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

Options

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

Answer

To find the quadratic polynomial of degree 2 with the given sum and product of its zeros, we can follow these steps:

### Step 1: Understand the standard form of a quadratic polynomial
The standard form of a quadratic polynomial is given by:
\[ P(x) = Ax^2 + Bx + C \]
where \( A \), \( B \), and \( C \) are constants.

### Step 2: Use the relationships for the sum and product of zeros
For a quadratic polynomial \( P(x) = Ax^2 + Bx + C \):
- The sum of the zeros \( \alpha + \beta \) is given by:
\[ \alpha + \beta = -\frac{B}{A} \]
- The product of the zeros \( \alpha \beta \) is given by:
\[ \alpha \beta = \frac{C}{A} \]

### Step 3: Set up equations based on the given values
From the problem, we know:
- The sum of the zeros \( \alpha + \beta = -\frac{1}{2} \)
- The product of the zeros \( \alpha \beta = -3 \)

Using these values, we can set up the following equations:
1. From the sum:
\[ -\frac{B}{A} = -\frac{1}{2} \]
This simplifies to:
\[ B = \frac{A}{2} \]

2. From the product:
\[ \frac{C}{A} = -3 \]
This simplifies to:
\[ C = -3A \]

### Step 4: Substitute \( B \) and \( C \) into the polynomial
Now we substitute \( B \) and \( C \) back into the standard form of the polynomial:
\[ P(x) = Ax^2 + \left(\frac{A}{2}\right)x + (-3A) \]
This can be rewritten as:
\[ P(x) = Ax^2 + \frac{A}{2}x - 3A \]

### Step 5: Factor out \( A \)
We can factor out \( A \) from the polynomial:
\[ P(x) = A\left(x^2 + \frac{1}{2}x - 3\right) \]

### Step 6: Choose a value for \( A \)
To simplify, we can choose \( A = 2 \) (this is a common choice to eliminate fractions):
\[ P(x) = 2\left(x^2 + \frac{1}{2}x - 3\right) \]
This gives us:
\[ P(x) = 2x^2 + x - 6 \]

### Final Polynomial
Thus, the required quadratic polynomial is:
\[ P(x) = 2x^2 + x - 6 \]