Final answer:
The function f(x) = -x^4 - 6x^3 - 9x^2 has two real zeros: x=0 and x=-3, both with a multiplicity of 2, found by factoring and using the quadratic formula as required.
Step-by-step explanation:
To determine the real zeros and the multiplicity of any repeated zeros of f(x) = -x^4-6x^3-9x^2, we can first factor out the greatest common factor (x^2) to simplify the polynomial. This gives us:
f(x) = x^2(-x^2 - 6x - 9).
Now, we can set f(x) to zero and solve for x:
0 = x^2(-x^2 - 6x - 9).
This equation has two parts, x^2 and -x^2 - 6x - 9. We can set each part equal to zero separately:
x^2 = 0 and -x^2 - 6x - 9 = 0.
The first part, x^2 = 0, gives us a zero at x = 0 with multiplicity 2 because the term is squared.
To solve the quadratic equation -x^2 - 6x - 9 = 0, we can use the quadratic formula. First, we multiply the equation by -1 to make it easier to work with:
x^2 + 6x + 9 = 0.
This is a perfect square trinomial, which factors to (x + 3)^2 = 0. Therefore, we have another zero at x = -3, with a multiplicity of 2.
The real zeros are x = 0 and x = -3, with multiplicities of 2 for each.