0
$\begingroup$

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.

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

$\endgroup$
  • $\begingroup$ I got the same asnwer as you Niki. It was just the first time I attempted, I made a careless cancelling mistake. $\endgroup$ – ri83ve Mar 4 '17 at 14:27
1
$\begingroup$

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 $x \not= -1$.

$\endgroup$
0
$\begingroup$

This seems like some homework question. Hint in the denominator expression make the denominator common and add; then remove the common factor of the top denominator and bottom denominator

Once you do that could easily simplify it.

$\endgroup$
0
$\begingroup$

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

where $x≠-1$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.