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The problem asks to find a quadratic polynomial \( P(x) \) that satisfies the following conditions:

1. \( P(x) \) is positive or zero for all real numbers.
2. \( P(1) = 0 \) and \( P(2) = 2 \).

To solve this, let's assume the general form of a quadratic polynomial:
\[
P(x) = a(x - 1)^2
\]
We choose this form because we know the polynomial must be non-negative (hence the square term) and must have a root at \( x = 1 \) (since \( P(1) = 0 \)).

Now, we use the second condition \( P(2) = 2 \) to determine the constant \( a \).

Substitute \( x = 2 \) into the polynomial:
\[
P(2) = a(2 - 1)^2 = a(1)^2 = a
\]
Given that \( P(2) = 2 \), we have:
\[
a = 2
\]

Thus, the quadratic polynomial is:
\[
P(x) = 2(x - 1)^2
\]

This polynomial satisfies all the given conditions. Let me know if you need further clarification!

### Additional related questions:
1. What happens if \( P(x) \) had more than one root?
2. How can you determine the vertex of the quadratic polynomial \( P(x) = 2(x - 1)^2 \)?
3. What are the key features of a polynomial that is always non-negative for all real numbers?
4. How would the solution change if \( P(1) = 1 \) instead of \( P(1) = 0 \)?
5. Can a quadratic polynomial have more than two roots?

### Tip:
Quadratic polynomials of the form \( a(x - h)^2 + k \) represent parabolas. If \( a > 0 \), the parabola opens upwards, ensuring that the polynomial remains non-negative.

Find a quadratic polynomial P(x) that satisfies the following properties: P(x) is non-negative for all real numbers, P(1) = 0, and P(2) = 2.

Learn how to find a quadratic polynomial P(x) that is non-negative for all real numbers with conditions P(1) = 0 and P(2) = 2. This problem covers key algebra concepts such as roots and non-negative functions, suitable for students in Grades 8-10.

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