Re: HSC 2013 4U Marathon
Am disappointed that no-one has answered Sy's conjugate root question!
Can you please criticise this proof? I am starting to have doubts about it.
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Start with the polynomial,
We need to make the lowest reducible rational polynomial that has that as a root, so we simply:
The above polynomial has the root
It is the simplest most reducible rational polynomial that has p + root(q) as a root.
Now consider:
Where a_1, a_2 ... are all rational polynomial functions of lowest reducibility.
If
is a root then among the series a_1, a_2... there must exist a_k for some k such that
is a root.
Since a_k is irreducible and rational, then that polynomial must be:
And that polynomial also has the root
Therefore that is also a root of P(x).
Is there a gap or something in this? Or is it all good?