Editing Horner's method (section)

=== Examples ===

Evaluate <math>f(x)=2x^3-6x^2+2x-1</math> for <math>x=3</math>.

We use [[synthetic division]] as follows:

  ''x''{{sub|0}}│   ''x''{{sup|3}}    ''x''{{sup|2}}    ''x''{{sup|1}}    ''x''{{sup|0}}
  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 {{mvar|x}}-value ({{val|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 <math>f(x)</math> on division by <math>x-3</math> is {{val|5}}.

But by the [[polynomial remainder theorem]], we know that the remainder is <math>f(3) </math>. Thus, <math>f(3) = 5</math>.

In this example, if <math>a_3 = 2, a_2 = -6, a_1 = 2, a_0 = -1</math> we can see that <math>b_3 = 2, b_2 = 0, b_1 = 2, b_0 = 5 </math>, the entries in the third row. So, synthetic division (which was actually invented and published by Ruffini 10 years before Horner's publication) is easier to use;  it can be shown to be equivalent to 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 <math>f(x)</math> on division by <math> x-3 </math>. 
The remainder is {{val|5}}. This makes Horner's method useful for [[polynomial long division]].

Divide <math>x^3-6x^2+11x-6</math> by <math>x-2</math>:

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

The quotient is <math>x^2-4x+3</math>.

Let <math>f_1(x)=4x^4-6x^3+3x-5</math> and <math>f_2(x)=2x-1</math>. Divide <math>f_1(x)</math> by <math>f_2\,(x)</math> using Horner's method.

   0.5 │ 4  −6   0   3  −5
       │     2  −2  −1   1
       └───────────────────────
         2  −2  −1   1  −4

The third row is the sum of the first two rows, divided by {{val|2}}. Each entry in the second row is the product of {{val|1}} with the third-row entry to the left. The answer is
<math display="block">\frac{f_1(x)}{f_2(x)}=2x^3-2x^2-x+1-\frac{4}{2x-1}.</math>