For each of the following,find a quadratic polynomial whose sum and product of the zeroes are as given respectively. Also,find the zeroes of these polynomials by factorisation.
Sum of zeroes $= -\frac{8}{3}$,Product of zeroes $= \frac{4}{3}$

(N/A) Given that,sum of zeroes $(S) = -\frac{8}{3}$ and product of zeroes $(P) = \frac{4}{3}$.
The general form of a quadratic polynomial is $f(x) = k(x^2 - Sx + P)$,where $k$ is a non-zero constant.
Substituting the values,we get $f(x) = x^2 - (-\frac{8}{3})x + \frac{4}{3} = x^2 + \frac{8}{3}x + \frac{4}{3}$.
To simplify,we can write the polynomial as $3x^2 + 8x + 4$.
To find the zeroes,we factorise $3x^2 + 8x + 4$:
$3x^2 + 6x + 2x + 4 = 3x(x + 2) + 2(x + 2) = (3x + 2)(x + 2)$.
Setting $f(x) = 0$,we get $(3x + 2)(x + 2) = 0$.
Therefore,$3x + 2 = 0 \implies x = -\frac{2}{3}$ and $x + 2 = 0 \implies x = -2$.
Hence,the zeroes of the polynomial are $-2$ and $-\frac{2}{3}$.