Horner's method has a variety of uses, and saves work when evaluating polynomials. It is sometimes called synthetic division.
We proceed by example:+Suppose we have the following equation:f(x)=x^{3}+4x^{2}+x-6
We wish to check whether -3 is a root of that equation, that is, to
find f(-3). Horner's method has the advantage that fewer calculations
are required. It also has the advantage of finding the reduced equation
(that is by dividing f(x) by x+3).
First werite down the coefficients of the terms:

1

4

1

-6

Next write the number to be evaluated, -3, as shown, and write 0 below
the first coefficient. Add the first coefficient, 1, to 0, and write
the result, 1 below:

1

4

1

-6

-3

0

1

Multiply the 1 by -3 and write the result below 4, and add (-3+4=1):

1

4

1

-6

-3

0

-3

1

1

Continue in this fashion until the cells are filled. The last cell below is 0, so -3 is a root of the equation.

1

4

1

-6

-3

0

-3

-3

6

1

1

-2

0

Furthermore, the last line shows the coefficients of the equation obtained by dividing f(x) by x+3 (1,1,-2), that is 1x^{2}+1x-2.
It is possible to continue to check whether the claim that -2 is a root or not:

1

4

1

-6

-3

0

-3

-3

6

1

1

-2

0

-2

0

-2

2

1

-1

0

We fill in the table as before, and because we end with 0, we are
assured that -2 is a root of the equation. As before, the result of
dividing by x+2 is the last line (1,-1), or 1x-1, which tells us that
the final root is x=1.
If the last cell isn't 0, then there is a remainder on dividing by the number.

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