# Help to simplify this complicated fraction

I require help to simplify this. I used the method to make the denominator a single fraction then multiply the top by the reciprocal but when it comes to cancelling, I'm not sure if i've done it right.

$\frac{\frac{1}{{x^2}+x+2}}{1+\frac{2}{x({x^2}+x+2)}}$

• I got the same asnwer as you Niki. It was just the first time I attempted, I made a careless cancelling mistake. – ri83ve Mar 4 '17 at 14:27

Assuming $x \not= 0$ and $x^2+x+2 \not= 0$ (the latter is always the case): $$\frac{\frac{1}{{x^2}+x+2}}{1+\frac{2}{x({x^2}+x+2)}} \cdot \frac{x(x^2+x+2)}{x(x^2+x+2)} = \frac{x}{x(x^2+x+2) + 2}$$ It is not possible to simplify this further, if not by factorizing the denominator. Please keep in mind the conditions I've put in the beginning.
EDIT: factorization leads to the final answer: $$\frac{x}{(x+1)(x^2+2)}$$ Only for non-zero $x$; now that you've got an explicit form of the fraction, you can put existence conditions. In this case, $x \not= 0$ and $x \not= -1$.
$\frac{\frac{1}{{x^2}+x+2}}{1+\frac{2}{x({x^2}+x+2)}}=\frac{x}{x(x^2+x+2) + 2} =\frac{x}{(x+1)•(x^2+2)}$
where $x≠-1$