Polynomial Division and Horner’s Method

Polynomial division divides one polynomial by another and gives a quotient and a remainder, and Horner’s method is a short scheme for dividing by x − r and for evaluating a polynomial at r. In brief: dividing x³ − 6x² + 11x − 6 by x − 1 gives the quotient x² − 5x + 6 with remainder 0. With Horner’s scheme, you bring down the coefficients 1, −6, 11, −6 and use r = 1 to get 1, −5, 6, 0.

Division is needed to factor polynomials of degree 3 and higher, to simplify rational expressions and to find slant asymptotes. The long division procedure parallels long division of numbers. Horner’s method (also called synthetic division) is a faster version for divisors of the form x − r. This guide shows both, with the check by multiplication.

How does polynomial long division work?

For polynomials f and g (g nonzero), there are unique polynomials q (quotient) and r (remainder) such that f = g · q + r, where the degree of r is less than the degree of g, or r = 0. Procedure:

1. Write both polynomials in descending order, with zero coefficients for missing powers.
2. Divide the leading term of the dividend by the leading term of the divisor. This is the next term of the quotient.
3. Multiply the divisor by that term and subtract the product from the dividend.
4. Repeat with the new polynomial until its degree is less than that of the divisor.

Example 1: (2x³ + 3x² − 5x + 4) ÷ (x + 2)

2x³ ÷ x = 2x². Multiply: 2x²(x + 2) = 2x³ + 4x². Subtract: (3x² − 4x²) − 5x + 4 = −x² − 5x + 4. Next: −x² ÷ x = −x. Multiply: −x(x + 2) = −x² − 2x. Subtract: −3x + 4. Next: −3x ÷ x = −3. Multiply: −3(x + 2) = −3x − 6. Subtract: 4 − (−6) = 10. The quotient is 2x² − x − 3 and the remainder is 10. So 2x³ + 3x² − 5x + 4 = (x + 2)(2x² − x − 3) + 10.

Example 2: a missing term

(x³ − 8) ÷ (x − 2): write x³ + 0x² + 0x − 8. The quotient is x² + 2x + 4 with remainder 0. This is the difference of cubes formula.

How does Horner’s method work?

To divide by x − r, use the coefficients only.

1. Write r on the left and the coefficients of the dividend (with zeros for missing powers) in a row.
2. Bring down the first coefficient.
3. Multiply it by r and add the result to the next coefficient.
4. Repeat across the row. The last number is the remainder, and the numbers before it are the coefficients of the quotient, with degree one lower than the dividend.

Example 3: (x³ − 6x² + 11x − 6) ÷ (x − 1)

r = 1. Coefficients: 1, −6, 11, −6. Bring down 1. Then 1 · 1 + (−6) = −5. Then −5 · 1 + 11 = 6. Then 6 · 1 + (−6) = 0. The quotient is x² − 5x + 6 and the remainder is 0, so x − 1 is a factor. Further factoring gives (x − 1)(x − 2)(x − 3).

Example 4: dividing by x + 2

For x + 2 use r = −2. With 2, 3, −5, 4: bring down 2. 2 · (−2) + 3 = −1. (−1)(−2) + (−5) = −3. (−3)(−2) + 4 = 10. The quotient is 2x² − x − 3 and the remainder 10, the same as in Example 1.

How do you evaluate a polynomial with Horner’s scheme?

The remainder when dividing by x − r equals the value p(r). So Horner’s scheme computes p(r) with only n multiplications and n additions, instead of computing all the powers. This is the content of the remainder theorem.

Example: p(x) = 3x⁴ − 2x² + x − 5 at x = 3. Coefficients: 3, 0, −2, 1, −5. Bring down 3. 3 · 3 + 0 = 9. 9 · 3 − 2 = 25. 25 · 3 + 1 = 76. 76 · 3 − 5 = 223. So p(3) = 223. Check directly: 3 · 81 − 18 + 3 − 5 = 243 − 20 = 223 ✓.

Horner’s scheme worked out in a table

The numbers 3, 9, 25, 76 are the coefficients of the quotient 3x³ + 9x² + 25x + 76, and 223 is the remainder.

How do you check the result?

Multiply the divisor by the quotient and add the remainder. You should get the original polynomial. Example 1: (x + 2)(2x² − x − 3) + 10 = 2x³ − x² − 3x + 4x² − 2x − 6 + 10 = 2x³ + 3x² − 5x + 4 ✓. Another check: set x equal to a simple number such as 1 on both sides.

Where does the method come from?

The scheme is named after William George Horner, who published it in 1819, but the same procedure was used earlier by the Chinese mathematician Qin Jiushao in the thirteenth century and, in related forms, by others. Paolo Ruffini described it in 1804, and in some countries the method is called Ruffini’s rule. Isaac Newton used a similar nested computation in the seventeenth century.

Where is it used today?

Computers evaluate polynomials by Horner’s method because it is fast and numerically stable. It is used to calculate values of approximations to functions, to find roots with Newton’s method and to factor polynomials step by step.

Which mistakes are common?

Forgetting zero coefficients

x³ − 8 has coefficients 1, 0, 0, −8. Skipping zeros gives a wrong result.

Wrong sign of r

For x + 2 use r = −2. For x − 5 use r = 5.

Using Horner’s scheme for a divisor that is not linear

The scheme works for x − r only. For divisors like x² + 1, use long division.

Mis-subtracting in long division

Subtract the whole product line, changing all signs. Write the subtraction as adding the opposite.