$f(x)=x^{2}+x-p(p+1)$

adding and subtracting $\mathrm{px}$, we get

$f(x)=x^{2}+p x+x-p x-p(p+1)$

$=x^{2}+(p+1) x-p x-p(p+1)$

$=x[x+(p+1)]-p[x+(p+1)]$

$=[x+(p+1)](x-p)$

$f(x)=0$

$\Rightarrow[\mathrm{x}+(\mathrm{p}+1)](\mathrm{x}-\mathrm{p})=0$

$\Rightarrow[\mathrm{x}+(\mathrm{p}+1)]=0$ or $(\mathrm{x}-\mathrm{p})=0$

$\Rightarrow \mathrm{x}=-(\mathrm{p}+1)$ or $\mathrm{x}=\mathrm{p}$

So, the zeroes of $f(x)$ are $-(p+1)$ and $p$.