find the value of the product of roots of this quadratic equation It is given that one of the roots of the quadratic equation : $x^2 + (p + 3)x - p^2 = 0$,
where $p$ is a constant, is negative of the other. The question is : find the value of the product of roots.

Without applying the factorization of a "difference of two squares" or Viete's relations, we can still use the information stated in the problem.  If we call the two roots of the quadratic equation $ \ r \ $ and $ \ -r \ \ , $ then we have
$$  r^2  \ +  \ (p + 3)·r  \ - \ p^2 \ \ = \ \ 0 $$
and $$ [-r]^2 \  +  \ (p + 3)·[-r]  \ - \ p^2 \ \ = \ \ r^2 \  -  \ (p + 3)·r  \ - \ p^2 \ \ = \ \ 0 \ \ . $$
This means that $ \ r^2  \ =  \ -(p + 3)·r  \ + \ p^2 \  = \ (p + 3)·r  \ + \ p^2 \ \Rightarrow \ 2·(p + 3)·r \ = \ 0 \ \ . $  So either $ \ r \ = \ 0 \ $ or $ \ p \ = \ -3 \ \ . $
But if $ \ r \ = \ 0 \ = \ -r \ \ , \  $  then $ \ 0^2  \ +  \ (p + 3)·0  \ - \ p^2 \ \ = \ \ 0  \ \ $  would require $ \ p \ = \ 0 \ \ , $ which would then make the quadratic equation $ \ x^2  \ +  \ 3·x  \   = \  0 \ \ . $  But that polynomial factors as $ \ x · (x +   3)  \   = \  0 \ \ , $  so we couldn't have both roots equal to zero.
Instead, it must be that $ \ p \ = \ -3 \ \ , $ making the equation $ \ x^2 \ + \ 0·x \ - \ (-3)^2 \ = \ x^2 \ - \ 9 \ = \ 0 \ \ , $ for which the roots are given by $ \ r^2 \ = \ 9 \ \Rightarrow \ r \ = \ +3 \ , \ -3 \ \ ; \ $ the product of the roots is thus $ \ -9 \ \ . $
Another way to arrive at this conclusion is that $ \ y \ = \ x^2  \ +  \ (p + 3)·x  \ - \ p^2  \ \ $  is the equation of an "upward-opening" parabola, for which we want the $ \ x-$intercepts to be   $ \ x \ = \ -r \ $ and $ \ x \ = \ r \ \ . $  Its axis of symmetry is located midway between these intercepts, so we have $ \ h \ = \ 0 \ $ in the "vertex form" of the parabola's equation, $ \ y \ = \ (x - 0)^2 \ - \ p^2 \ \ . $  (The vertex is definitely "below" the $ \ x-$axis at $ \ (0 \ , \ -p^2) \ \ , $ so we know these $ \ x-$intercepts exist.)  The equation of the parabola is therefore $ \ y \ = \ x^2 \ - \ p^2 \ \ , \ $ making $ \ p + 3 \ = \ 0 \ \ $ and the rest of the argument above follows.
