If one of the zeroes of a quadratic polynomial of the form `x^2 + ax + b` is the negative of the other, then it

Options

A. has no linear term and the constant term is negative
B. has no linear term and the constant term is positive
C. can have a linear term but the constant term is negative
D. can have a linear term but the constant term is positive

Explanation

To solve the problem step by step, we need to analyze the given quadratic polynomial \( x^2 + ax + b \) and the condition that one of its zeroes (roots) is the negative of the other.

### Step-by-Step Solution:

1. **Identify the Roots**:
Let the roots of the polynomial be \( \alpha \) and \( \beta \). According to the problem, one root is the negative of the other. We can express this as:
\[ \alpha = -\beta \]

2. **Use the Sum of Roots Formula**:
For a quadratic polynomial \( x^2 + ax + b \), the sum of the roots is given by:
\[ \alpha + \beta = -\frac{a}{1} = -a \]
Substituting \( \alpha = -\beta \) into the sum of roots, we get:
\[ -\beta + \beta = -a \implies 0 = -a \]
This implies:
\[ a = 0 \]

3. **Use the Product of Roots Formula**:
The product of the roots for the polynomial is given by:
\[ \alpha \cdot \beta = \frac{b}{1} = b \]
Substituting \( \alpha = -\beta \) into the product of roots, we have:
\[ (-\beta) \cdot \beta = b \implies -\beta^2 = b \]
Rearranging gives us:
\[ \beta^2 = -b \]

4. **Analyze the Implications**:
Since \( \beta^2 \) is a square of a real number, it must be non-negative:
\[ \beta^2 \geq 0 \]
Therefore, for \( -b \) to be non-negative, we must have:
\[ -b \geq 0 \implies b \leq 0 \]

5. **Conclusion**:
From the above analysis, we conclude that:
- The coefficient \( a \) must be \( 0 \) (indicating there is no linear term).
- The constant term \( b \) must be less than or equal to \( 0 \) (indicating that it is either negative or zero).

Thus, the final conclusion is that the polynomial has no linear term and the constant term is negative or zero.