Question

The polynomial of degree 4, P(x) has a root multiplicity 2 at x=4 and roots multiplicity 1 at x=0 and x=-4 and it goes through the point (5, 18) how do you find a formula for p(x)?

Answer 1

The polynomial is `P(x)=2/5x(x-4)^2(x+4)`

Explanation:

If the polynomial has a root of multiplicity 2 at `x=4`, the `(x-4)^2`
is a factor

Multiplicity 1 at `x=0`, then `x` is a factor

Multiplicity 1 at `x=-4`, then `(x+4)` is a factor

So `P(x)=Ax(x-4)^2(x+4)`

As it pases through `(5,18)` so
`18=A*5*(5-4)^2*(5+4)`

So `A=18/5*1/9=2/5`

The polynomial is `P(x)=2/5x(x-4)^2(x+4)`