The zeros of the polynomial `p(x)=3x^(2)-1` are

Options

A. `(1)/(3)and3`
B. `(1)/(sqrt(3))andsqrt(3)`
C. `(-1)/(sqrt(3))andsqrt(3)`
D. `(1)/(sqrt(3))and(-1)/(sqrt(3))`

Answer

To find the zeros of the polynomial \( p(x) = 3x^2 - 1 \), we need to set the polynomial equal to zero and solve for \( x \).

### Step-by-Step Solution:

1. **Set the polynomial equal to zero:**
\[ 3x^2 - 1 = 0 \]

2. **Add 1 to both sides:**
\[ 3x^2 = 1 \]

3. **Divide both sides by 3:**
\[ x^2 = \frac{1}{3} \]

4. **Take the square root of both sides:**
\[ x = \pm \sqrt{\frac{1}{3}} \]

5. **Simplify the square root:**
\[ x = \pm \frac{\sqrt{1}}{\sqrt{3}} = \pm \frac{1}{\sqrt{3}} \]

6. **Rationalize the denominator:**
\[ x = \pm \frac{1}{\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}} = \pm \frac{\sqrt{3}}{3} \]

### Final Answer:
The zeros of the polynomial \( p(x) = 3x^2 - 1 \) are:
\[ x = \frac{\sqrt{3}}{3} \quad \text{and} \quad x = -\frac{\sqrt{3}}{3} \]