Question

The polynomial of degree 4, P ( x ) , has a root of multiplicity 2 at x = 1 and roots of multiplicity 1 at x = 0 and x = − 1 . It goes through the point ( 5 , 288 ) .

Answer 1

P ( x ) = (3/5)*[(x-1)^2]*(x)*(x + 1 )

Explanation:

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

sub P(5)=288

288 = a (4)^2(5) (6)

a = 3/5