Find the zeroes of the quadratic polynomial `7y^(2)-(11)/(3)y-(2)/(3)` and verify the relation between the zeroes and the coefficients.

Answer

To find the zeroes of the quadratic polynomial \( 7y^2 - \frac{11}{3}y - \frac{2}{3} \) and verify the relation between the zeroes and the coefficients, we will follow these steps:

### Step 1: Rewrite the Polynomial
The polynomial can be rewritten as:
\[ 7y^2 - \frac{11}{3}y - \frac{2}{3} = 0 \]
To eliminate the fractions, we can multiply the entire equation by 3 (the least common multiple of the denominators):
\[ 3(7y^2) - 3\left(\frac{11}{3}y\right) - 3\left(\frac{2}{3}\right) = 0 \]
This simplifies to:
\[ 21y^2 - 11y - 2 = 0 \]

### Step 2: Factor the Quadratic Polynomial
Next, we need to factor the quadratic equation \( 21y^2 - 11y - 2 = 0 \). We look for two numbers that multiply to \( 21 \times -2 = -42 \) and add up to \( -11 \).

The two numbers that satisfy these conditions are \( -14 \) and \( 3 \). We can split the middle term:
\[ 21y^2 - 14y + 3y - 2 = 0 \]

### Step 3: Group the Terms
Now, we group the terms:
\[ (21y^2 - 14y) + (3y - 2) = 0 \]
Factoring out common terms gives us:
\[ 7y(3y - 2) + 1(3y - 2) = 0 \]

### Step 4: Factor Out the Common Binomial
Now we can factor out the common binomial \( (3y - 2) \):
\[ (3y - 2)(7y + 1) = 0 \]

### Step 5: Find the Zeroes
Setting each factor to zero gives us the zeroes:
1. \( 3y - 2 = 0 \) leads to \( y = \frac{2}{3} \)
2. \( 7y + 1 = 0 \) leads to \( y = -\frac{1}{7} \)

Thus, the zeroes of the polynomial are:
\[ y_1 = \frac{2}{3}, \quad y_2 = -\frac{1}{7} \]

### Step 6: Verify the Relation Between Zeroes and Coefficients
For a quadratic polynomial of the form \( ay^2 + by + c \):
- The sum of the roots \( y_1 + y_2 = -\frac{b}{a} \)
- The product of the roots \( y_1 \cdot y_2 = \frac{c}{a} \)

Here, \( a = 21 \), \( b = -11 \), and \( c = -2 \).

**Sum of the roots:**
\[ y_1 + y_2 = \frac{2}{3} - \frac{1}{7} \]
To add these, we find a common denominator (21):
\[ \frac{2}{3} = \frac{14}{21}, \quad -\frac{1}{7} = -\frac{3}{21} \]
Thus,
\[ y_1 + y_2 = \frac{14}{21} - \frac{3}{21} = \frac{11}{21} \]
Now, calculate \( -\frac{b}{a} \):
\[ -\frac{-11}{21} = \frac{11}{21} \]
So, the sum of the roots matches.

**Product of the roots:**
\[ y_1 \cdot y_2 = \frac{2}{3} \cdot -\frac{1}{7} = -\frac{2}{21} \]
Now, calculate \( \frac{c}{a} \):
\[ \frac{-2}{21} \]
So, the product of the roots also matches.

### Conclusion
The zeroes of the polynomial \( 7y^2 - \frac{11}{3}y - \frac{2}{3} \) are \( \frac{2}{3} \) and \( -\frac{1}{7} \), and we have verified that the relations between the zeroes and the coefficients hold true.