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Polynomial Long Division

Date: 03/17/2004 at 18:01:48
From: Angela
Subject: Long divison!!

Use long divion to divide (2x - 3) into 4x^4 - x^2 - 2x + 1.  I really
need help in doing this.

Date: 03/18/2004 at 11:03:29
From: Doctor Roberts
Subject: Re: Long divison!!

Dear Angela,

There are three things that tend to give students problems when it
comes to polynomial long division.  I'll address all three, but if 
your confusion is about something else, let me know.

First, when you write the expressions down to divide them, make sure
to include _every_ power of x.  If there is a power missing in the
given problem, fill it in with a 0 to hold that place.  For example,
if you were dividing x - 1 into 2x^3 + x - 5, you'd write it like this:
  x - 1 ) 2x^3 + 0x^2 + 1x - 5  

Adding the zero term keeps the columns in order, as we will see when
we do the problem shortly.  Also, since the answer may include the
second power of x, this provides a column to write that part of the
answer.  And note that nothing really changed about the polynomial,
since zero times x^2 is zero.

Adding zero terms is sometimes needed in the divisor as well.  For
example, if you had been dividing by x^2 - 1 instead of x - 1, you
would write the divisor as x^2 + 0x - 1.

OK, let's start doing this division, and I'll point out the other two
common problems as we come to them.  To start the division, look only
at the first term of the divisor and the first term of the dividend,
so that gives us 'x' and '2x^3'.  What times 'x' will make it 2x^3? 
Or, maybe you prefer to think about what you would get if you divide
2x^3 by x.  Either way, the answer is 2x^2.

So that's the first term of the quotient, or answer.  Let's put it on
top in the x^2 column (good thing we added that 0x^2!):

  x - 1 ) 2x^3 + 0x^2 + 1x - 5  

Now multiply the 2x^2 times the (x - 1) and write the answers in the
correct columns:

  x - 1 ) 2x^3 + 0x^2 + 1x - 5  
          2x^3 - 2x^2

As with regular long division, once you have determined how many times
the divisor goes in, written that number up top, and multiplied, the
next step is to subtract.  And this is the second common mistake that
people make with polynomial long division.  Remember that you are
subtracting the whole expression of 2x^3 - 2x^2.  The subtraction will
change the signs on _both_ of those terms since

  -(2x^3 - 2x^2) is equal to -2x^3 + 2x^2.

So what we get when we subtract is:

  x - 1 ) 2x^3 + 0x^2 + 1x - 5  
         -2x^3 + 2x^2
             0 + 2x^2

Can you see why we got +2x^2?  What you really had for the subtraction
was 0x^2 - (-2x^2) which became 0x^2 + 2x^2 which is just 2x^2.  Some
people find it helpful to immediately change the signs after they
multiply and then just add, while others prefer to do the subtraction
and just remember that they are subtracting and think about what
happens if you subtract a negative.  Whatever makes the most sense to
you is fine, but in my experience this is the place where most 
students make their mistakes in the long division process.

Continuing, we can "bring down" the next term, which would be the 1x,
and then proceed the same way as the first cycle.  What's 2x^2 divided
by x?  Sounds like 2x to me!  Put that up top in the x column and
multiply it by x - 1, then subtract:

                 2x^2 + 2x
  x - 1 ) 2x^3 + 0x^2 + 1x - 5  
         -2x^3 + 2x^2
                 2x^2 + 1x
                 2x^2 - 2x

Note that this time I did not change the signs and add, I just 
remembered that when I subract I have 1x - (-2x) which is 1x + 2x or
3x.  Now bring down the -5 and continue by dividing 3x by x to get 3:

                 2x^2 + 2x + 3
  x - 1 ) 2x^3 + 0x^2 + 1x - 5  
         -2x^3 + 2x^2
                 2x^2 + 1x
                 2x^2 - 2x
                        3x - 5
                        3x - 3
                           - 2

Again, this time I had -5 - (-3) which is -5 + 3 or -2.  Since there
is nothing left to "bring down", and because the degree of the
remainder (there is no x in it, so it's degree zero) is less than the
degree of the divisor (with the x^1 it's degree one) we're finished
and the -2 is the remainder.

This brings us to the third idea that sometimes causes problems--what
to do with the remainder.  To figure what to do, let's look at how
remainders work with regular division.
Say we want to divide 3 )5.  The answer is 1 rem. 2, since 3 goes 
into 5 one time with 2 left over.  Another way to write this is 
  3 )5 = 1 + 2/3.  

The remainder is divided by the divisor and added to the answer. 
Checking the answer, 3(1 + 2/3) = 3 + 6/3 = 3 + 2 = 5.

So, in synthetic division, if there's a remainder, just write it as a 
fraction, with the remainder as the numerator and the divisor as the 
denominator, and add it to the rest of the answer.  If it's negative, 
as in our example, you can write it one of two ways:

                   -2                            2
  2x^2 + 2x + 3 + -----   OR   2x^2 + 2x + 3 - -----
                  x - 1                        x - 1

Note that the two choices are either to write it as adding a negative
(on the left) or subtracting a positive (on the right).  Either is
fine, though the right hand one is slightly more elegant.

So, the three things to keep in mind for long division are:

  1. Write the divisor and dividend in descending order of powers of
     the variable and fill in zero terms where needed.

  2. Be careful in the subtraction step, especially when subtracting
     a negative.

  3. If you have a reminder, write it as a fraction by putting it over
     the divisor.

Can you take a shot at your problem now?  If you get stuck, write back
and show me what you've done and I'll give you some help.

- Doctor Roberts, The Math Forum 
Associated Topics:
High School Polynomials
Middle School Division

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