If `x^(2)-3x+2` is a factor of `x^(4)-px^(2)+q=0`, then p,q are

Options

A. 2,3
B. 4,5
C. 5,4
D. 0,0

Answer

To solve the problem, we need to determine the values of \( p \) and \( q \) given that \( x^2 - 3x + 2 \) is a factor of \( x^4 - px^2 + q = 0 \).

### Step-by-Step Solution:

1. **Identify the Roots of the Factor**:
The quadratic \( x^2 - 3x + 2 \) can be factored as:
\[ (x - 1)(x - 2) = 0 \]
Therefore, the roots are \( x = 1 \) and \( x = 2 \).

2. **Substitute the Roots into the Polynomial**:
Since \( x^2 - 3x + 2 \) is a factor of \( x^4 - px^2 + q \), both roots must satisfy the equation \( x^4 - px^2 + q = 0 \).

- **For \( x = 1 \)**:
\[ 1^4 - p(1^2) + q = 0 \implies 1 - p + q = 0 \implies q = p - 1 \quad \text{(Equation 1)} \]

- **For \( x = 2 \)**:
\[ 2^4 - p(2^2) + q = 0 \implies 16 - 4p + q = 0 \implies q = 4p - 16 \quad \text{(Equation 2)} \]

3. **Set the Two Equations for \( q \) Equal**:
From Equation 1 and Equation 2, we have:
\[ p - 1 = 4p - 16 \]
Rearranging gives:
\[ -1 + 16 = 4p - p \implies 15 = 3p \implies p = 5 \]

4. **Substitute \( p \) Back to Find \( q \)**:
Substitute \( p = 5 \) into Equation 1:
\[ q = 5 - 1 = 4 \]

5. **Final Values**:
Thus, we find:
\[ p = 5, \quad q = 4 \]

### Summary:
The values of \( p \) and \( q \) are:
\[ \boxed{(5, 4)} \]