Find abcd and prove they are not integers
A good thought I guess, but quite 'brute force-ish'. And how would you exactly go about finding all 4 solutions =p
Since '1' is the constant term, this means that at least one of the root(s) must be a factor of 1.
So we test the following:
P(1) = ...
P(-1) = ...
We then observe that neither of them is equal to 0 ie: none of the zeroes are integers.
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Generally when trying to determine is a polynomial P(x) has integral zeroes or not, we substitute the factors of the CONSTANT term.
So say the constant term is 6, we would then test:
P(1)
P(-1)
P(6)
P(-6)
P(2)
P(-2)
P(3)
P(-3)
If none of them are equal to 0, then it has no integral zeroes.
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Now suppose the polynomial is NOT monic ie: has a leading coefficient.
We then test the factors of Constant term/Leading coefficient.
So suppose we have the polynomial P(x) =2x^3 - blah blah + 6, then we would test factors of 3 because 6/2 = 3.
So we can deduce from this that suppose the constant term is PRIME and the polynomial is NOT monic, then it has no integral zeroes.