Horner's Method - Examples

Examples

Evaluate

for

We use synthetic division as follows:

x₀│ x³ x² x¹ x⁰ 3 │ 2 −6 2 −1 │ 6 0 6 └──────────────────────── 2 0 2 5

The entries in the third row are the sum of those in the first two. Each entry in the second row is the product of the x-value (3 in this example) with the third-row entry immediately to the left. The entries in the first row are the coefficients of the polynomial to be evaluated. Then the remainder of on division by is 5.

But by the remainder theorem, we know that the remainder is . Thus

In this example, if we can see that, the entries in the third row. So, synthetic division is based on Horner's method.

As a consequence of the polynomial remainder theorem, the entries in the third row are the coefficients of the second-degree polynomial, the quotient of on division by . The remainder is 5. This makes Horner's method useful for polynomial long division.

Divide by :

2 │ 1 -6 11 -6 │ 2 -8 6 └──────────────────────── 1 -4 3 0

The quotient is .

Let and . Divide by using Horner's method.

2 │ 4 -6 0 3 │ -5 ────┼──────────────────────┼─────── 1 │ 2 -2 -1 │ 1 │ │ └──────────────────────┼─────── 2 -2 -1 1 │ -4

The third row is the sum of the first two rows, divided by 2. Each entry in the second row is the product of 1 with the third-row entry to the left. The answer is