For each of the following find a quadratic polynomial whose sum and product respectively of the zeroes are as given .Also find the zeroes of these polynomials by factorisation.
`(-8)/(3),(4)/(3)`

Answer

To find a quadratic polynomial whose sum and product of the zeroes are given as \(-\frac{8}{3}\) and \(\frac{4}{3}\) respectively, we can follow these steps:

### Step 1: Write the general form of the quadratic polynomial
The general form of a quadratic polynomial based on the sum and product of its zeroes is given by:
\[ p(x) = x^2 - (sum \ of \ zeroes) \cdot x + (product \ of \ zeroes) \]

### Step 2: Substitute the values of the sum and product
Here, the sum of the zeroes is \(-\frac{8}{3}\) and the product of the zeroes is \(\frac{4}{3}\). Substituting these values into the polynomial gives:
\[ p(x) = x^2 - \left(-\frac{8}{3}\right)x + \frac{4}{3} \]
This simplifies to:
\[ p(x) = x^2 + \frac{8}{3}x + \frac{4}{3} \]

### Step 3: Eliminate the fraction by multiplying through by 3
To eliminate the fractions, we can multiply the entire polynomial by 3:
\[ p(x) = 3x^2 + 8x + 4 \]

### Step 4: Factor the polynomial
Now we need to factor the polynomial \(3x^2 + 8x + 4\). We look for two numbers that multiply to \(3 \cdot 4 = 12\) and add up to \(8\). The numbers \(6\) and \(2\) fit this requirement.

So, we can rewrite the middle term:
\[ 3x^2 + 6x + 2x + 4 \]

### Step 5: Group the terms
Now, we group the terms:
\[ (3x^2 + 6x) + (2x + 4) \]
Factoring out the common terms:
\[ 3x(x + 2) + 2(x + 2) \]

### Step 6: Factor out the common binomial
Now we can factor out the common binomial \((x + 2)\):
\[ p(x) = (3x + 2)(x + 2) \]

### Step 7: Find the zeroes of the polynomial
To find the zeroes, we set each factor equal to zero:
1. \(3x + 2 = 0\)
\[ 3x = -2 \implies x = -\frac{2}{3} \]
2. \(x + 2 = 0\)
\[ x = -2 \]

### Conclusion
The zeroes of the polynomial \(3x^2 + 8x + 4\) are:
\[ x = -\frac{2}{3} \quad \text{and} \quad x = -2 \]