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Horner's Method or Scheme

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)=x3+4x2+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 1x2+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.


Ken Ward's Mathematics Pages


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