From question number 1 to 16, find zeroes of the given quadratic polynomials and verify the relation between zeroes and coefficients :
`3t^(2)+5t`

Explanation

To find the zeroes of the quadratic polynomial \(3t^2 + 5t\) and verify the relation between the zeroes and coefficients, we will follow these steps:

### Step 1: Write the polynomial in standard form
The given polynomial is:
\[ 3t^2 + 5t = 0 \]
This can be compared with the standard form of a quadratic equation, which is:
\[ at^2 + bt + c = 0 \]
Here, \(a = 3\), \(b = 5\), and \(c = 0\).

### Step 2: Identify the sum and product of the roots
According to Vieta's formulas:
- The sum of the roots (\(\alpha + \beta\)) is given by:
\[ -\frac{b}{a} = -\frac{5}{3} \]
- The product of the roots (\(\alpha \cdot \beta\)) is given by:
\[ \frac{c}{a} = \frac{0}{3} = 0 \]

### Step 3: Factor the polynomial
To find the roots, we can factor the polynomial:
\[ 3t^2 + 5t = t(3t + 5) = 0 \]
Setting each factor to zero gives us the roots.

### Step 4: Solve for the roots
1. From \(t = 0\):
\[ t = 0 \]
2. From \(3t + 5 = 0\):
\[ 3t = -5 \implies t = -\frac{5}{3} \]

Thus, the roots are:
\[ \alpha = 0, \quad \beta = -\frac{5}{3} \]

### Step 5: Verify the relations
1. **Sum of the roots**:
\[ \alpha + \beta = 0 + \left(-\frac{5}{3}\right) = -\frac{5}{3} \]
This matches our earlier calculation of \(-\frac{b}{a}\).

2. **Product of the roots**:
\[ \alpha \cdot \beta = 0 \cdot \left(-\frac{5}{3}\right) = 0 \]
This matches our earlier calculation of \(\frac{c}{a}\).

### Conclusion
Both relations between the zeroes and coefficients have been verified:
- Sum of roots: \(-\frac{5}{3}\)
- Product of roots: \(0\)