I'd like help with long division of polynomials

Date: 4 Mar 1995 19:13:28 -0500
From: Sandip Mody
Subject: polynomial help!

I need help as soon as possible on step by step instructions on 
the following:

(x^4+4y^4)/(x^2-2xy+2y^2)

Please show me carefully! If possible please give other examples.  
Thank you.

Date: 5 Mar 1995 00:30:49 -0500
From: Dr. Ken
Subject: Re: polynomial help!

Hello there!

Here's the key to doing all these types of problems:  FACTOR!!

It's just like finding out what 20160/5040 is: you have to find out 
what the factors of the numerator are, and then find out what the 
factors of the denominator are, and then cancel the ones that are 
the same (to get 4 in this case).

In your case, you should find factors of the numerator and 
denominator, and you should also notice that someone gave you 
this problem with the expectation that you might be able to 
solve it.  So if you're trying to factor the polynomials, don't just 
try random factors, try ones that have a good chance of cancelling, 
because most likely that's what's going to happen if it's going to 
simplify.

So for starters, try factoring the smaller of the two polynomials, 
the one in the denominator.  A little trial-and-error will probably 
convince you that it won't factor nicely.  So that suggests that if 
there's going to be any cancelling, the whole denominator is going 
to have to cancel with part of the numerator (i.e., you won't just 
get part of the denominator cancelling).

So you can try to factor the numerator into something times 
x^2 - 2xy + 2y^2.  If trial-and-error doesn't give you the factor 
you're looking for, you can use long division.  Don't give up until 
you find the factor:  it does factor.

Here are a couple of similar problems for you to chew on:

(x^8 + 4x^8)/(x^4 + 2x^2 y^2 + 2y^4)

(x^6 + 1)/(x^2 + 1)

(1 + x^6)/(x^4 - x^2 +1)

-Ken "Dr." Math

Date: 5 Mar 1995 10:19:47 -0500
From: Sandip Mody
Subject: Re: polynomial help!

Sorry if you misunderstood me, but I needed help on how to solve 
(x^4 + 4y^4) / (x^2 - 2xy + 2y^2) by doing long divison.  I am also not 
very clear on how to factor polynomials. 

Date: 5 Mar 1995 12:57:28 -0500
From: Dr. Ken
Subject: Re: polynomial help!

Ah.

This will be more difficult to show over the network, but I'll give it a
shot.  To divide x^2 - 2xy + 2y^2 into x^4 + 4y^4, you set it up like a
normal division problem:

                  _________________________________
x^2 - 2xy + 2y^2  )  x^4 + 4y^4

Then to solve it, you ask yourself "what do I need to multiply the first
term in the divisor (the part on the outside) by to get the first term in
the dividend (the part on the inside)?"  Well, to get x^4 from x^2, you need
to multiply by x^2.  So we multiply the divisor by x^2, and subract it from
the dividend, and write that like this:

                  ___x^2___________________________
x^2 - 2xy + 2y^2  )  x^4 + 4y^4
                   -(x^4 - 2x^3 y + 2x^2 y^2)
                    -------------------------
                           2x^3 y - 2x^2 y^2 + 4y^4

Now, notice that there's a 4y^4 in the dividend, too.  The only possible way
to get rid of that is to multiply the divisor by 2y^2.  So let's do that,
and see what happens:

                  ___x^2_+_2y^2____________________
x^2 - 2xy + 2y^2  )  x^4 + 4y^4
                   -(x^4 - 2x^3 y + 2x^2 y^2)
                    -------------------------
                           2x^3 y - 2x^2 y^2 + 4y^4
                        -(-4x y^3 + 2x^2 y^2 + 4y^4)
                           ------------------------
                           2x^3 y + 4x y^3 - 4x^2 y^2

Now we ask ourselves: is there anything (i.e. anything simple enough to
figure out relatively easy) we can multiply the divisor by to get this
latest result?  Actually, we should ask ourselves that at every step, but
the answer up until now was always no.  But now, sure, we can.  Notice that
2x^3 y + 4x y^3 - 4x^2 y^2 = 2xy(x^2 - 2xy + 2y^2).  So we need to multiply
the divisor by 2xy.  That looks like this:

                  ___x^2_+_2y^2___+_2xy____________
x^2 - 2xy + 2y^2  )  x^4 + 4y^4
                   -(x^4 - 2x^3 y + 2x^2 y^2)
                    -------------------------
                           2x^3 y - 2x^2 y^2 + 4y^4
                        -(-4x y^3 + 2x^2 y^2 + 4y^4)
                           ------------------------
                           2x^3 y + 4x y^3 - 4x^2 y^2
                         -(2x^3 y - 4x^2 y^2 + 4x y^3)
                           --------------------------
                                                    0

Viola!  So the answer is x^2 + 2xy + 2y^2.

Factoring polynomials is another story, albeit a related one.  Basically,
High School Algebra consists of learning lots of little tricks about how to
factor polynomials, and there's no one method that will always work.  If you
have specific questions that puzzle you, send them in and we'll try to help
you.

-Ken "Dr." Math

Date: 7 Mar 1995 16:39:35 -0500
From: Sandip Mody
Subject: Dividing Polynomials

Sorry to bother you again, but do you know how to divide (x^2-64y^2) / (x-4y)
and (a^4 + a^2b^2 + b^4) / (a^2 - ab + b^2) using long divison.  I am really
having trouble with this and would appreciate the help.  Thank you very much.

Date: 7 Mar 1995 23:00:18 -0500
From: Dr. Ken
Subject: Re: Dividing Polynomials

Hello there!

Here's one thing you may want to keep in mind when you're dividing
polynomials that have two or more variables in them (the hard kind).  Take a
look at the terms in the dividend, and see whether you can find terms that
are only divisible by one term of the divisor.  For instance, in your first
problem, you have an x^2 and a -64y^2.  Now look at the terms in the
divisor.  There's nothing (nothing easy, at least) that you can multiply the
-4y by to get x^2, right?  So when you're trying to get rid of the x^2,
you're going to have to use the x term in the divisor, and multiply by
another x.  Similarly, to get rid of the -64y^2, you're going to have to use
the -4y, and multiply by 16y.  So this problem would start out like this:

     __x____+_16y_______________
x-4y ) x^2  - 64y^2
     -(x^2  -  4xy)
       -----------
       4xy  - 64y^2
    -(16xy  - 64y^2)       
      -------------
     -12xy

So the answer would be x + 16y, with a remainder of -12xy (are you sure this
wasn't supposed to be x-8y in the denominator?  That would have made it come
out evenly).

In the second problem, use the same technique.  The only way you're going to
get rid of the a^4 term on top is to use the a^2 term in the bottom, and the
only way to get rid of the b^4 term is to use the b^2 term in the bottom.
So the first two things you should have in your quotient are what you
multiply by to get a^4 and b^4 (namely, a^2 and b^2).

               __a^2__+_b^2_________________________
a^2 - ab + b^2 ) a^4 + a^2b^2 + b^4
               -(a^4 - a^3b + a^2b^2)
                 -------------------
                       a^3b  +  b^4
                     -(a^2b^2 - ab^3 + b^4)
                       -------------------
                       a^3b - a^2b^2 + ab^3


At this point, you should say to yourself, "okay, this term has either got
to die in this round, or I'm in big trouble."  And luckily, it does: we can
multiply ab times the dividend to get what we want.  So our answer (the
quotient) is a^2 + b^2 + ab.

I hope this helps you out, and I hope this method of starting out is
helpful to you.

-Ken "Dr." Math