Question

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Answer 1

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

${x}^{3} + 2 {x}^{2} - 3x = 0$

$x({x}^{2} + 2 x - 3) = 0$

$x((x-1)(x+3)) = 0$

At x = 0, p(x) = 0.

At x = 1, p(x) = 0

At x = -3, p(x) = 0

So, roots are 0,1 and -3

Thus,
$\alpha \beta \gamma = (0)(1)(-3) = 0$