The polynomial of degree $4, P(x)$, has a root of multiplicity 2 at $x=2$ and roots of multiplicity 1 at $x=0$ and $x=-4$. It goes through the point $(5,40.5)$. Find a formula for $P(x)$. \[ P(x)= \]

Step 1 ：We are given a polynomial of degree 4, P(x), with a root of multiplicity 2 at x=2 and roots of multiplicity 1 at x=0 and x=-4. It also passes through the point (5,40.5).

Step 2 ：We can write the polynomial in the form P(x) = a*(x-2)^2*x*(x+4).

Step 3 ：We can find the value of 'a' by substitifying the point (5,40.5) into the equation. This gives us 40.5 = a*(5-2)^2*5*(5+4).

Step 4 ：Solving this equation gives us a value of a = 0.1.

Step 5 ：Substituting a = 0.1 back into the equation gives us the polynomial P(x) = 0.1x(x - 2)^2(x + 4).

Step 6 ：\(\boxed{P(x) = 0.1x(x - 2)^2(x + 4)}\) is the final answer.