Harristotelian Logic

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.

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.