Complete The Synthetic Division Problem Below 2 7 5

Mastering Synthetic Division: A Step-by-Step Guide with the Problem 2 7 5

Synthetic division is a powerful, streamlined method for dividing polynomials, specifically when the divisor is a linear factor of the form (x - c). It transforms the often cumbersome process of long division into a cleaner, tabular format that is faster and less prone to error. This technique is not just a computational trick; it is fundamentally connected to the Remainder Theorem and is an essential tool for factoring polynomials, finding zeros, and evaluating polynomial functions efficiently. To solidify your understanding, we will walk through the complete process using a specific example: dividing the polynomial represented by the coefficients 2, 7, 5 by the binomial (x - 2). This corresponds to the problem of dividing 2x² + 7x + 5 by x - 2.

Understanding the Setup: What Do the Numbers 2 7 5 Represent?

Before performing any operation, we must correctly interpret the given numbers. The sequence 2, 7, 5 are the coefficients of the dividend polynomial, listed in descending order of their corresponding exponents. This means:

• The 2 is the coefficient of the x² term (2x²).
• The 7 is the coefficient of the x term (7x).
• The 5 is the constant term (5, which can be thought of as 5x⁰).

Therefore, our dividend is the quadratic polynomial 2x² + 7x + 5. The divisor is given implicitly by the context of a typical synthetic division setup. When we see a problem phrased as "complete the synthetic division problem below 2 7 5" with a divisor like (x - 2), the number 2 from (x - 2) is the synthetic divisor or 'c' value. We use the opposite of the constant term from the binomial. For (x - 2), c = 2. For (x + 3), c = -3.

The Step-by-Step Synthetic Division Process

Let’s solve: (2x² + 7x + 5) ÷ (x - 2).

Step 1: Draw the Synthetic Division "Box" and Place the Values

1. Write the value of c = 2 in a small box to the left.
2. To the right of the box, write the coefficients of the dividend in order: 2, 7, 5.
3. Leave a blank row below the coefficients for calculations.

Your setup should look like this:

    2 | 2   7   5
      |___________

Step 2: Bring Down the Leading Coefficient

• Simply bring the first coefficient (the leading 2) straight down below the line. This becomes the first coefficient of your quotient polynomial.

    2 | 2   7   5
      |___________
        |__2__

Step 3: Multiply and Add (The Core Loop)

This is the repeating cycle:

1. Multiply the number you just wrote below the line (the 2) by the synthetic divisor (2). 2 * 2 = 4.
2. Write this product (4) under the next coefficient in the dividend row (7).
3. Add the numbers in that column vertically: 7 + 4 = 11.
4. Write the sum (11) below the line.

    2 | 2   7   5
      |     4
      |___________
        |__2__11__

Step 4: Repeat the Multiply-Add Cycle

1. Take your new result below the line (11) and multiply it by the synthetic divisor (2). 11 * 2 = 22.
2. Write this product (22) under the next coefficient (5).
3. Add the numbers in that column: 5 + 22 = 27.
4. Write the final sum (27) below the line.

    2 | 2   7   5
      |     4   22
      |___________
        |__2__11__27

Step 5: Interpret the Final Row

The numbers below the line now have specific meanings, read from left to right:

• The first number (2) is the coefficient of the x term in the quotient.
• The second number (11) is the constant term in the quotient.
• The final number (27) is the remainder.

Since our original dividend was a 2nd-degree polynomial (quadratic) and we divided by a 1st-degree polynomial (linear), our quotient will be a 1st-degree polynomial (linear). The remainder is a constant.

Writing the Final Answer

The result of the division is expressed as: Quotient + (Remainder / Divisor)

From our work:

• Quotient: 2x + 11 (from coefficients 2 and 11)
• Remainder: 27
• Divisor: (x - 2)

Therefore: (2x² + 7x + 5) ÷ (x - 2) = 2x + 11 + 27/(x - 2)

You can verify this using the Remainder Theorem, which states that the remainder of a polynomial f(x) divided by (x - c) is equal to f(c). Here, f(x) = 2x² + 7x + 5 and c = 2. f(2) = 2(2)² + 7(2) + 5 = 2(4) + 14 + 5 = 8 + 14 + 5 = 27. This matches our remainder perfectly, confirming the calculation.

The Profound Connection: The Remainder Theorem and Factor Theorem

Synthetic division is the computational engine for two of the most important theorems in algebra.

• The Remainder Theorem: As shown above, it provides an ultra-fast way to evaluate a polynomial at a specific value. Instead of substituting and calculating, you perform synthetic division with that value. The final remainder is the function value.
• The Factor Theorem: This is a direct corollary. If the remainder from synthetic division is zero, then (x - c) is an exact factor of the polynomial, and c is a root (or zero) of the function. For example, if we had divided 2x² + 7x + 5 by (x + 1), we would use c = -1. If the remainder were 0, it would prove that x =

−1 is a root of the polynomial. Performing the synthetic division with ( c = -1 ):

   -1 | 2   7