Find the quadratic polynomial whose zeroes are `-(1)/(3)` and `(2)/(5)`. Verify the relation between coefficients and zeroes of the polynomial.

Explanation

To find the quadratic polynomial whose zeroes are \(-\frac{1}{3}\) and \(\frac{2}{5}\), we can follow these steps:

### Step 1: Identify the zeroes
Let the zeroes of the polynomial be:
\[ \alpha = -\frac{1}{3}, \quad \beta = \frac{2}{5} \]

### Step 2: Use the formula for the quadratic polynomial
The quadratic polynomial can be expressed in terms of its zeroes as:
\[ p(x) = k(x - \alpha)(x - \beta) \]
where \(k\) is a constant. For simplicity, we can take \(k = 1\).

### Step 3: Substitute the zeroes into the polynomial
Substituting \(\alpha\) and \(\beta\):
\[ p(x) = (x - (-\frac{1}{3}))(x - \frac{2}{5}) = (x + \frac{1}{3})(x - \frac{2}{5}) \]

### Step 4: Expand the polynomial
Now we will expand the expression:
\[ p(x) = (x + \frac{1}{3})(x - \frac{2}{5}) \]
Using the distributive property (FOIL method):
\[ p(x) = x^2 - \frac{2}{5}x + \frac{1}{3}x - \frac{1}{15} \]

### Step 5: Combine like terms
Combine the \(x\) terms:
\[ p(x) = x^2 + \left(-\frac{2}{5} + \frac{1}{3}\right)x - \frac{1}{15} \]

### Step 6: Find a common denominator for the coefficients
To combine \(-\frac{2}{5}\) and \(\frac{1}{3}\), we find the least common multiple of 5 and 3, which is 15:
\[ -\frac{2}{5} = -\frac{6}{15}, \quad \frac{1}{3} = \frac{5}{15} \]
Thus,
\[ -\frac{2}{5} + \frac{1}{3} = -\frac{6}{15} + \frac{5}{15} = -\frac{1}{15} \]

### Step 7: Write the polynomial in standard form
Now we can write the polynomial as:
\[ p(x) = x^2 - \frac{1}{15}x - \frac{1}{15} \]

### Step 8: Clear the fractions
To eliminate the fractions, multiply the entire polynomial by 15:
\[ 15p(x) = 15x^2 - x - 1 \]
Thus, the quadratic polynomial is:
\[ p(x) = 15x^2 - x - 1 \]

### Step 9: Verify the relation between coefficients and zeroes
1. **Sum of the zeroes**:
\[ \alpha + \beta = -\frac{1}{3} + \frac{2}{5} \]
Finding a common denominator (15):
\[ -\frac{1}{3} = -\frac{5}{15}, \quad \frac{2}{5} = \frac{6}{15} \]
Thus,
\[ \alpha + \beta = -\frac{5}{15} + \frac{6}{15} = \frac{1}{15} \]

According to the relation:
\[ -\frac{b}{a} = -\frac{-1}{15} = \frac{1}{15} \]

2. **Product of the zeroes**:
\[ \alpha \beta = -\frac{1}{3} \cdot \frac{2}{5} = -\frac{2}{15} \]
According to the relation:
\[ \frac{c}{a} = \frac{-1}{15} \]

Both relations are verified.

### Final Answer:
The quadratic polynomial is:
\[ p(x) = 15x^2 - x - 1 \]