Question

select the correct answer from each drop - down menu.
the graph of the function $y = x^{2}(x^{2}-6x + 9)$ has zeros of , so the function has distinct real zeros and complex zeros.
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Explanation:

Step1: Factor the quadratic

First, factor the quadratic \(x^2 - 6x + 9\). We know that \(x^2 - 6x + 9=(x - 3)^2\) (since \((a - b)^2=a^2-2ab + b^2\) with \(a = x\) and \(b = 3\), so \(x^2-6x + 9=(x - 3)^2\)).
So the function \(y=x^2(x^2 - 6x + 9)\) can be written as \(y=x^2(x - 3)^2\).

Step2: Find the zeros

To find the zeros of the function, we set \(y = 0\). So we solve the equation \(x^2(x - 3)^2=0\).
Using the zero - product property (if \(ab = 0\), then either \(a = 0\) or \(b = 0\)), we have:

• For \(x^2=0\), we get \(x = 0\) (with multiplicity 2).
• For \((x - 3)^2=0\), we get \(x=3\) (with multiplicity 2).

So the zeros of the function are \(x = 0\) (multiplicity 2) and \(x = 3\) (multiplicity 2).

Step3: Determine the number of distinct real zeros

The distinct real zeros are the values of \(x\) that make the function zero, regardless of multiplicity. So the distinct real zeros are \(x = 0\) and \(x = 3\), so there are 2 distinct real zeros.

Step4: Determine the number of complex zeros (including real zeros, since real numbers are a subset of complex numbers)

The degree of the polynomial \(y=x^2(x^2 - 6x + 9)=x^4-6x^3 + 9x^2\) is 4. By the fundamental theorem of algebra, a polynomial of degree \(n\) has exactly \(n\) complex zeros (counting multiplicities). The total number of complex zeros (counting multiplicities) is 4. But if we are talking about non - real complex zeros, since all zeros are real (\(x = 0\) and \(x = 3\)), the number of non - real complex zeros is 0. But if we consider complex zeros in the sense of all zeros (real and non - real), we have 4 complex zeros. However, the question seems to be about distinct real zeros and then the number of complex zeros (maybe non - real? But in the context of the problem, let's re - evaluate.

Wait, the first blank: the zeros are \(x = 0\) (multiplicity 2) and \(x = 3\) (multiplicity 2), so we can write the zeros as \(x = 0\) and \(x = 3\) (or more precisely, \(x = 0\) (with multiplicity 2) and \(x = 3\) (with multiplicity 2)).

The second blank: the number of distinct real zeros is 2 (since 0 and 3 are the distinct real roots).

The third blank: The degree of the polynomial is 4. The total number of complex zeros (counting multiplicities) is 4. The number of real zeros (counting multiplicities) is \(2 + 2=4\) (since both 0 and 3 have multiplicity 2). So the number of non - real complex zeros is \(4-4 = 0\). But if we consider complex zeros in the sense of all zeros (real and non - real), the number of complex zeros is 4. But since all zeros are real, the number of non - real complex zeros is 0.

Answer:

First drop - down: \(x = 0\) (multiplicity 2) and \(x = 3\) (multiplicity 2) (or \(0\) and \(3\))
Second drop - down: 2
Third drop - down: 0 non - real (or 4 if including real zeros as complex)

But based on the context of a typical high - school problem:

The graph of the function \(y=x^2(x^2 - 6x + 9)\) has zeros of \(\boldsymbol{x = 0}\) (with multiplicity 2) and \(\boldsymbol{x = 3}\) (with multiplicity 2), so the function has \(\boldsymbol{2}\) distinct real zeros and \(\boldsymbol{0}\) non - real complex zeros (or 4 complex zeros if including real ones, but usually in this context, complex zeros refer to non - real, so 0).