Find the zeros of the polynomial

Here, $p(x)=x^{2}+2 x-195$

Let $p(x)=0$

$=>\mathrm{x}^{2}+(15-13) \mathrm{x}-195=0$

$=>\mathrm{x}^{2}+15 \mathrm{x}-13 \mathrm{x}-195=0$

$=>\mathrm{x}(\mathrm{x}+15)-13(\mathrm{x}+15)=0$

$=>(\mathrm{x}+15)(\mathrm{x}-13)=0$

$=>\mathrm{x}=-15,13$

Hence, the zeroes are $-15$ and 13 .