p(x) is a polynomial of degree n. q(x) is a polynomial of degree 2. p(x) = p''(x)q(x) and the roots of p(x) are not all equal. Show that the roots of p(x) are all distinct.
Solution
Suppose not, so for some α and n > 1, we have p(x) = (x - α)nr(x) with r(α) ≠ 0. Then (x - α)nr(x) = ( n(n - 1)(x - α)n-2r(x) + 2n(x - α)n-1r'(x) + (x - α)nr''(x) ) q(x). Dividing by (x - α)n-2, we get (x - α)2r(x) = ( n(n - 1)r(x) + 2n(x - α)r'(x) + (x - α)2r''(x) ) q(x). Setting x = α, we conclude that q(α) = 0 and hence (x - α) divides q(x). Dividing through by (x - α) and repeating the argument, we find that q(x) has a second factor (x - α), in other words q(x) = c(x - α)2. Dividing that out too, we get: r(x) = c n(n - 1) r(x) + c 2 n(x - α)r'(x) + c(x - α)2r''(x). Setting x = α gives c = 1/(n(n - 1)). Hence r''(x)/r'(x) = -2n/(x - α). Integrating: r'(x) = A/(x - α)2n, which is impossible since r(x) is supposed to be a polynomial.
© John Scholes
jscholes@kalva.demon.co.uk
14 Dec 1999