Give the polynomial of degree 2 with sum and product of its zeros as `-1/2` and -3 respectively:

Options

A. `(x^2-x+6)=0`
B. `(2x^2+x-6)=0`
C. `(x^2+x-3)=0`
D. `(x^2-2x+3)=0`

Answer

To find a polynomial of degree 2 with the given sum and product of its zeros, we can use the standard form of a quadratic polynomial, which is:

\[ P(x) = x^2 - (sum \, of \, zeros) \cdot x + (product \, of \, zeros) \]

### Step 1: Identify the sum and product of the zeros
We are given:
- Sum of the zeros \( S = -\frac{1}{2} \)
- Product of the zeros \( P = -3 \)

### Step 2: Substitute the values into the polynomial formula
Using the formula for the polynomial, we substitute the values of the sum and product:

\[ P(x) = x^2 - \left(-\frac{1}{2}\right)x + (-3) \]

### Step 3: Simplify the polynomial
Now we simplify the expression:

\[ P(x) = x^2 + \frac{1}{2}x - 3 \]

### Step 4: Clear the fraction (optional)
To eliminate the fraction, we can multiply the entire polynomial by 2:

\[ P(x) = 2(x^2 + \frac{1}{2}x - 3) \]
\[ P(x) = 2x^2 + x - 6 \]

### Final Polynomial
Thus, the polynomial of degree 2 is:

\[ P(x) = 2x^2 + x - 6 \]