Expert: Scott A Wilson - 2/10/2009

Question
QUESTION: hi

i have a problem with horner method,and want you to just give me a short description about it,and tell me what does it do
i search it through internet i found this addres"http://en.wikipedia.org/wiki/Horner_scheme#Examples"
can you plZ describe for me the example(in this that i gave you)?
(i don't know how it write the second rows "6 0 6" in the synthetic diagram )

thanx
Bita

ANSWER: Hormer's method is for dividing one polynomial by another.

3 |   2    -6     2    -1
  |         6     0     6    
  |----------------------
      2     0     2     5
Note that you are dividing by  x+3, so the put a three in the upper left hand corner.  Multiply the first two on the bottom by 3, and that's where the first 6 comes from.  -6+6 = 0, which is at the bottom below the -6 and 6.  3*0 = 0, and that is added two 2, which gives the 2nd to last column, 2 0 2.  Take the 2 at the bottom times 3 and you get 6.  Add this to the last column, so you get -1, 6, 5 in the last column since -1+6=5.  Since a zero was not gotten (a 5 was), it is not divisible by x-3.

Let's try another.
x^4 + 4x^3 + 6x^2 + 4x + 1 divided by x+1.  Again, since we are trying to use x+1, we put a -1 in the upper left hand side.

-1 | 1  4  6  4 1
  | 0 -1 -3 -3 1
  |--------------
    1  3  3  1 0

Since a 0 was gotten on the bottom, it is divisible by x-1 (0 is the remainder.  I'm not sure if the numbers look like they're in columns, but they really go a -1 in the first column, 1 0 1 in the next, 4 -1 3 in the next, 6 -3 3 in the next, 4 -3 1 in the next, and 1 1 0 in the last column.

Be sure you put a 0 in if the problem is missing a power of x.  For example, if the polynomial we were trying to divide into was
x^4 + 3x^3 + 5x + 7, the top row would have a 1 3 0 5 7 in it.
The 0 is for the x².


---------- FOLLOW-UP ----------

QUESTION: thanx for your answer

but could you plz tell me in dividing one polynomial by another for example in your first example we ourself should find x+3 for dividing or the question tells us this polynomial?

thanx
Bita

Answer
I hope a better way would be to put the variable that is
being divided above the problem, like
3
|   2    -6     2    -1
|         6     0     6    
|----------------------
    2     0     2     5
Now do these line up?  I notice that this language that does
this reduces the size of spaces to be less than a character
width.  It's almost like we type in courier but is translated
in Arial or something like that.  I have found that if I type
a tab in a word processing language and then highlight the
space with that tab in it that I can paste it into here.
To cut it is control-X and to paste it is control-V.  I'm
from those good old days when we didn't have menus at the top.
In fact, in high school, all we had for a computer was a TTY
(that's teletype) hooked to a main frame around 100 miles away.
The paper was a continuous roll with little pockets on the side
to guide it through the printer.  The printers weren't able to
handle standard paper.  The computers use to take up an entired
room.  Calculators were first introduced when I was in junior
high.  They were $50 for a simple one that could add, subtract,
multilpy, divide, take squareroots, and had one memory.  There
weren't any more advanced than that.  By the way, minimum wage
was a little over a buck and hour, so of course they weren't
required yet.  In college, my first computer had no hard drive
and was almost $2,000.  It came with a dot matrix printer.  Back
then, they didn't think you could make a computer print anywhere
near as good as a typewriter worked.  Yes, they used to use
typewriters, and they even had what was known as white-out to
correct the errors.  Just like a few centuries ago, it took
several months to get a message from here to europe, but now
it can be done in seconds if you have an e-mail address.  Back
when my dad was in WWII (before he was even married), he told
me that letters would arrive from the front in packs of a
month at a time.  But enough about those good old days.  They
weren't that good.

Not that what this problem is doing is saying the
2x^3 - 6x^2 + 2x - 1 is (x-3)(2x²+2), remainder 5.

Multiplying the fuction out gives
2x^3 - 6x^2 + 2x - 6, which if we add in the 5 remainder,
is the function back again.


There were a couple of errors in your letter that I noticed.
First of all, the problem just done was dividing by (x-3),
not (x+3); you always put up there what the value of x to be
tested is.  In this case, we're dividing by x-3, or testing
to see if x=3 is a choice that makes it 0.

If you were dividing by x+3, the following would be done
-3
|   2    -6     2    -1
|        -6    36   102    
|----------------------
    2   -12   -34   101

On the second problem, the 1 becomes a -1, the 3 becomes a -3,
the next 3 also becomes a -3, but the last 1 should become a -1.  
Right now it is down as 1, but apparently a -1 was added anyway
since a 0 was gotten at the bottom.  I think that was a typo.