Question

Simplify........................​

Answer 1

Answer:

$\frac {x^{3}-4x^{2}-20x+32}{x^{3}+2x^{2}-4x-8}$

Step-by-step explanation:

make the denominator same

$\frac{(x^{2}-4)(x-2)}{(x^{2}-4)(x+2) }-\frac{(2x+20)(x+2)}{(x^{2}-4)(x+2)}$

then expand

$\frac {x^{3}-2x^{2}-4x+8}{x^{3}+2x^{2}-4x-8} - \frac {2x^{2}+24x+40}{x^{3}+2x^{2}-4x-8}$

join

$\frac {x^{3}-4x^{2}-20x+32}{x^{3}+2x^{2}-4x-8}$

i think but not sure

Answer 2

Answer:

$\frac{x - 2}{x + 2} - \frac{2x + 20}{ {x}^{2} - {2}^{2} } \\ \frac{x - 2}{x + 2} - \frac{2x + 20}{(x + 2)(x - 2)} \\ = \frac{ {(x - 2)}^{2} - (2x + 20)}{(x + 2)(x - 2)} \\ = \frac{ {x}^{2} - 4x + 4 - 2x - 20 }{(x - 2)(x + 2)} \\ = \frac{ {x}^{2} - 6x - 16 }{(x2)(x + 2)} \\ = \frac{(x - 8)(x + 2)}{(x + 2)(x - 2)} \\ = \frac{(x - 8)}{(x - 2)}$