$f(x)=x^{2}-3 x-m(m+3)$

adding and subtracting $\mathrm{mx}$,

$f(x)=x^{2}-m x-3 x+m x-m(m+3)$

$=x[x-(m+3)]+m[x-(m+3)]$

$=[x-(m+3)](x+m)$

$f(x)=0 \Rightarrow[x-(m+3)](x+m)=0$

$\Rightarrow[\mathrm{x}-(\mathrm{m}+3)]=0$ or $(\mathrm{x}+\mathrm{m})=0$

$\Rightarrow \mathrm{x}=\mathrm{m}+3$ or $\mathrm{x}=-\mathrm{m}$

So, the zeroes of $\mathrm{f}(\mathrm{x})$ are $-\mathrm{m}$ and $+3$.