If ` f(x) = 2x^4 - 13 x^2 + ax +b` is divisible by ` x^2 -3x +2 ` then `(a,b)=`

Options

A. `(-9 ,-2)`
B. `(6,4)`
C. `(9,2)`
D. `(2,9)`

Answer

To solve the problem, we need to find the values of \( a \) and \( b \) such that the polynomial \( f(x) = 2x^4 - 13x^2 + ax + b \) is divisible by \( x^2 - 3x + 2 \).

### Step-by-Step Solution:

1. **Identify the roots of the divisor**:
The polynomial \( x^2 - 3x + 2 \) can be factored as \( (x - 1)(x - 2) \). Thus, the roots are \( x = 1 \) and \( x = 2 \).

2. **Evaluate \( f(1) \)**:
Substitute \( x = 1 \) into \( f(x) \):
\[ f(1) = 2(1)^4 - 13(1)^2 + a(1) + b = 2 - 13 + a + b = a + b - 11 \]
Since \( f(x) \) is divisible by \( x^2 - 3x + 2 \), \( f(1) = 0 \):
\[ a + b - 11 = 0 \implies a + b = 11 \quad \text{(Equation 1)} \]

3. **Evaluate \( f(2) \)**:
Substitute \( x = 2 \) into \( f(x) \):
\[ f(2) = 2(2)^4 - 13(2)^2 + a(2) + b = 2(16) - 13(4) + 2a + b \]
Simplifying this gives:
\[ f(2) = 32 - 52 + 2a + b = 2a + b - 20 \]
Since \( f(x) \) is divisible by \( x^2 - 3x + 2 \), \( f(2) = 0 \):
\[ 2a + b - 20 = 0 \implies 2a + b = 20 \quad \text{(Equation 2)} \]

4. **Solve the system of equations**:
We have the following two equations:
- \( a + b = 11 \) (Equation 1)
- \( 2a + b = 20 \) (Equation 2)

To eliminate \( b \), subtract Equation 1 from Equation 2:
\[ (2a + b) - (a + b) = 20 - 11 \]
This simplifies to:
\[ a = 9 \]

5. **Substitute \( a \) back to find \( b \)**:
Substitute \( a = 9 \) into Equation 1:
\[ 9 + b = 11 \implies b = 11 - 9 = 2 \]

6. **Final result**:
The values of \( a \) and \( b \) are:
\[ (a, b) = (9, 2) \]

### Conclusion:
Thus, the final answer is \( (a, b) = (9, 2) \).