Find a cubic polynomial whose zeroes are `(1)/(2)`,1 and -1.

Explanation

To find a cubic polynomial whose zeroes are \( \frac{1}{2} \), \( 1 \), and \( -1 \), we can follow these steps:

### Step 1: Write the factors based on the zeroes
The zeroes of the polynomial are \( \frac{1}{2} \), \( 1 \), and \( -1 \). Therefore, the factors of the polynomial can be written as:
\[ \left(x - \frac{1}{2}\right), \quad (x - 1), \quad (x + 1) \]

### Step 2: Form the polynomial by multiplying the factors
The polynomial \( P(x) \) can be expressed as:
\[ P(x) = \left(x - \frac{1}{2}\right)(x - 1)(x + 1) \]

### Step 3: Simplify the factors
First, we can simplify \( (x - 1)(x + 1) \):
\[ (x - 1)(x + 1) = x^2 - 1 \]
Now, substitute this back into the polynomial:
\[ P(x) = \left(x - \frac{1}{2}\right)(x^2 - 1) \]

### Step 4: Distribute \( \left(x - \frac{1}{2}\right) \)
Now we will distribute \( \left(x - \frac{1}{2}\right) \) across \( (x^2 - 1) \):
\[ P(x) = x(x^2 - 1) - \frac{1}{2}(x^2 - 1) \]
This simplifies to:
\[ P(x) = x^3 - x - \frac{1}{2}x^2 + \frac{1}{2} \]

### Step 5: Combine like terms
Now, we combine the terms:
\[ P(x) = x^3 - \frac{1}{2}x^2 - x + \frac{1}{2} \]

### Step 6: Multiply by 2 to eliminate the fraction
To express the polynomial in standard form without fractions, multiply the entire polynomial by 2:
\[ P(x) = 2\left(x^3 - \frac{1}{2}x^2 - x + \frac{1}{2}\right) = 2x^3 - x^2 - 2x + 1 \]

### Final Result
Thus, the cubic polynomial whose zeroes are \( \frac{1}{2} \), \( 1 \), and \( -1 \) is:
\[ P(x) = 2x^3 - x^2 - 2x + 1 \]