Find zeroes of the polynomial `2t^(2)+3t` and verify the relation between zeroes and coefficients.

Explanation

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

### Step 1: Write the polynomial
We start with the polynomial:
\[ p(t) = 2t^2 + 3t \]

### Step 2: Factor the polynomial
To find the zeroes, we need to factor the polynomial. We can factor out \(t\) from the expression:
\[ p(t) = t(2t + 3) \]

### Step 3: Set the polynomial equal to zero
To find the zeroes, we set the polynomial equal to zero:
\[ t(2t + 3) = 0 \]

### Step 4: Solve for \(t\)
This equation gives us two factors that can be set to zero:
1. \(t = 0\)
2. \(2t + 3 = 0\)

For the second factor, we solve for \(t\):
\[ 2t + 3 = 0 \implies 2t = -3 \implies t = -\frac{3}{2} \]

Thus, the zeroes of the polynomial are:
\[ t = 0 \quad \text{and} \quad t = -\frac{3}{2} \]

### Step 5: Verify the relation between zeroes and coefficients
The general form of a quadratic polynomial is:
\[ ax^2 + bx + c \]
For our polynomial \(2t^2 + 3t + 0\), we have:
- \(a = 2\)
- \(b = 3\)
- \(c = 0\)

#### Sum of the zeroes
The sum of the zeroes \(t_1 + t_2\) should equal \(-\frac{b}{a}\):
\[ t_1 + t_2 = 0 + \left(-\frac{3}{2}\right) = -\frac{3}{2} \]
Calculating \(-\frac{b}{a}\):
\[ -\frac{b}{a} = -\frac{3}{2} \]
Both sides are equal, thus verifying the first relation.

#### Product of the zeroes
The product of the zeroes \(t_1 \cdot t_2\) should equal \(\frac{c}{a}\):
\[ t_1 \cdot t_2 = 0 \cdot \left(-\frac{3}{2}\right) = 0 \]
Calculating \(\frac{c}{a}\):
\[ \frac{c}{a} = \frac{0}{2} = 0 \]
Again, both sides are equal, thus verifying the second relation.

### Conclusion
The zeroes of the polynomial \(2t^2 + 3t\) are \(t = 0\) and \(t = -\frac{3}{2}\), and the relations between the zeroes and coefficients have been verified.

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