Prove that polynomial p(x)=ax^2+bx+c with integer coefficients that do not simultaneously equal zero does not exist, such that p(2^(1/3))=0?
2 Answers
See below.
Explanation:
Given
we have
but
See explanation...
Explanation:
Note that
#x^3-2#
Suppose
Note that
Further note that
Then
#x^2+b/ax+c/a#
and of:
#x^3+b/ax^2+c/ax#
So it must be a zero of:
#(x^3+b/ax^2+c/ax) - (x^3-2) = b/ax^2+c/ax+2#
and thus a zero of the monic polynomial:
#x^2+c/bx+(2a)/b#
So it must also be a zero of:
#(x^2+b/ax+c/a)-(x^2+c/bx+(2a)/b) = (b/a-c/b)x+(c/a-(2a)/b)#
SInce
#{ (b/a-c/b = 0), (c/a-(2a)/b = 0) :}#
and hence:
#{ ((b/a)^2-c/a = 0), ((b/a) (c/a) -2 = 0) :}#
Hence:
#(b/a)^3 = 2#
So:
#b/a = 2^(1/3)#
which is not possible since