Question

The polynomial of degree 4, P ( x ) has a root of multiplicity 2 at x = 4 and roots of multiplicity 1 at x = 0 and x = − 1 . It goes through the point ( 5 , 9 ) . Find a formula for P ( x ) .

Answer 1

To have roots as described, that means we have the following factors: From multiplicity 2 at x=1 has (x-1)^2 as its factor From multiplicity 1 at x=0 has x as a factor From multiplicity 1 at x = -4 has a factor of x+4 Putting these together we get that P(x) = A (x) (x+4) (x-1)^2 Multiply these out and find P(x) = A (x^2 + 4x) (x^2 - 2x + 1) A ( x^4 - 2x^3 + x^2 + 4x^3 - 8x^2 + 4x ) Combine like terms and find P(x) = A (x^4 + 2x^3 - 7x^2 + 4x) To find A, we use the point they gave us (5, 72) P(5) = A [ (5)^4 + 2(5)^3 - 7(5)^2 + 4(5) ] = 72 A [ 625 + 250 - 175 + 20 ] = 72 A [ 720 ] = 72 Divide both sides by 720 and find that A = 0.1 Final answer: P(x) = 0.1 ( x^4 + 2x^3 - 7x^2 + 4x) or P(x) = 0.1 x^4 + 0.2 x^3 - 0.7x^2 + 0.4x