
Victoria V. answered 10/26/19
20+years teaching PreCalculus & all Surrounding Topics
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