A monic polynomial P of degree 5 satisfies P(1)=1, P(2)=4, P(3)=9, P(4)=16, P(5)=73. Find P(6).
+2401 P(x) = x^2 for when x = {1, 2, 3, 4}.
F(x) = P(x) - x^2
F(5) = 73 - 25
F(5) = 48
Since P(x) is a monic polynomial with a degree of 5, so is F(x).
F(x) = 0 for when x = {1, 2, 3, 4, 5}
F(x) = (x-a)(x-1)(x-2)(x-3)(x-4)
F(5) = 24(5 - a)
a = 3
F(x) = (x-3)(x-1)(x-2)(x-3)(x-4)
(x-3)(x-1)(x-2)(x-3)(x-4) = P(x) - x^2
P(x) = (x-3)(x-1)(x-2)(x-3)(x-4) + x^2
P(6) = 396
Did I get it right?
=^._.^=
