Find the zeroes of the quadratic polynomial $t^{2}-15$ and verify the relationship between the zeroes and the coefficients.

(N/A) Given polynomial: $p(t) = t^{2}-15$.
To find the zeroes,set $p(t) = 0$:
$t^{2}-15 = 0$
$t^{2} = 15$
$t = \pm \sqrt{15}$
So,the zeroes are $\alpha = \sqrt{15}$ and $\beta = -\sqrt{15}$.
Comparing $t^{2}-15$ with the standard form $at^{2}+bt+c$,we get $a=1, b=0, c=-15$.
Verification:
Sum of zeroes: $\alpha + \beta = \sqrt{15} + (-\sqrt{15}) = 0$.
From coefficients: $\frac{-b}{a} = \frac{-0}{1} = 0$.
Thus,$\alpha + \beta = \frac{-b}{a}$ is verified.
Product of zeroes: $\alpha \cdot \beta = (\sqrt{15})(-\sqrt{15}) = -15$.
From coefficients: $\frac{c}{a} = \frac{-15}{1} = -15$.
Thus,$\alpha \cdot \beta = \frac{c}{a}$ is verified.