and divided by a linear expression, the quotient is quadratic. Thus, the quotient is 2x² - 6x + 13.
    • The remainder is -33.

11. Write the solution:

    2x³ - 5x + 6 = (x + 3)(2x² - 6x + 13) - 33

    (2x³ - 5x + 6) / (x + 3) = 2x² - 6x + 13 - 33/(x + 3)

    Therefore, 2x³ - 5x + 6 divided by x + 3 is 2x² - 6x + 13 with a remainder of -33.

Example 3: Dividing by a Fraction

Divide 4x² - x + 5 by x - (1/2).

1. Identify k and the coefficients:

   • The divisor is x - (1/2), so k = 1/2.
   • The coefficients of the dividend 4x² - x + 5 are 4, -1, and 5.

2. Set up the synthetic division:

   1/2 | 4  -1   5
       |
       |_________
   

3. Bring down the first coefficient:

   1/2 | 4  -1   5
       |
       |_________
         4
   

4. Multiply and carry:

   1/2 | 4  -1   5
       |  2
       |_________
         4
   

5. Add down:

   1/2 | 4  -1   5
       |  2
       |_________
         4   1
   

6. Multiply and carry:

   1/2 | 4  -1   5
       |  2  1/2
       |_________
         4   1
   

7. Add down:

   1/2 | 4  -1   5
       |  2  1/2
       |_________
         4   1  11/2
   

8. Interpret the result:

   • The coefficients of the quotient are 4 and 1. Since we started with a quadratic and divided by a linear expression, the quotient is linear. Thus, the quotient is 4x + 1.
   • The remainder is 11/2.

9. Write the solution:

   4x² - x + 5 = (x - 1/2)(4x + 1) + 11/2

   (4x² - x + 5) / (x - 1/2) = 4x + 1 + (11/2)/(x - 1/2)

   Therefore, 4x² - x + 5 divided by x - 1/2 is 4x + 1 with a remainder of 11/2.

Why Does Synthetic Division Work?

Synthetic division is a condensed form of polynomial long division. To understand why it works, let's compare the two methods with a simple example: dividing x² + 3x - 10 by x - 2.

Polynomial Long Division

          x + 5
    x - 2 | x² + 3x - 10
            -(x² - 2x)
            ---------
                  5x - 10
                  -(5x - 10)
                  ---------
                        0

Synthetic Division

2 | 1   3  -10
  |   2   10
  |_________
    1   5   0

Notice the parallels:

• The '2' in synthetic division comes from the x - 2 divisor, just as in long division.
• The '1' (coefficient of x in the quotient) is found in both methods.
• The '5' (constant term in the quotient) is also found in both methods.
• The '0' remainder is consistent in both.

The key is that synthetic division streamlines the process by:

• Focusing solely on the coefficients.
• Eliminating the need to explicitly write the variable x.
• Using addition instead of subtraction (by changing the sign of k).

Essentially, synthetic division performs the same calculations as long division, but in a more compact and efficient manner. It leverages the structure of polynomial division to simplify the steps and reduce the amount of writing required.

Common Mistakes to Avoid

• Forgetting the Zero Placeholder: When a term is missing in the dividend (e.g., x² in x³ + 5x - 3), you must include a zero as its coefficient. Failing to do so will result in an incorrect answer.
• Incorrectly Identifying k: Remember that synthetic division works with the form x - k. If the divisor is x + 3, then k = -3, not 3. Pay close attention to the sign.
• Misinterpreting the Result: The numbers on the bottom row represent the coefficients of the quotient and the remainder. Be sure to assign the correct powers of x to the quotient coefficients.
• Using Synthetic Division with Non-Linear Divisors: Synthetic division is designed for linear divisors of the form x - k. It cannot be used directly with quadratic or higher-degree divisors. You would need to use polynomial long division in those cases.
• Arithmetic Errors: Synthetic division involves a series of multiplications and additions. Careless arithmetic errors are a common source of mistakes. Double-check your calculations, especially when dealing with negative numbers or fractions.

Advanced Applications of Synthetic Division

Beyond simple polynomial division, synthetic division has several advanced applications:

• Finding Roots of Polynomials: If you know a value k is a root of a polynomial P(x), then P(k) = 0. This means that when you perform synthetic division with k, the remainder will be zero. This allows you to factor the polynomial and find other roots.
• Testing Potential Roots (Rational Root Theorem): The Rational Root Theorem provides a list of possible rational roots for a polynomial. You can use synthetic division to quickly test these potential roots. If the remainder is zero, you've found a root.
• Evaluating Polynomials (Remainder Theorem): The Remainder Theorem states that if you divide a polynomial P(x) by x - k, the remainder is equal to P(k). Therefore, synthetic division can be used to efficiently evaluate polynomials at specific values.
• Factoring Polynomials: When you find a root k using synthetic division (remainder is zero), you've effectively factored the polynomial. The quotient from the synthetic division represents the other factor.
• Solving Polynomial Equations: By finding roots and factoring, synthetic division can help solve polynomial equations.

Synthetic Division vs. Polynomial Long Division

When to Use Which Method

• Synthetic Division: Use when dividing by a linear expression of the form x - k. It's faster and more efficient in these cases.
• Polynomial Long Division: Use when dividing by a polynomial of degree two or higher, or when you need a more detailed understanding of the division process.

Conclusion

Synthetic division is a valuable tool for simplifying polynomial division when the divisor is linear. By understanding the steps, avoiding common mistakes, and practicing with various examples, you can master this technique and apply it to a wide range of algebraic problems. From finding roots to evaluating polynomials, synthetic division offers a powerful shortcut that can save you time and effort. Remember to pay attention to the details, especially the signs and the placement of zero placeholders. With practice, you'll find synthetic division to be an indispensable tool in your mathematical arsenal.