The polynomial of degree 4, P(x), has a root of multiplicity 2 at x=0 and x=-1. It goes through the point of (5,21). Find a formula for P(x).

Answer 1

P(x) = a(x^2)(x+1)^2 = y

plug in the point (5,21) and solve for a

a(5^2)(5+1)^2 = 21

a = 21/(25)(36) = 7/(25)(12) = 7/300

P(x) = (7/300)(x^2)(x+1)^2

P(x) = (7/300)(x^4 +2x^3 +x^2)

P(x) = 7x^4/300 + 7x^3/150 + 7x^2/300 is a 4th degree polynomial with roots 0 and -1, each with multiplicity 2, going through the point (5,21)