Complex numbers and polynomials - University College Cork

d and q divides a. But ad is odd. Hence both of a, d are odd and so p, q areodd. Next note thatap 3 + bp 2 q + cpq 2 + dq 3 = 0.The first and last terms are odd, so their sum is even, and since bc is evenone of b, c is even. Hence one of bp 2 q, cpq 2 is even. But their sum is even.Hence both are even. Hence b, c are even. Finally, if the roots are rational wecan denote them by p 1 /q 1 , p 2 /q 2 , p 3 /q 3 , where, for k = 1, 2, 3, p k , q k are oddintegers, with p k , q k having no common factors. Thenp 1q 1+ p 2q 2+ p 3q 3= − b a ,whence we derive (?) a contradiction. Thus at least one root is irrational.7. Suppose m, n ≥ 1 andf(x) = x n + · · · + a 1 x + a 0 , g(x) = b m x m + · · · + b 1 x + b 0 , b m ≠ 0,are two polynomials with integer coefficients and share a common real root.Prove that the common root is an integer. This is false! For instance x 2 −2, x(x 2 − 2) share a common real root, which is irrational.To see what’s involved, let n = 3, m = 2 and let x denote the common root.Then0 = x 3 + a 2 x 2 + a 1 x + a 0 = b 2 x 2 + b 1 x + b 0 .Thenwhence0 = b 2 x 3 + a 2 b 2 x 2 + a 1 b 2 x + a 0 b 2 = b 2 x 3 + b 1 x 2 + b 0 x,0 = (a 2 b 2 − b 1 )x 2 + (a 1 b 2 − b 0 )x + a 0 b 2 = b 2 x 2 + b 1 x + b 0 .Now eliminate the x 2 term and conclude that[b 2 (a 1 b 2 − b 0 ) − b 1 (a 2 b 2 − b 1 )]x + [a 0 b 2 2 − b 0 (a 2 b 1 − b 1 )] = 0.It follows from this that ifthen[b 2 (a 1 b 2 − b 0 ) − b 1 (a 2 b 2 − b 1 )] = 0,[a 0 b 2 2 − b 0 (a 2 b 1 − b 1 )] = 0.Otherwise, x is rational, and so a rational root of f. By the rational rootstheorem, it must then be an integer.g32