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Math Lesson 11.3.1 - The Synthetic Division Method of Polynomials

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Welcome to our Math lesson on The Synthetic Division Method of Polynomials, this is the first lesson of our suite of math lessons covering the topic of Solutions for Polynomial Equations, you can find links to the other lessons within this tutorial and access additional Math learning resources below this lesson.

The Synthetic Division Method of Polynomials

The synthetic division method is a method used to divide polynomials in a much simpler way compared to the long division method explained in tutorial 11.2. The condition to perform a synthetic division is to have a linear polynomial as a divisor, i.e. the divisor Q(x) of the polynomial P(x) must have the form mx ± n, where m is a coefficient and n a constant.

As stated in the previous tutorial, we may write the division of two polynomials as

P(x)/Q(x) = S(x) + R(x)/Q(x)

where S(x) is the quotient polynomial and R(x) is the remainder polynomial.

For example, if P(x) = x2 + 4x + 7, we can write it as

P(x) = x2 + 4x + 3 + 4
= (x2 + 4x + 3) + 4
= (x + 1)(x + 3) + 4

Taking Q(x) = x + 1 yields

P(x)/Q(x) = (x + 1)(x + 3) + 4/x + 1
= (x + 1)(x + 3)/x + 1 + 4/x + 1
= x + 3 + 4/x + 1

where

S(x) = x + 3 and R(x) = 4/(x + 1)

The procedure to perform the synthetic division is as follows:

Step 1: Write down the coefficients and the constant of the original polynomial P(x) at the place where we write the dividend when using the long division method. In addition, we write the zero of the divisor Q(x) in the place where we put the divisor in the long division method. In our example, we have P(x) = x2 + 4x + 7 and Q(x) = x + 1. Thus, we write 1, 4 and 7 in the upper part of the scheme and -1 aside (because if x + 1 = 0 then x = -1).

Math Tutorials: Solutions for Polynomial Equations Example

Step 2: Bring the leading coefficient of the original polynomial to the bottom row.

Math Tutorials: Solutions for Polynomial Equations Example

Step 3: Multiply the leading coefficient of the polynomial with the zero of the divisor and write down the result in the middle row below the second coefficient.

Math Tutorials: Solutions for Polynomial Equations Example

Step 4: Add the numbers in the second column and write down the result in the bottom row, next to the leading coefficient.

Math Tutorials: Solutions for Polynomial Equations Example

Step 5: Repeat the procedure explained in step 3 but this time using the new number.

Math Tutorials: Solutions for Polynomial Equations Example

Step 6: Repeat the procedure explained in step 4 but this time using the new number.

Math Tutorials: Solutions for Polynomial Equations Example

Step 7: Since there are no more operations to do, the last number obtained gives the remainder of the division (here it is 4). Hence, the remainder is R(x) = 4. On the other hand, the number preceding it represents the constant of the quotient (here it is 3) and the number preceding it, the last coefficient of the quotient polynomial and so on. Hence, the quotient is S(x) = x + 3. Finally, since -1 is a zero of the divisor, we have Q(x) = x + 1. Therefore, since the general form of a polynomial factorisation is

P(x) = Q(x) ∙ S(x) + R(x)

then, we have in the specific case

x2 + 4x + 7 = (x + 1)(x + 3) + 4

The following figure shows the Synthetic Division Method of Polynomials represented schematically.

Math Tutorials: Solutions for Polynomial Equations Example

Example 1

Divide the polynomials below using the Synthetic Division Method and write the result as a product of factors including the remainder (if exists).

  1. P(x) = 2x3 - 3x2 - x + 1 and Q(x) = 2x - 1
  2. P(x) = 4x3 - 2x2 - 5x + 3 and Q(x) = 3x - 3

Solution 1

  1. The coefficients of the dividend P(x) are 2, -3, -1 and 1, where 2 is the leading coefficient. On the other hand, the zero of the divisor Q(x) is calculated as
    2x - 1 = 0
    2x = 1
    x = 1/2
    Hence, using the Synthetic Division Method yields Math Tutorials: Solutions for Polynomial Equations Example The last number at the bottom row gives the remainder, which here is 0. The numbers in the middle row represent the coefficients of the quotient polynomial from the highest to the lowest. They are 1, -1 and -1, which means the coefficient preceding x2 is 1, the coefficient preceding x is -1 and the constant of the quotient polynomial is -1. Hence, we have
    P(x)/Q(x) = 2x3 - 3x2 - x + 1/2x - 1
    = 1/2 (2x2 - 2x - 2) + 0/2x - 1
    = x2 - x - 1
  2. The coefficients of the dividend P(x) are 4, -2, -5 and 3, where 4 is the leading coefficient. On the other hand, the zero of the divisor Q(x) is calculated as
    3x - 3 = 0
    3x = 3
    x = 1
    Hence, using the Synthetic Division Method yields Math Tutorials: Solutions for Polynomial Equations Example The coefficients of the quotient Q(x) therefore are 4 and 2 and the constant is -3. Moreover, there is no remainder in this division. Hence, we obtain
    P(x)/Q(x) = 4x3 - 2x2 - 5x + 3/3x - 3
    = 4x2 + 2x - 3 /3/ + 0/3x - 3
    = 4/3 x2 + 2/3 x - 3/3
    = 4/3 x2 + 2/3 x - 1
    (The quotient is divided by 3 because 3x - 3 = 3(x - 1), where 1 was the zero of the quotient polynomial we used in the synthetic division method.)

More Solutions for Polynomial Equations Lessons and Learning Resources

Polynomials Learning Material
Tutorial IDMath Tutorial TitleTutorialVideo
Tutorial
Revision
Notes
Revision
Questions
11.3Solutions for Polynomial Equations
Lesson IDMath Lesson TitleLessonVideo
Lesson
11.3.1The Synthetic Division Method of Polynomials
11.3.2The Rational Zero (Root) Theorem
11.3.3How to list all Possible Zeroes (Roots) of a Polynomial Equation
11.3.4How to identify the Zeroes of the Polynomial from the List of Possible Zeroes
11.3.5How to Solve Polynomial Equations

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