Define the function G(x) as the sum f(x) + f(3x) + ... + f((2m-1)x) - (2m-1). Because f(x) is a monotonic polynomial, G(x) is also a strictly monotonic function. Evaluate G at x = 1, which gives G(1) = f(1) + f(3) + ... + f(2m-1) - (2m-1). If G(1) is positive, then since G(x) goes to negative infinity as x approaches negative infinity, it must cross the x-axis exactly once to become positive. If G(1) is negative, it must cross the x-axis exactly once to become positive as x approaches positive infinity. If G(1) is zero, x = 1 is the single root. In all possible scenarios, the continuous function G(x) crosses zero exactly once, guaranteeing at least one real root.