Find the polynomial whose sum and product of the zeros are `-1/2` and `1/2` respectively.

Answer

To find the polynomial whose sum and product of the zeros are \(-\frac{1}{2}\) and \(\frac{1}{2}\) respectively, we can follow these steps:

### Step 1: Identify the sum and product of the zeros
Given:
- Sum of the zeros, \( S = -\frac{1}{2} \)
- Product of the zeros, \( P = \frac{1}{2} \)

### Step 2: Use the relationships for a quadratic polynomial
For a quadratic polynomial of the form \( ax^2 + bx + c = 0 \):
- The sum of the zeros (\( \alpha + \beta \)) is given by \( -\frac{b}{a} \)
- The product of the zeros (\( \alpha \beta \)) is given by \( \frac{c}{a} \)

### Step 3: Set up equations based on the sum and product
From the sum of the zeros:
\[ -\frac{b}{a} = -\frac{1}{2} \implies b = \frac{a}{2} \]

From the product of the zeros:
\[ \frac{c}{a} = \frac{1}{2} \implies c = \frac{a}{2} \]

### Step 4: Substitute \( b \) and \( c \) into the polynomial
Now substituting \( b \) and \( c \) into the polynomial \( ax^2 + bx + c \):
\[ ax^2 + \left(\frac{a}{2}\right)x + \left(\frac{a}{2}\right) \]

### Step 5: Factor out \( a \)
Factoring out \( a \):
\[ a \left( x^2 + \frac{1}{2}x + \frac{1}{2} \right) \]

### Step 6: Choose a value for \( a \)
To simplify, we can choose \( a = 2 \):
\[ 2 \left( x^2 + \frac{1}{2}x + \frac{1}{2} \right) = 2x^2 + x + 1 \]

### Step 7: Write the polynomial
Thus, the polynomial is:
\[ 2x^2 + x + 1 \]

### Final Answer
The polynomial whose sum and product of the zeros are \(-\frac{1}{2}\) and \(\frac{1}{2}\) respectively is:
\[ 2x^2 + x + 1 \]

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