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This is a problem involving polynomial division and the Chinese Remainder Theorem (CRT) in polynomials.

We are given the following information:

1. A polynomial \( f(x) \) is divided by \( x - p \), and the remainder is \( p^2 \).
2. The same polynomial \( f(x) \) is divided by \( x - q \), and the remainder is \( q^2 \).
3. We are asked to find the remainder when \( f(x) \) is divided by \( x^2 - (p + q)x + pq \), which can be factored as \( (x - p)(x - q) \).

### Solution:

Let the polynomial be \( f(x) \). We know the following:

- When divided by \( x - p \), the remainder is \( p^2 \). Thus, by the Remainder Theorem:
  \[
  f(p) = p^2
  \]
- When divided by \( x - q \), the remainder is \( q^2 \). Hence:
  \[
  f(q) = q^2
  \]

Now, we need to find the remainder when \( f(x) \) is divided by \( (x - p)(x - q) = x^2 - (p + q)x + pq \). Since the divisor is quadratic, the remainder will be a linear polynomial of the form:
\[
R(x) = ax + b
\]
We need to determine the coefficients \( a \) and \( b \).

Using the information that \( f(p) = p^2 \) and \( f(q) = q^2 \), substitute these into the remainder form:
- At \( x = p \):
  \[
  R(p) = ap + b = p^2
  \]
  Thus, \( ap + b = p^2 \) (Equation 1).

- At \( x = q \):
  \[
  R(q) = aq + b = q^2
  \]
  Hence, \( aq + b = q^2 \) (Equation 2).

We now have the system of equations:
1. \( ap + b = p^2 \)
2. \( aq + b = q^2 \)

### Solving the system:

Subtract Equation 1 from Equation 2:
\[
(aq + b) - (ap + b) = q^2 - p^2
\]
This simplifies to:
\[
a(q - p) = (q - p)(q + p)
\]
Since \( p \neq q \), divide by \( q - p \) (non-zero):
\[
a = q + p
\]

Substitute \( a = q + p \) into Equation 1:
\[
(p(q + p)) + b = p^2
\]
\[
pq + p^2 + b = p^2
\]
Thus, \( b = -pq \).

### Final remainder:
The remainder when the polynomial \( f(x) \) is divided by \( (x - p)(x - q) \) is:
\[
R(x) = (q + p)x - pq
\]

### Conclusion:
The remainder is \( (p + q)x - pq \).

---

Would you like further clarification on any part of the solution? Here are some related questions:

1. What is the Chinese Remainder Theorem in the context of polynomials?
2. How does the Remainder Theorem apply to polynomial division?
3. Why does the remainder of dividing by a quadratic take a linear form?
4. Can this method be extended to higher-degree polynomials?
5. How can we verify the solution using a specific example?

**Tip:** Always ensure the divisor is factored correctly when using polynomial division to avoid errors in calculating the remainder.

When a polynomial is divided by x - p, the remainder is p^2. When the same polynomial is divided by x - q, the remainder is q^2. What is the remainder when the polynomial is divided by x^2 - (p + q)x + pq (p ≠ q)?

This problem explores the division of a polynomial when the remainders are known for divisions by x - p and x - q. Learn how to find the remainder when the polynomial is divided by a quadratic expression x^2 - (p + q)x + pq, applying the Remainder Theorem and the Chinese Remainder Theorem. Suitable for Grades 10-12.

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