For `x^(2) + 2x + 5` to be a factor of `x^(4) + alphax^(2) + beta` the values of `alpha `and `beta` respectively are

Options

A. 2,5
B. 5,25
C. 6,25
D. 5,2

Answer

To find the values of \( \alpha \) and \( \beta \) such that \( x^2 + 2x + 5 \) is a factor of \( x^4 + \alpha x^2 + \beta \), we can use polynomial long division.

### Step-by-Step Solution:

1. **Set Up the Division**:
We need to divide \( x^4 + \alpha x^2 + \beta \) by \( x^2 + 2x + 5 \).

2. **First Division**:
Divide the leading term \( x^4 \) by the leading term \( x^2 \):
\[ \frac{x^4}{x^2} = x^2 \]
This means we will multiply \( x^2 \) by the entire divisor \( x^2 + 2x + 5 \).

3. **Multiply and Subtract**:
Multiply \( x^2 \) by \( x^2 + 2x + 5 \):
\[ x^2(x^2 + 2x + 5) = x^4 + 2x^3 + 5x^2 \]
Now, subtract this from \( x^4 + \alpha x^2 + \beta \):
\[ (x^4 + \alpha x^2 + \beta) - (x^4 + 2x^3 + 5x^2) = -2x^3 + (\alpha - 5)x^2 + \beta \]

4. **Second Division**:
Now, divide the leading term \( -2x^3 \) by the leading term \( x^2 \):
\[ \frac{-2x^3}{x^2} = -2x \]
Multiply \( -2x \) by \( x^2 + 2x + 5 \):
\[ -2x(x^2 + 2x + 5) = -2x^3 - 4x^2 - 10x \]
Subtract this from the previous result:
\[ (-2x^3 + (\alpha - 5)x^2 + \beta) - (-2x^3 - 4x^2 - 10x) = ((\alpha - 5) + 4)x^2 + (10)x + \beta \]
Simplifying gives:
\[ (\alpha - 1)x^2 + 10x + \beta \]

5. **Third Division**:
Now, divide the leading term \( (\alpha - 1)x^2 \) by the leading term \( x^2 \):
\[ \frac{(\alpha - 1)x^2}{x^2} = \alpha - 1 \]
Multiply \( \alpha - 1 \) by \( x^2 + 2x + 5 \):
\[ (\alpha - 1)(x^2 + 2x + 5) = (\alpha - 1)x^2 + 2(\alpha - 1)x + 5(\alpha - 1) \]
Subtract this from the previous result:
\[ ((\alpha - 1)x^2 + 10x + \beta) - ((\alpha - 1)x^2 + 2(\alpha - 1)x + 5(\alpha - 1)) = (10 - 2(\alpha - 1))x + (\beta - 5(\alpha - 1)) \]

6. **Set Remainder to Zero**:
For \( x^2 + 2x + 5 \) to be a factor, the remainder must be zero. Thus, we set:
\[ 10 - 2(\alpha - 1) = 0 \quad \text{and} \quad \beta - 5(\alpha - 1) = 0 \]

7. **Solve for \( \alpha \)**:
From the first equation:
\[ 10 - 2\alpha + 2 = 0 \implies 12 = 2\alpha \implies \alpha = 6 \]

8. **Solve for \( \beta \)**:
Substitute \( \alpha = 6 \) into the second equation:
\[ \beta - 5(6 - 1) = 0 \implies \beta - 25 = 0 \implies \beta = 25 \]

### Final Answer:
The values of \( \alpha \) and \( \beta \) are \( \alpha = 6 \) and \( \beta = 25 \).