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Factoring a Fifth-Degree Polynomial

Date: 08/15/2002 at 14:42:36
From: Mike Cooper
Subject: Factoring this specific problem

Hi and thanks for your help.

The problem is simplify this rational expression:

   x^5+x^4-4x^3-4x^2
   -----------------
     x^5+4x^4+4x^3

I have no problem with the denominator but I am working from a 
teachers' manual and it shows the factorization of the numerator as 
follows:

   x^5+x^4-4x^3-4x^2 =
   x^2[(x^3+x^2-4x-4)] =  this I understand.

   x^2[(x^3-4x)+(x^2-4)] = this is a bit confusing. Why is the first 
                           term combined with the third and the second 
                           with the fourth, and why the plus sign 
                           between the two? Anyway, it continues:

   x^2[x(x^2-4)+(x^2-4)] =  I get this factorization based on the 
                            previous one with the plus, which I hope 
                            you can help me understand.

Now here is another part I need help with:

   x^2[(x^2-4)(x+1)] 

Here I get lost. Where did the x go, and where does this (x+1) come 
from, and what happened to the plus? 

I do these all the time, but this one for some reason is not coming to
me. It ends,

   x^2(x-2)(x+2)(x+1) which I get from the previous line. 

Help!


Date: 08/15/2002 at 16:06:04
From: Doctor Peterson
Subject: Re: Factoring this specific problem

Hi, Mike.

Factoring fifth degree polynomials is often more magic than method; 
it requires tricky thinking and a bit of luck! I'll first explain the 
way they did it, and then suggest what to my mind is a much better 
approach. 

You are factoring x^5 + x^4 - 4x^3 - 4x^2 and get this far:

    x^2(x^3 + x^2 - 4x - 4)

Ignore the x^2 on the outside; we need to factor

    x^3 + x^2 - 4x - 4

Now they are factoring by grouping, looking for two pairs of terms 
that have a common factor. They seem to see that you have two 1's and 
two -4's, so they separate it into two pairs of terms, each with a 1 
and a 4: x^3 - 4x and x^2 - 4. They get this just by changing the 
order, switching the two middle terms around to get

    x^3 - 4x + x^2 - 4

and then put in the parentheses to group the pairs:

    (x^3 - 4x) + (x^2 - 4)

Now they factor out the common factor x in the first pair:

    x(x^2 - 4) + (x^2 - 4)

Now, as they had hoped, they see that this is the sum of two multiples 
of (x^2 - 4), so they can use the distributive property in reverse to 
combine these: x times A plus 1 times A is (x+1) times A:

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

I would have written it as (x + 1)(x^2 - 4), but it's the same thing. 
And I would have seen different pairs; two 1's and two 4's to me imply 
that we can pair neighbors like this:

    x^3 + x^2 - 4x - 4 = (x^3 + x^2) + (-4x - 4)
                       = x^2(x + 1) - 4(x + 1)
                       = (x^2 - 4)(x + 1)
                       = (x + 2)(x - 2)(x + 1)

That's no less work, but more obvious to me than swapping terms.

Now, let's see how we could simply your original expression without 
having to see how to factor that cubic, which is a hard job.

Here's the problem:

    x^5 + x^4 - 4x^3 - 4x^2
    -----------------------
       x^5 + 4x^4 + 4x^3

When your goal is to simplify something, you hope to find common 
factors. So if you find a factor of the denominator, you can try it 
out in the numerator and see whether it works. It's not obvious that 
starting in the denominator is a good idea until we first factor out 
powers of x:

    x^2(x^3 + x^2 - 4x - 4)
    -----------------------
    x^3(x^2 + 4x + 4)

Now we see that the denominator has a quadratic, which we can factor 
by standard methods:

    x^2(x^3 + x^2 - 4x - 4)
    -----------------------
    x^3(x + 2)^2

Okay, if the numerator is worth factoring, it's only because there's 
something that will cancel. So the only factor worth finding is x+2. 
Therefore, we just try dividing x^3+x^2-4x-4 by x+2 to see if it is a 
factor, and we find that

    x^3 + x^2 - 4x - 4 = (x + 2)(x^2 - x - 2)

So our expression is now

    x^2(x + 2)(x^2 - x - 2)
    -----------------------
    x^3(x + 2)^2

At this point we can cancel:

    (x^2 - x - 2)
    -------------
    x(x + 2)

Now all that's left is to factor the numerator; we might as well see 
if we can cancel the remaining x+2, but it turns out that it doesn't 
go. Factoring, we get

    (x - 2)(x + 1)
    -------------
    x(x + 2)

That's the answer, without ever factoring a cubic. Or, since I'm not 
convinced that factored form is simpler, you could just stop with

    x^2 - x - 2
    -----------
     x^2 + 2x

I think this approach is a lot more intelligent than forcing yourself 
to factor a cubic from scratch. But I suppose you have to learn to 
factor them anyway, at least for tests, so the exercise was worth 
something.

If you have any further questions, feel free to write back.

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


Date: 08/15/2002 at 16:12:13
From: Doctor Ian
Subject: Re: Factoring this specific problem

Hi Mike,

Ultimately, they want to combine the terms this way because it will 
make things easier to factor. Here's what's going on. (Note that 
x^0 = 1. Note also that when you have a bunch of additions, it doesn't 
matter what order you do them in, or how you group them.)

    x^3 + x^2 - 4x - 4

  = x^3 + x^2 - 4x^1 - 4x^0         Make exponents explicit.

  = x^3 + x^2 + -4x^1 + -4x^0       Change to additions.

  = x^3 + -4x^1 + x^2 + -4x^0       Change order of additions. 

  = (x^3 + -4x^1) + (x^2 + -4x^0)   Change order of additions. 
  
  = (x^3 - 4x^1) + (x^2 - 4x^0)     Change back to subtractions.

  = (x^3 - 4x) + (x^2 - 4)



>x^2[(x^2-4)(x+1)] 

Okay, let's slow it down again:

    x(x^2 - 4) + (x^2 - 4)

  = x(x^2 - 4) + 1(x^2 - 4)       Make the 1 explicit.
  
    a(   c   ) + b(   c   )

    (a + b)(   c   )

  = (x + 1)(x^2 - 4)              Apply distributive property.


I think that last one threw you because you may not be used to 
thinking of the distributive property as applying to terms bigger 
than a single variable.

Does this help?

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


Date: 08/15/2002 at 20:49:53
From: Mike Cooper
Subject: Thank you

Thanks so much for your help. Sometimes obvious things look like Greek 
to me as I try to help the kids with their algebra, and you always 
save the day. Your help is invaluable.

Mike
Associated Topics:
High School Polynomials
Middle School Factoring Expressions

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