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We are given the expansion of \((p + qx)^5\) where \(p\) is a positive constant and \(q\) is a negative constant. The coefficient of \(x^2\) is 1280, and the coefficient of \(x^4\) is 2560. We need to find the values of \(p\) and \(q\).

### Step 1: Apply the Binomial Theorem
The expansion of \((p + qx)^5\) is given by:
\[
(p + qx)^5 = \sum_{r=0}^{5} \binom{5}{r} p^{5-r} (qx)^r
\]
Expanding the terms relevant to \(x^2\) and \(x^4\):

- The coefficient of \(x^2\) comes from the term where \(r = 2\):
  \[
  \text{Coefficient of } x^2 = \binom{5}{2} p^{3} q^{2}
  \]
- The coefficient of \(x^4\) comes from the term where \(r = 4\):
  \[
  \text{Coefficient of } x^4 = \binom{5}{4} p^{1} q^{4}
  \]

### Step 2: Use the Coefficients Provided
Substituting the given coefficients:

1. For the \(x^2\) term:
   \[
   \binom{5}{2} p^3 q^2 = 1280
   \]
   Since \(\binom{5}{2} = 10\), we have:
   \[
   10 p^3 q^2 = 1280
   \]
   Simplifying, we get:
   \[
   p^3 q^2 = 128
   \quad \text{(Equation 1)}
   \]

2. For the \(x^4\) term:
   \[
   \binom{5}{4} p q^4 = 2560
   \]
   Since \(\binom{5}{4} = 5\), we have:
   \[
   5 p q^4 = 2560
   \]
   Simplifying, we get:
   \[
   p q^4 = 512
   \quad \text{(Equation 2)}
   \]

### Step 3: Solve the System of Equations
From Equation 2:
\[
p = \frac{512}{q^4}
\]
Substitute into Equation 1:
\[
\left(\frac{512}{q^4}\right)^3 q^2 = 128
\]
Simplifying:
\[
\frac{512^3}{q^{12}} q^2 = 128
\]
\[
\frac{512^3}{q^{10}} = 128
\]
\[
512^3 = 128 \cdot q^{10}
\]
Since \(512 = 2^9\) and \(128 = 2^7\), this simplifies to:
\[
(2^9)^3 = 2^7 \cdot q^{10}
\]
\[
2^{27} = 2^7 \cdot q^{10}
\]
\[
q^{10} = 2^{20}
\]
\[
q = \pm 2^2 = \pm 4
\]
Given \(q\) is negative:
\[
q = -4
\]

Substituting \(q = -4\) into Equation 2:
\[
p \cdot (-4)^4 = 512
\]
\[
p \cdot 256 = 512
\]
\[
p = 2
\]

### Final Answer:
\[
p = 2, \quad q = -4
\]

Would you like further details or have any questions?

Here are 5 related questions for further exploration:
1. How would the coefficients change if the power of the expansion was 6 instead of 5?
2. What happens to the values of \(p\) and \(q\) if both were negative?
3. How would you approach the problem if only the coefficient of \(x^3\) was given?
4. What is the significance of the binomial theorem in polynomial expansion?
5. Can you generalize this method for any power \(n\) and coefficients?

**Tip:** When solving for variables in binomial expansions, ensure you carefully match each term's coefficient with the corresponding expansion term to avoid errors.

In the expansion of (p + qx)^5, where p is a positive constant and q is a negative constant, the coefficient of x^2 is 1280 and the coefficient of x^4 is 2560. Hence find the values of p and q.

Learn how to find the values of p and q from the binomial expansion of (p + qx)^5 where p is positive and q is negative. The problem involves finding the coefficients of x^2 and x^4, leading to a solution using algebraic methods. Suitable for students in Grades 10-12.

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