Show that the polynomial has as a root the number 1, with multiplicity 3.
the polynomial becomes:
So g(y), that is f(y+1)=f(x) has as a factor the , but not as a power greater than y = x -1.
Thus, 1 is a root of f(x) with multiplicity 3.
Find the polynomial of 4th degree with rational coefficients, which has as roots the numbers i and .
Since the desired polynomial f(x) has rational
coefficients the Theorems 5 and 6 will apply for its irrational
and complex roots.
Therefore the numbers -i, are two other roots the polynomial.
Hence f(x) has the following form:
where a, a non zero rational number.
After doing the proper operations we get:
and one of the asked polynomials is, for example the one we get for a=1:
Prove that the polynomial equation , does not have integer solutions.
If the equations
has integer solutions, these will be divisors of
according to Theorem 8, hence they will be the numbers 1, -1,
But if , then:
Thus the polynomial f(x) does not have integer solutions.