Find zeroes of the polynomial `5t^(2)+12x+7` by factorisation mothod and verify the relation between zeroes and coefficients of the polynomial.

Explanation

To find the zeroes of the polynomial \(5t^2 + 12t + 7\) using the factorization method, we will follow these steps:

### Step 1: Write the polynomial in standard form
The given polynomial is already in standard form:
\[ 5t^2 + 12t + 7 = 0 \]

### Step 2: Factor the polynomial
We need to factor the polynomial \(5t^2 + 12t + 7\). We will look for two numbers that multiply to \(5 \times 7 = 35\) (the product of the coefficient of \(t^2\) and the constant term) and add up to \(12\) (the coefficient of \(t\)).

The two numbers that satisfy these conditions are \(5\) and \(7\).

### Step 3: Rewrite the middle term
We can rewrite the polynomial by splitting the middle term using the two numbers found:
\[ 5t^2 + 5t + 7t + 7 = 0 \]

### Step 4: Group the terms
Now, we can group the terms:
\[ (5t^2 + 5t) + (7t + 7) = 0 \]

### Step 5: Factor by grouping
Now, factor out the common factors from each group:
\[ 5t(t + 1) + 7(t + 1) = 0 \]

### Step 6: Factor out the common binomial
Now, we can factor out the common binomial \((t + 1)\):
\[ (5t + 7)(t + 1) = 0 \]

### Step 7: Set each factor to zero
Now, we set each factor to zero to find the zeroes of the polynomial:
1. \(5t + 7 = 0\)
2. \(t + 1 = 0\)

### Step 8: Solve for \(t\)
From \(5t + 7 = 0\):
\[ 5t = -7 \implies t = -\frac{7}{5} \]

From \(t + 1 = 0\):
\[ t = -1 \]

### Step 9: List the zeroes
The zeroes of the polynomial \(5t^2 + 12t + 7\) are:
\[ t_1 = -\frac{7}{5}, \quad t_2 = -1 \]

### Step 10: Verify the relation between zeroes and coefficients
The sum of the zeroes \(t_1 + t_2\) is:
\[ -\frac{7}{5} + (-1) = -\frac{7}{5} - \frac{5}{5} = -\frac{12}{5} \]
According to Vieta's formulas, the sum of the zeroes is given by \(-\frac{b}{a}\):
\[ -\frac{12}{5} = -\frac{12}{5} \quad \text{(where \(b = 12\) and \(a = 5\))} \]

The product of the zeroes \(t_1 \cdot t_2\) is:
\[ -\frac{7}{5} \cdot (-1) = \frac{7}{5} \]
According to Vieta's formulas, the product of the zeroes is given by \(\frac{c}{a}\):
\[ \frac{7}{5} = \frac{7}{5} \quad \text{(where \(c = 7\) and \(a = 5\))} \]

Thus, we have verified the relation between the zeroes and the coefficients of the polynomial.