Find the zeroes of the following polynomial by the factorisation method and verify the relationship between the zeroes and the coefficients of the polynomial:
$5 t^{2}+12 t+7$

(A) Let $f(t) = 5 t^{2}+12 t+7$.
To find the zeroes,we factorise the polynomial by splitting the middle term:
$5 t^{2}+12 t+7 = 5 t^{2}+5 t+7 t+7$
$= 5 t(t+1)+7(t+1)$
$= (5 t+7)(t+1)$
The zeroes are found by setting $f(t) = 0$:
$5 t+7 = 0 \implies t = -7/5$
$t+1 = 0 \implies t = -1$
Thus,the zeroes are $\alpha = -7/5$ and $\beta = -1$.
Verification:
Sum of zeroes $= \alpha + \beta = -7/5 - 1 = -12/5$.
From the polynomial $5 t^{2}+12 t+7$,the coefficient of $t$ is $12$ and the coefficient of $t^{2}$ is $5$.
Sum of zeroes $= -(\text{coefficient of } t) / (\text{coefficient of } t^{2}) = -12/5$.
Product of zeroes $= \alpha \cdot \beta = (-7/5) \cdot (-1) = 7/5$.
From the polynomial,the constant term is $7$ and the coefficient of $t^{2}$ is $5$.
Product of zeroes $= (\text{constant term}) / (\text{coefficient of } t^{2}) = 7/5$.
Hence,the relationship is verified.