Drexel dragonThe Math ForumDonate to the Math Forum

Ask Dr. Math - Questions and Answers from our Archives
_____________________________________________
Associated Topics || Dr. Math Home || Search Dr. Math
_____________________________________________

Synthetic Division

Date: 05/06/2003 at 22:23:17
From: Noel
Subject: Algebra

   (x^2-6x+9)/(x-3)


Date: 05/07/2003 at 03:38:15
From: Doctor Luis
Subject: Re: Algebra

Hi Noel,

Here's how you divide a polynomial by (x-a):

First draw a grid like this, showing the powers of x and with the 
coefficients below. 

         x^2     x    1
      -------------------
   3 |   +1     -6    +9
   --|-------------------
     |

If a power of x is missing, make sure you put a 0 as the coefficient.
On the side, put the negative of the number you're dividing by. If 
you're dividing by x-3, put -(-3)=+3. If you're dividing by x+1 put -1 
there.

Done drawing it? Good.

First, bring down the 1. (the first number inside the box)

         x^2     x    1
      -------------------
   3 |   +1     -6    +9
   --|-------------------
     |    1

Next, multiply the number you brought down by the number on the side. 
Since I brought down a 1, I multiply 3*1 = 3. Add this number to the 
second number (here it's -6), and write that number down (3-6 = -3) in 
the next spot.

         x^2     x    1
      -------------------
   3 |   +1     -6    +9
   --|-------------------
     |    1     -3

Now repeat this process. Multiply the 3 times -3 and add it to the 9. 
3*(-3)+9 = 0.

         x^2     x    1
      -------------------
   3 |   +1     -6    +9
   --|-------------------
     |    1     -3     0

This last number should be a zero (for our problem it is), if the
polynomial x-3 divides x^2-6x+9. If it isn't, then it represents a 
remainder.

This is how you read the answer back.

         x^2     x    1
      -------------------
   3 |   +1     -6    +9
   --|--------------|----
     |    1     -3  |  0
     |--------------|----
          x      1

You start counting one power of x less on the bottom row. Since we 
started with x^2, the next lower power is x. You can see the answer 
is x-3. Remember that the last number is actually a remainder 
(technically, it represents the -1 power).

Therefore,

  x^2 - 6x + 9
  ------------- = x - 3
     x - 3

This process is called synthetic division.

Let's try a more difficult example that'll teach you more about 
synthetic division.

How about dividing x^3 + 3x + 8 by x+1 ? I won't draw the powers of x 
this time, because you already know what the numbers in the box mean. 
You can imagine them to be there in your mind. Note that the 0 is 
there because there's no x^2 power.

    |
 -1 | 1   0   3   8
 ---|-----------|----
    | 1  -1   4 | 4

See how that worked? Bring down the 1. Multiply by -1 and add to 0, 
and you get -1. Multiply -1 by -1 and add to 3, you get 4. Multiply 4 
by -1 and add to 8, and you get 4. This time, we did get a remainder.

Reading the table back, we see that the solution to our problem is

  x^3 + 3x + 8                     4
 -------------- = x^2 - x + 4 + --------
     x + 1                       x + 1

See what happens when you get a remainder term? There's an extra 
4/(x+1) term at the end. That's how you write the remainder. The thing 
at the bottom of the remainder is just the thing you're dividing by.

I hope this helped! Let us know if you have any other questions.

- Doctor Luis, The Math Forum
  http://mathforum.org/dr.math/ 
Associated Topics:
High School Polynomials

Search the Dr. Math Library:


Find items containing (put spaces between keywords):
 
Click only once for faster results:

[ Choose "whole words" when searching for a word like age.]

all keywords, in any order at least one, that exact phrase
parts of words whole words

Submit your own question to Dr. Math

[Privacy Policy] [Terms of Use]

_____________________________________
Math Forum Home || Math Library || Quick Reference || Math Forum Search
_____________________________________

Ask Dr. MathTM
© 1994-2013 The Math Forum
http://mathforum.org/dr.math/