341This one has been doing the rounds
As a counter example: Consider f(x) = x-i
f(x2+x+1) = x2+x+1-i = (x-i)(x+1+i) = f(x) g(x)
But f is of odd degree
f(x) and g(x) are two polynomials such that f(x2+x+1)=f(x)g(x). Show that f(x) is a polynomial of even order.(i.e. highest power of x in f(x) is even).
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5 Answers
1let f() have a real root c0.
then c1=c02+c0+1 is also a root.
check c1≥3/4.
define ci=ci-12+ci-1+1
then all ci 's are real roots of f. also c2>1 implying ci>c()i-1 for all i>2..
but f can have only finitely many roots. so f doesn't have any real root.
1gr8 work ith_power..
For those who are still wrestling with the problem ...remember that odd power polynomials must have at least one root...(try why?) ...so just show that its impossible for f(x) to have any real root..
Good work guys..Its time for me to increase the difficulty of questions :)
1i never said my proof is complete. it is not given f & g are REAL POLYNOMIALS!!!!!!!
341341In fact, to prove rigorously that f cannot have any real root, you will have to prove that it is not possible to make a ring of roots,
ro, r1,...., rn satisfying
r1 = r02+r0+1
r2 = r12+r1+1
.
.
.
r0 = rn2+rn+1
as this give rise to the absurdity that both
r0<r1<r2<...<rn and
r0> rn hold simultaneously
Which means that f(x) has all roots non-real
Now, if we add the condition that f(x) has all coefficients real, we can state that f(x) is of even power.
(If all roots are non-real then they occur in conjugate pairs) and so there are an even number of roots. which means the degree of f is even)