Since P(≤ for each k=0.1 we have Q(-)≤0 when k is odd and C(x)≥0 when k is even Since Q is continuous, the Intermediate Value Theorem implies that the e polynomia Q(a) has at least one zero between i, and i'i+1 eac j=0,1,.,n-1. Thus Q has at least n zeros in interval [-1, 1. But the degree of Q(ar)is less than n, So Q=0. This implies that Pn= Tn