The polynomials a(x), b(x), c(x), d(x) satisfy a(x5) + x b(x5) + x2c(x5) = (1 + x + x2 + x3 + x4) d(x). Show that a(x) has the factor (x -1).
Solution
Take k to be a complex 5th root of 1, so that 1 + k + k2 + k3 + k4 = 0. Putting x = k, k2, k3, k4 in the given equation we get:
a(1) + k b(1) + k2c(1) = 0 a(1) + k2b(1) + k4c(1) = 0 a(1) + k3b(1) + k c(1) = 0 a(1) + k4b(1) + k3c(1) = 0Multiplying by -k , -k2, -k3, -k4 respectively, we get
-k a(1) - k2b(1) - k3c(1) = 0 -k2a(1) - k4b(1) - k c(1) = 0 -k3a(1) - k b(1) - k4c(1) = 0 -k4a(1) - k3b(1) - k2c(1) = 0Adding all eight equations gives 5 a(1) = 0. Hence a(x) has the root x = 1 and hence the factor (x - 1).
© John Scholes
jscholes@kalva.demon.co.uk
11 Jul 2002