Long Division of Polynomials - DOC by DgjB6vi

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									            Long Division of Polynomials



Long division of polynomials is an infrequently used technique in algebra.
Nevertheless, it requires some attention as it may appear in many circumstances.
Let us, while working an example, establish a routine to long divide one polynomial
by another. The only skills that are needed for long division of polynomials are
subtracting polynomials and multiplying a polynomial by a monomial. Both of these
techniques are elementary algebraic skills.



Example 1:

x  3 x 2  7x  12



In order to perform a long division involving polynomials you only have to cycle
through a 3 step routine as often as necessary. But before we discuss this routine, we
need to know that the x-3 is called the divisor, and that the x 2  7x  12 is called the
dividend.



This routine is:

1)      Ask yourself by what term should you multiply the leading term in the divisor
        to get exactly the leading term of the dividend.

2)      Multiply the entire divisor by that term and put the result underneath the
        dividend. Put that term above the top line, quotient line, in the problem.

3)      Subtract the result of the multiplication in step 2 from the dividend - do not
        forget to change signs - and bring down the next term from the dividend.


Repeat these three steps until the result of step 3 is just a number, be it 0 or not.




                                                                                            1
In the first cycle of example 1 we have:


1)     You have to multiply by x
              x
2)      x  3 x 2  7x  12
             x 2  3x



              x
        x  3 x 2  7 x  12
3)
           x  3x
             2


                4x  12



Repeating the cycle, we now have:



1)     We must multiply by -4

              x4
        x  3 x 2  7 x  12

           x  3x
             2
2)
                4 x  12
                   4 x  12
              


              x4
        x  3 x 2  7 x  12

           x  3x
             2
3)
                4 x  12

              4 x  12
                        0



The answer is x-4.




                                           2
Example 2:

4x  3 20x 2  7x  6


The first cycle of three steps results in:

1)      Multiply by 5x

                5x
2)      4x  3 20x 2  7x  6
               20x 2  15x



                5x
        4x  3 20x 2  7x  6
3)
            20x  15x
                2


                   8x  6


The second cycle gives:

1)      Multiply by 2

                5x  2
        4x  3 20x 2  7x  6

            20x  15x
                2
2)
                   8x  6
                         8x  6



                5x  2
        4x  3 20x 2  7x  6

            20x  15x
                2
3)
                   8x  6

                 8x  6
                             0


The answer is 5x+2.




                                             3
Please note that in the final form of the previous examples, the x2 terms, the x terms
and the terms without x, are all nicely lined up underneath each other. This format
makes life easy with long dividing. In example 3 we first have a situation where we
first have to make an arrangement to make this alignment possible.


Example 3:

x  2 x3  8

Please note that there are no x2 terms and no x terms in the dividend. In order to make
the vertical alignment possible I insert two terms in the dividend so that the problem
looks like:

x  2 x 3  0x 2  0x  8


Now we can start with the first cycle and get:

1)     Multiply by x2.

               x2
2)      x  2 x 3  0x 2  0x  8
               x 3  2x 2



               x2
        x  2 x 3  0x 2  0x  8
3)
           x  2x
             3     2


                2x 2  0x



The second cycle gives:

1)     Multiply by 2x

               x 2  2x
        x  2 x 3  0x 2  0x  8
2)             x 3  2x 2
          
                    2x 2  0x
                    2x 2  4x




                                                                                         4
              x 2  2x
        x  2 x 3  0x 2  0x  8

           x  2x
             3     2
3)
                2x 2  0x

                  2x 2  4x
                          4x  8


The third, and last cycle, gives:

1)     Multiply by 4

              x 2  2x  4
        x  2 x 3  0x 2  0x  8

           x  2x
             3     2


2)              2x 2  0x
                   2x 2  4x
              
                          4x  8
                          4x  8



              x 2  2x  4
        x  2 x 3  0x 2  0x  8
              x 3  2x 2
          
3)                 2x 2  0x
              _    2x 2  4x
                           4x  8
                          4x  8
                               0



The answer is x 2  2x  4 .




Let us now look at an example in which the remainder does not come out to be zero.




                                                                                     5
Example 4:

2x  3 6x 2  3x  5


The first cycle gives us:

1)     Multiply by 3x

               3x
2)      2x  3 6x 2  3x  5
               6x 2  9x



               3x
        2x  3 6x 2  3x  5
3)
            6x  9x
               2


                  6x  5


The second cycle yields:

1)     Multiply by 3

               3x  3
        2 x  3 6 x 2  3x  5

            6x  9x
               2
2)
                  6x  5
                       6x  9



               3x  3
        2 x  3 6 x 2  3x  5

            6x  9x
               2
3)
                  6x  5
                  6x  9
                            14


                             14
The answer is 3x  3 
                            2x  3



                                     6

								
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