http://z5lcip2ba3ec6v5za5snjzncpvr6ocyvqnkox2alerut6ruozpjtvkqd.onion/math/catalog
Quick proof. For the first statement: P(x) = Q(x) (x - c) + R so P(c) = Q(c) (c - c) + R = R For the second: P(x) = Q(x) (x - c) + P(c) [tex: \frac { P(x) - P(c) } { x - c } = Q(x) ] [tex: P'(c) = \lim {x \to \infty } \frac { P(x) - P(c) } { x - c } = Q(c) ] Thread 112859 in /math/ P: 11, last 5 months ago...