Abel's theorem in problems and solutions - School of Mathematics

174 Problems of Chapter 2
Since is not a real number, Since and are roots of
the equation in the decomposition (cf., 269)
we must find the factors and We
can write:
In this way the polynomial is divisible by the polynomial of second
degree whose coefficients are real (cf., the solution
of Problem 211).
272. Let be the given polynomial. If the degree of the polynomial
is higher than 2 then the equation has, by the
fundamental Theorem of algebra of complex numbers, a root If is
a real number we divide by We obtain
(cf., 268). If is not a real number the polynomial is divisible by
a polynomial of second degree with real coefficients (cf., 271). In both
cases the quotient is a polynomial, with real coefficients, which results,
for example, from the Euclidean algorithm (cf., §2.1). This quotient is
again divisible by some polynomial of the first or second degree with real
coefficients, etc.. This procedure ends when the quotient obtained has
first or second degree. So we have obtained the required decomposition.
273.
Answer. 1 is a root of order 3, –1 is a root of order 2.
274. Compare the coefficients of the terms of the same degree in the
two members of the given equalities. Let
Thus
it is easy to see that in the case (a), for every the coefficients of in
the two members of the equality are equal to
and that in the case (b) the coefficients of in the two members are
equal to
c) For brevity let us use the summation symbol The symbol
(respectively, means that one has to consider the expression
which lies on the right of this symbol for (respectively,