2019/20 AHSMC Part 1 Problem 14

Problem 14 of 16

$p(x)$ is a sixth degree polynomial such that $p(k) = 2k^2 + 1$ for $k = \pm 1$ , $k = \pm 2$ , $k = \pm 3$ . If $p(0) = -71$ then $p(\sqrt{3})$ is

Show solution

Solution

Let $q(x) = 2x^2 + 1 - p(x)$ . $q(x)$ is a sixth degree polynomial such that $q(\pm 1) = 0$ , $q(\pm 2) = 0$ , $q(\pm 3) = 0$ hence $q(x) = c(x-1)(x+1)(x-2)(x+2)(x-3)(x+3)$ , $x \in \mathbb{R}$ , with $c$ a non-zero real constant. Therefore $p(x) = 2x^2 + 1 - c(x-1)(x+1)(x-2)(x+2)(x-3)(x+3) = 2x^2 + 1 - c(x^2-1)(x^2-4)(x^2-9)$ and since $p(0) = -71$ we obtain $c = -2$ , and then $p(\sqrt{3}) = 31$ . The answer is (D).

University of Alberta

Answer choices

Solution