| It's a way of reducing the bookkeeping needed when dividing
a polynomial by another (perhaps restricted to dividing by a
monic first degree polynomial, i.e., x +/- something. I only
know of it with this restriction).
x^2 - x + 3 remainder -1
--------------------
x - 2 |x^3 - 3x^2 + 5x - 7
x^3 - 2x^2
----------
- x^2 + 5x
- x^2 + 2x
-----------
3x - 7
3x - 6
------
- 1
That was "long division." This is "synthetic division":
1 -1 3 remainder -1 ==> x^2 - x + 3 remainder -1
------------
2 | 1 -3 5 -7
2
--
-1 -2
--
3 6
--
-1
where the result of the addition is just brought up to be the
coefficient of the answer, and is then multiplied by the "2"
and added to the next term. I'm in a hurry so I'll skip the
explanation :-) and answer any questions tomorrow.
Dan
|
| Just a side note. What makes synthetic division unique to division
by polynomials of the form x-a, is the fact that the division can be done
in three rows as follows:
Divide x^4 - x^3 - 10x^2 + 15x +2 by x-2
this can be written as
1 -1 -10 15 2
2 2 2 -16 -2
----------------------
|
1 1 -8 -1 | 0 2 is a factor of the divisor.
In this method you start from the left, carry down the 1, then multiply by
the factor which is 2, and add. now take the result and mutiply by the factor
and add to the next entry. Continue, the last digit is the remainder, the
remainder is a fraction r/(x-f) . f is the factor and r is the remainder.
TOM.
|
| re .2,
Aha, now I remember! :-) Thanks.
In .1 I only got rid of the x's and pointed out that certain
numbers were copied from here to there. The real method as
in .2 goes further and only shows that "copied" value once,
putting it where it is to be used.
Redoing that example, Divide x^4 - x^3 - 10x^2 + 15x +2 by x-2
Set up the coefficients
1 -1 -10 15 2
and the value "a" -- remember you are dividing by x-a, not x+a
1 -1 -10 15 2
2
Now put in the horizontal line
1 -1 -10 15 2
2
----------------------
Okay, we bring down the 1 ... well, actually we are adding
it to the implied zero beneath it ...
1 -1 -10 15 2
2
----------------------
1
Now, multiply that last "sum" by the "a" value (here, 2) and
put it where it will be added in the next step
1 -1 -10 15 2
2 2
----------------------
1
Okay, do the addition again (the first addition was with the
implied zero)
1 -1 -10 15 2
2 2
----------------------
1 1
Now keep repeating that multiplication of the last sum by a,
writing it under the next term, adding them and putting the
sum under the line ...
1 -1 -10 15 2 1 -1 -10 15 2
2 2 2 2 2 2 -16
---------------------- ==> ---------------------- ==>
1 1 -8 1 1 -8 -1
1 -1 -10 15 2 1 -1 -10 15 2
2 2 2 -16 -2 2 2 2 -16 -2
---------------------- or ----------------------
1 1 -8 -1 0 1 1 -8 -1 | 0
The result is x^3 + x^2 - 8x -1 with a remainder of 0.
Dan
P.S. The remainder when dividing by x-a is the value of the
polynomial when x = a.
|