If x+a is a factor of `2x^(2)+2ax+5x+10`, find the value of a.

Explanation

To find the value of \( a \) such that \( x + a \) is a factor of the polynomial \( 2x^2 + 2ax + 5x + 10 \), we can use polynomial long division or the Remainder Theorem. Here, we will use the Remainder Theorem, which states that if \( x + a \) is a factor of a polynomial, then substituting \( x = -a \) into the polynomial should yield a result of zero.

### Step-by-Step Solution:

1. **Substitute \( x = -a \) into the polynomial**:
\[ P(x) = 2x^2 + 2ax + 5x + 10 \]
Substitute \( x = -a \):
\[ P(-a) = 2(-a)^2 + 2a(-a) + 5(-a) + 10 \]

2. **Calculate \( P(-a) \)**:
\[ P(-a) = 2a^2 - 2a^2 - 5a + 10 \]
Simplifying this gives:
\[ P(-a) = 0 - 5a + 10 = -5a + 10 \]

3. **Set the expression equal to zero**:
Since \( P(-a) \) must equal zero for \( x + a \) to be a factor:
\[ -5a + 10 = 0 \]

4. **Solve for \( a \)**:
Rearranging the equation gives:
\[ -5a = -10 \]
Dividing both sides by -5:
\[ a = \frac{10}{5} = 2 \]

5. **Conclusion**:
The value of \( a \) is \( 2 \).