Since the polynomial of degree 4, P(x), has a root of multiplicity 2 at x=3, we can write one factor of the polynomial as (x-3)^2. Similarly, since P(x) has a root of multiplicity 1 at x=0 and x=-1, we can write the other factors as x and (x+1), respectively. Thus, the polynomial can be written as:
P(x) = k(x-3)^2 x(x+1)
where k is some constant.
To find the value of k, we use the fact that P(5) = 12. Substituting x=5 into the equation for P(x), we get:
P(5) = k(2)^2 (5)(6) = 120k
Setting this equal to 12, we have:
120k = 12
k = 1/10
So the polynomial is:
P(x) = (1/10)(x-3)^2 x(x+1)
Therefore, the polynomial of degree 4, P(x), with a root of multiplicity 2 at x=3 and a root of multiplicity 1 at x=0 and x=-1, and which goes through point (5,12), is given by:
P(x) = (1/10)(x-3)^2 x(x+1)
Note that the point (5,12) is not a root of the polynomial, but rather a point on the graph of the polynomial.