Question

$\frac{ {x}^{2} - 9 }{x + 3} \\ \\ \frac{ {x}^{3} + 4 {x}^{2} + 4x }{x + 2}$

Answer 1

$\frac{ {x}^{2} - 9 }{x + 3}=\frac{(x-3)(x+3)}{x + 3}=x-3$

$x \neq -3$

$\frac{ x^3 + 4 x^2 + 4x }{x + 2} =\frac{x( x^2 + 4 x + 4) }{x + 2} =\frac{x(x+2) ^2}{x + 2} =x(x+2)$

$x \neq -2$