Long Division with Polynomials

Date: 12/11/2007 at 05:39:44
From: Kim
Subject: Synthetic Division (2x^6 + 5x^4 - x^3 + 1) / (x^2 + x + 1)

How do you divide (2x^6 + 5x^4 - x^3 + 1) / (x^2 + x + 1) either by 
synthetic or long division?  Please help!!

I know how to divide binomials by long and synthetic division but not 
trinomials, as this problem is.  I'm guessing you can solve the 
problem by factoring the x^2 + x + 1 and dividing the two binomials,
but I'm not sure.  I know how to set up 2x^6 + 5x^4 - x^3 + 1 for
synthetic division though: 2 0 5 -1 0 0 1.  That's about all I can do.
Thanks for helping.

Date: 12/11/2007 at 09:23:07
From: Doctor Peterson
Subject: Re: Synthetic Division (2x^6 + 5x^4 - x^3 + 1) / (x^2 + x + 1)

Hi, Kim.

Synthetic division is used primarily for division by binomials of the 
form x-a.  The method can be extended, but then it loses its 
simplicity.  (It could also be used, as you suggest, by factoring 
first; but in this case that would require complex numbers!)

Long division works well for trinomial divisors; you do all the same 
work you would do for a binomial.  There's really nothing new except 
the size of the problem.  Let's do an example that's just a little 
simpler than yours: I'll divide 3x^4-x^3+3x-1 by x^2-x+1.

              ______________________________
  x^2 - x + 1 ) 3x^4 -  x^3 + 0x^2 + 3x - 1

Notice that I put in a place-holder term 0x^2 to make room for an 
x^2 column in my work.  Now I first divide the leading terms: 3x^4 / 
x^2 = 3x^2.  That will be the first term of the quotient.  Then I 
multiply that by the divisor and write that under the proper columns 
of the dividend:

              ________________3x^2__________
  x^2 - x + 1 ) 3x^4 -  x^3 + 0x^2 + 3x - 1
                3x^4 - 3x^3 + 3x^2

The next step is to subtract (remembering to subtract EACH TERM, not 
add them), and bring down the next term from the dividend to make a 
new dividend for the next step:

              ________________3x^2__________
  x^2 - x + 1 ) 3x^4 -  x^3 + 0x^2 + 3x - 1
                3x^4 - 3x^3 + 3x^2
                ------------------
                       2x^3 - 3x^2 + 3x

Now I repeat the process, dividing leading terms, 2x^3 / x^2 = 2x. 
That goes in the quotient, then I multiply, subtract, and bring down:

              ________________3x^2_+ 2x_____
  x^2 - x + 1 ) 3x^4 -  x^3 + 0x^2 + 3x - 1
                3x^4 - 3x^3 + 3x^2
                ------------------
                       2x^3 - 3x^2 + 3x
                       2x^3 - 2x^2 + 2x
                       ----------------
                              -x^2 +  x - 1

Now I repeat again for the third term of the quotient, -1:

              ________________3x^2_+ 2x_- 1_
  x^2 - x + 1 ) 3x^4 -  x^3 + 0x^2 + 3x - 1
                3x^4 - 3x^3 + 3x^2
                ------------------
                       2x^3 - 3x^2 + 3x
                       2x^3 - 2x^2 + 2x
                       ----------------
                              -x^2 +  x - 1
                              -x^2 +  x - 1
                              -------------
                                          0

The remainder turns out to be 0.  If it hadn't been, I would have 
stopped at this point anyway.

Now, can you do the same for your problem?  If you have trouble, 
write back and show me your work in the same format I did (but all 
in one piece, not step by step).

Here is an example of a more complicated division problem:

  Polynomial Long Division
    http://mathforum.org/library/drmath/view/56437.html 


- Doctor Peterson, The Math Forum
  http://mathforum.org/dr.math/ 

Date: 12/11/2007 at 14:31:01
From: Kim
Subject: Thank you (Synthetic Division (2x^6 + 5x^4 - x^3 + 1) / (x^2
+ x + 1))

Thank you for your help!  Everything makes perfect sense now.  So this
is what I got when I worked out the original problem.  I *think* it's
correct.

                                                            5x + 2
                                                          -----------
             ______________2x^4_-_2x^3_+_5x^2_- 4x_-_1_+_(x^2_+_x_+_1)
x^2 + x + 1 )2x^6 +  0^5 + 5x^4 + 1x^3 + 0x^2 + 0x + 1
             2x^6 + 2x^5 + 2x^4
             ------------------
                   -2x^5 + 3x^4 -  x^3
                   -2x^5 - 2x^4 - 2x^3
                    ------------------
                           5x^4 +  x^3 + 0x^2
                           5x^4 + 5x^3 + 5x^2
                           ------------------
                                 -4x^3 - 5x^2 + 0x
                                 -4x^3 - 4x^2 - 4x
                                  ----------------
                                         -x^2 + 4x + 1
                                         -x^2 -  x - 1
                                        -------------
                                                5x + 2
Thanks again!

Date: 12/11/2007 at 22:29:13
From: Doctor Peterson
Subject: Re: Thank you (Synthetic Division (2x^6 + 5x^4 - x^3 + 1) /
(x^2 + x + 1))

Hi, Kim.

We can check the results of the division by multiplying:

  p = d*q + r

where p is the dividend, d is the divisor, q is the quotient, and r is
the remainder.  Here we go:

    q:                2x^4 - 2x^3 + 5x^2 - 4x - 1
    d:                               x^2 +  x + 1
        -----------------------------------------
                      2x^4 - 2x^3 + 5x^2 - 4x - 1
               2x^5 - 2x^4 + 5x^3 - 4x^2 - x
        2x^6 - 2x^5 + 5x^4 - 4x^3 -  x^2
        -----------------------------------------
        2x^6 + 0x^5 + 5x^4 - 1x^3 + 0x^2 - 5x - 1
    r:                                     5x + 2
        -----------------------------------------
        2x^6 + 0x^5 + 5x^4 - 1x^3 + 0x^2 + 0x + 1

should match

    d:  2x^6 + 0x^5 + 5x^4 - 1x^3 + 0x^2 + 0x + 1

And it does!  It's always a good idea to check your answers on these
problems because it's very easy to make a sign mistake with all the
calculations and subtractions.  Good work!


- Doctor Peterson, The Math Forum
  http://mathforum.org/dr.math/