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.