I need help factoring this rational function? thanks!?

Question

r(x)= [x^4+x^3-7x^2-13x-6] / [x^6-6x^5+15x^4-30x^3+15x^2-6x+1]

r(x)= [x^4+x^3-7x^2-13x-6] / [x^6-6x^5+15x^4-30x^3+15x^2-6x+1]

r(x)=
[x^4+x^3-7x^2-13x-6] / [x^6-6x^5+
15x^4-30x^3+15x^2-6x+1]

Answer 1

Are you certain the denominator isn't:

x^6 - 6x^5 + 15x^4 - 20x^3 + 15x^2 - 6x +1 ?

Because that denominator has a very simple factorization: (x-1)^6. Whereas the polynomial you actually gave for the denominator doesn't have any factors I'd be inclined to call simple, and I'll bet even money that there's no way of expressing them using any finite sequence of elementary operations and nth roots. I think you should double-check your input.

Edit: well, it's a very good thing nobody took me up on that bet - it seems I am completely wrong. See the question at http://answers.yahoo.com/question/index;_ylt=Aqz3W3s3NPcz.j__CQYxK0bsy6IX?qid=20061127003230AA2oHuA

I will give a summary here:

As mentioned, (x-1)^6 = x^6 - 6x^5 + 15x^4 - 20x^3 + 15x^2 - 6x +1. So logically (assuming that really isn't a typo), your original denominator is (x-1)^6 - 10x^3. We need to find the roots of this equation, so set:

(x-1)^6 - 10x^3 = 0
(x-1)^6 - 10√x^6 = 0
Since x=0 is not a solution, we may divide by √x^6. so:
(√x-1/√x)^6 - 10 = 0

Let t=√x-1/√x. Then this polynomial is:

t^6-10=0

The solutions of this polynomial are precisely the complex sixth roots of 10, which are:

t=√(∛(10))
t=√(∛(10)) * (1/2 + i√3/2)
t=√(∛(10)) * (-1/2 + i√3/2)
t=-√(∛(10))
t=√(∛(10)) * (-1/2 - i√3/2)
t=√(∛(10)) * (-1/2 + i√3/2)

We now set √x-1/√x equal to each of these roots in turn. In so doing, we find the possible roots of x. Let u=√x. Then:

u-1/u=t
u²-tu-1=0
u = (t±√(t²+4))/2
x = u² = (t² ± t√(t²+4) + 2)/2

x=(∛10 ± √(∛100+4∛10) + 2)/2

x=(∛10 * (-1/2 + i√3/2) ± √(∛(10)) * (1/2 + i√3/2)√(∛10 * (-1/2 + i√3/2) + 4) + 2)/2
= (∛10 * (-1/2 + i√3/2) ± √(∛100 * (-1/2 - i√3/2) + 4∛10 * (-1/2 + i√3/2)) + 2)/2
= (∛10 * (-1/2 + i√3/2) ± √(-(∛100 + 4∛10)/2 + (4∛10 - ∛100)i√3/2) + 2)/2

x=(∛10 * (-1/2 - i√3/2) ± √(∛(10)) * (-1/2 + i√3/2)√(∛10 * (-1/2 - i√3/2) + 4) + 2)/2
= (∛10 * (-1/2 - i√3/2) ± √(∛100 * (-1/2 + i√3/2) + 4∛10 * (-1/2 - i√3/2)) + 2)/2
= (∛10 * (-1/2 - i√3/2) ± √(-(∛100 + 4∛10)/2 + (∛100 - 4∛10)i√3/2) + 2)/2

Thus the factors of the denominator are:

(x - (∛10 + √(∛100+4∛10) + 2)/2) * (x - (∛10 - √(∛100+4∛10) + 2)/2) * (x - (∛10 * (-1/2 + i√3/2) + √(-(∛100 + 4∛10)/2 + (4∛10 - ∛100)i√3/2) + 2)/2) * (x - (∛10 * (-1/2 + i√3/2) - √(-(∛100 + 4∛10)/2 + (4∛10 - ∛100)i√3/2) + 2)/2) * (x - (∛10 * (-1/2 - i√3/2) + √(-(∛100 + 4∛10)/2 + (∛100 - 4∛10)i√3/2) + 2)/2) * (x - (∛10 * (-1/2 - i√3/2) - √(-(∛100 + 4∛10)/2 + (∛100 - 4∛10)i√3/2) + 2)/2).

And yes, that is one ugly polynomial.

Answer 2

[x4+x3-7x2-13x-6] = (x+1)^2*(x+2)*(x-3)
[x6-6x5+15x4-20x3+15x2-6x+1 - 10x]
= (x-1)^6-10x
= ((x-1)^3+sqrt(10x)) * ((x-1)^3-sqrt(10x))
= go on!

Answer 3

Please provide the complete denominator.