Are the following statements True or False Justify your answers.
If two of the zeroes of a cubic polynomial are zero then it does not have linear and constant terms.

Answer

To determine whether the statement "If two of the zeroes of a cubic polynomial are zero, then it does not have linear and constant terms" is true or false, we will analyze the situation step by step.

### Step-by-Step Solution:

1. **Understanding the Polynomial**:
A cubic polynomial can be expressed in the general form:
\[ P(x) = ax^3 + bx^2 + cx + d \]
where \( a, b, c, \) and \( d \) are constants, and \( a \neq 0 \).

2. **Identifying the Zeroes**:
Let’s assume the cubic polynomial has three zeroes: \( \alpha, \beta, \gamma \). According to the statement, two of these zeroes are zero. We can set:
\[ \alpha = 0, \quad \beta = 0, \quad \text{and let } \gamma = r \text{ (where } r \text{ is any real number)}. \]

3. **Forming the Polynomial**:
The polynomial can be expressed in terms of its zeroes:
\[ P(x) = a(x - \alpha)(x - \beta)(x - \gamma) \]
Substituting the values of \( \alpha \) and \( \beta \):
\[ P(x) = a(x - 0)(x - 0)(x - r) = a(x)(x)(x - r) = ax^2(x - r) \]

4. **Expanding the Polynomial**:
Now, we expand \( P(x) \):
\[ P(x) = ax^2(x - r) = ax^3 - arx^2 \]

5. **Identifying the Coefficients**:
From the expanded polynomial \( P(x) = ax^3 - arx^2 \), we can see that:
- The coefficient of \( x^3 \) is \( a \).
- The coefficient of \( x^2 \) is \( -ar \).
- The coefficient of \( x \) (linear term) is \( 0 \) (since there is no \( x \) term present).
- The constant term \( d \) is also \( 0 \) (since there is no constant term present).

6. **Conclusion**:
Since both the linear term and the constant term are zero, we conclude that the statement is **True**.

### Justification:
The statement is true because if two of the zeroes of a cubic polynomial are zero, the polynomial simplifies to a form that lacks both linear and constant terms.