To find a possible formula for the polynomial P(x), we can start by considering its roots.
We are given that P(x) has roots of multiplicity 2 at x = 4 and x = 0, and a root of multiplicity 1 at x = -1. This means that (x - 4) and x are factors of P(x), and (x + 1) is a factor of the derivative of P(x).
Since the leading coefficient of P(x) is 1, we can write the polynomial in factored form as:
P(x) = (x - 4)^2 * x * (x + 1) * Q(x)
where Q(x) is a polynomial of degree 1 or 2 (to fulfill the degree requirement of 5 for P(x)) with a leading coefficient of 1.
Now, we need to determine Q(x). Since P(x) is a polynomial of degree 5, Q(x) must be a quadratic polynomial. Let's assume Q(x) = ax^2 + bx + c, where a, b, and c are constants.
Expanding the factored form of P(x), we have:
P(x) = (x - 4)^2 * x * (x + 1) * (ax^2 + bx + c)
Multiplying this out, we get:
P(x) = a x^5 + (8a - a)x^4 + (14a - 4a + b)x^3 + (-9a + 14b - 4b + c)x^2 + (-4c)x
Equating the coefficients of the corresponding powers of x with zero, we can determine the values of a, b, and c.
For the coefficient of x^5, we have a = 1.
For the coefficient of x^4, we have 8a - a = 0, which gives 7a = 0. Therefore, a = 0.
For the coefficient of x^3, we have 14a - 4a + b = 0, which simplifies to 10a + b = 0. Since a = 0, we have b = 0.
For the coefficient of x^2, we have -9a + 14b - 4b + c = 0, which simplifies to c = 0.
For the coefficient of x, we have -4c = 0, which satisfies our assumption that c = 0.
Therefore, we have Q(x) = ax^2 + bx + c = 0x^2 + 0x + 0 = 0.
Substituting these values into the factored form of P(x), we get:
P(x) = (x - 4)^2 * x * (x + 1) * (0)
Finally, simplifying further, we have:
P(x) = (x - 4)^2 * x * (x + 1)
So, a possible formula for P(x) is P(x) = (x - 4)^2 * x * (x + 1).