Re: Gotcha


<jstevh@xxxxxxxxx> wrote in message
news:1177821822.884887.303000@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
Start with

x^2 - 6x + 35 = 0

and solve using the quadratic formula to get

x = (3 +/- sqrt(-26)

and now comes a simple yet clever trick--multiply the last two
coefficients of the original polynomial with one solution for x and
I'll use x = 3 + sqrt(-26), so I have

x^2 - 6(3+sqrt(-26))x + 35(3 + sqrt(-26)) = 0

where the purpose here is to make sure that BOTH roots must have 3 +
sqrt(-26) as a factor, but notice that one root must have 7 as a
factor as well, as the last coefficient has 7 as a factor, so this is
just a tricky way of forcing one root to have 7 as a factor without
forcing them both.

Now I can group my radicals to one side and everything else on the
other to get

sqrt(-26)(-6x + 35) = -x^2 + 18x - 105

and square both sides to get

(-26)(36x^2 -420x + 1225) = x^4 - 36x^3 + 534x^2 - 3780x + 11025

which is

x^4 - 36x^3 +1470x^2 - 14700x + 42875 = 0

where at least one root must have 7 as a factor.

I'm confused by your conclusion. I suppose your conclusion results because
42875 = 7*7*7*125, so somewhere in the process of forming your equation P(x)
= x^4 - 36x^3 +1470x^2 - 14700x + 42875 = 0
from it's roots, there must exist equations that have 7 as a factor. There
are, but not in the 4 primary roots of P(x).

The roots of your P(x) are:
r1 = 9 + 3i*sqrt(26) +sqrt[-258 + 19i*sqrt(26)]
r2 = 9 + 3i*sqrt(26) -sqrt[-258 + 19i*sqrt(26)]
r3 = 9 - 3i*sqrt(26) +sqrt[-258 - 19i*sqrt(26)]
r4 = 9 - 3i*sqrt(26) -sqrt[-258 - 19i*sqrt(26)]

and P(x) = (x-r1)*(x-r2)*(x-r3)*(x-r4)
But none of the roots r1, r2, r3, r4 appear to have 7 as a factor.

However combining these in 2 steps:
(x-r1)*(x-r2) = x^2 -(18 + 6i*sqrt(26)*x + 105 + 35i*sqrt(26)
and
(x-r3)*(x-r4) = x^2 -(18 - 6i*sqrt(26)*x + 105 - 35i*sqrt(26)
and both equations of this intermediate step have factors of 7 in the
constant terms. Multipying these two intermediate steps yield P(x) again.

To conclude, although you have not shown that 7 is a factor of at least one
of the roots r1,r2,r3,r4, I at least leave you with the comfy feeling that
somewhere in the mix, those factors of 7 get taken care of, even if they are
not a factor of the roots of P(x).

KeithK

But that polynomial is irreducible over Q.

So you see, I've been right all along and you people have been
fighting very valuable and important mathematics.

You've been fighting against the discipline of mathematics itself in a
futile attempt to destroy the future of humanity.

You have been fighting history.


James Harris



.

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