Let \(f(x)=t^{2}-15\)
Now, if we recall the identity
$$
\left(a^{2}-b^{2}\right)=(a-b)(a+b)
$$
Using this identity, we can write
$$
t^{2}-15=(t-\sqrt{15})(x+\sqrt{15})
$$
So, the value of \(t^{2}-15\) is zero when \(t=\sqrt{15}\) or \(t=-\sqrt{15}\)
Therefore, the zeroes of \(t^{2}-15\) are \(\sqrt{15}\) and \(-\sqrt{15}\).
Verification
Now,
Sum of zeroes \(=\alpha+\beta=\sqrt{15}+(-\sqrt{15})=0\) or
$$
=-\frac{\text { Coefficient of } t}{\text { Coefficient of } t^{2}}=-\frac{0}{1}=0
$$
Product of zeroes \(=\alpha \beta=(\sqrt{1} 5)(-\sqrt{15})=-15\) or
$$
=\frac{\text { Constant term }}{\text { Coefficient of } \mathrm{t}^{2}}=\frac{-15}{1}=-15
$$
So, the relationship between the zeroes and the coefficients is verified.
