The graph of a polynomial function has zeros of 0 (multiplicity 2), 1 (multiplicity 2), and 5/2 (multiplicity 2). Write a function in standard form that could represent this function.

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The zeros given are $$0$$ (multiplicity 2), $$1$$ (multiplicity 2), and $$\frac{5}{2}$$ (multiplicity 2). This tells us that a possible factorization of the polynomial is

$$
f(x)= a\, x^2 (x-1)^2 \left(x-\frac{5}{2}\right)^2,
$$

where $$a$$ is a constant. For simplicity, we can choose $$a=1$$.

Thus, we have

$$
f(x)= x^2 (x-1)^2 \left(x-\frac{5}{2}\right)^2.
$$

We now expand the expression step by step.

1. Write the factors $$(x-1)^2$$ and $$\left(x-\frac{5}{2}\right)^2$$ explicitly:

$$
   (x-1)^2 = x^2 - 2x + 1,
   $$
   $$
   \left(x-\frac{5}{2}\right)^2 = x^2 - 5x + \frac{25}{4}.
   $$

2. Multiply $$(x^2 - 2x + 1)$$ and $$\left(x^2 - 5x + \frac{25}{4}\right)$$:

$$
   \begin{aligned}
   (x^2 - 2x + 1)&\left(x^2 - 5x + \frac{25}{4}\right) \
   &= x^2\cdot x^2 + x^2\cdot (-5x) + x^2\cdot \frac{25}{4} \
   &\quad {} + (-2x)\cdot x^2 + (-2x)\cdot (-5x) + (-2x)\cdot \frac{25}{4} \
   &\quad {} + 1\cdot x^2 + 1\cdot (-5x) + 1\cdot \frac{25}{4} \
   &= x^4 - 5x^3 + \frac{25}{4}x^2 - 2x^3 + 10x^2 - \frac{50}{4}x + x^2 - 5x + \frac{25}{4}.
   \end{aligned}
   $$

3. Combine like terms:

- The $$x^4$$ term: $$x^4$$.
   - The $$x^3$$ terms: $$-5x^3 - 2x^3 = -7x^3$$.
   - The $$x^2$$ terms: $$\frac{25}{4}x^2 + 10x^2 + x^2$$. Write $$10x^2 + x^2 = 11x^2 = \frac{44}{4}x^2$$. Then, 
     
     $$
     \frac{25}{4}x^2 + \frac{44}{4}x^2 = \frac{69}{4}x^2.
     $$
     
   - The $$x$$ terms: $$-\frac{50}{4}x - 5x$$. Write $$-5x = -\frac{20}{4}x$$. Then,
     
     $$
     -\frac{50}{4}x - \frac{20}{4}x = -\frac{70}{4}x = -\frac{35}{2}x.
     $$
     
   - The constant term: $$\frac{25}{4}$$.

So, the product becomes:

$$
   (x-1)^2\left(x-\frac{5}{2}\right)^2 = x^4 - 7x^3 + \frac{69}{4}x^2 - \frac{35}{2}x + \frac{25}{4}.
   $$

4. Now multiply this result by $$x^2$$:

$$
   \begin{aligned}
   f(x) &= x^2\left(x^4 - 7x^3 + \frac{69}{4}x^2 - \frac{35}{2}x + \frac{25}{4}\right) \
   &= x^6 - 7x^5 + \frac{69}{4}x^4 - \frac{35}{2}x^3 + \frac{25}{4}x^2.
   \end{aligned}
   $$

Thus, one possible function in standard form that has the given zeros is

$$
f(x)= x^6 - 7x^5 + \frac{69}{4}x^4 - \frac{35}{2}x^3 + \frac{25}{4}x^2.
$$

This is the answer.
A possible polynomial function with zeros at 0 (multiplicity 2), 1 (multiplicity 2), and 5/2 (multiplicity 2) is: $$ f(x) = x^6 - 7x^5 + \frac{69}{4}x^4 - \frac{35}{2}x^3 + \frac{25}{4}x^2 $$

The graph of a polynomial function has zeros of 0 (multiplicity 2), 1 (multiplicity 2), and 5/2 (multiplicity 2). Write a function in standard form that could represent this function.

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