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 ) .

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

Step-by-step 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

problem:

f(x) = - 2/3x + 5; f(5/2)

substitute 5/2 in for x:

f(5/2) = -2/3(5/2) + 5

simplify the right side:

f(5/2) = -5/3 + 5

f(5/2) = 10/3

f(5/2) = 10/3

i hope this !