Associated Topics || Dr. Math Home || Search Dr. Math

### 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

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/
```
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
Math Forum Home || Math Library || Quick Reference || Math Forum Search