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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 ) .

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Really226idn

Answer:

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

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