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Determine

the real zeros and state the multiplicity of any repeated zeros of f(x)= -x^4-6x^3-9x^2

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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.

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