Question

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

Answer 1

$\frac{x^3+x^2-9x-9}{x^4+4x^3+3x^2} = \frac{x^2(x+1)-9(x+1)}{(x^4+x^3)+(3x^3+3x^2)}= \frac{(x+1)(x^2-9)}{x^3(x+1)+3x^2(x+1)}= \frac{(x+1)(x^2-9)}{x^2(x+1)(x+3)} = \\ \\ = \frac{(x+1)(x+3)(x-3)}{x^2(x+1)(x+3)}= \frac{x-3}{x^2}~. \\ \\ \underline{Observatie}:~x \notin \{ -3~;~-1~;~0 \}$

Answer 2

$\frac{x^3+x^2-9x-9}{x^4+4x^3+3x^2}= \\\\ =\frac{x^2(x+1)-9(x+1)}{x^2(x^2+4x+3)}= \\\\ =\frac{(x+1)(x^2-9)}{x^2(x^2+x+3x+3)}= \\\\ =\frac{(x+1)(x+3)(x-3)}{x^2[x(x+1)+3(x+1)]}= \\\\ = \frac{(x+1)(x+3)(x-3)}{x^2(x+3)(x+1)}= \\\\ =\boxed{\frac{x-3}{x^2}}$