0
$\begingroup$

I am given the following sum:

$$\frac{x}{2x+1} + \frac{3}{x^2} + \frac{1}{x}$$

In order to add these fractions, I must find a common denominator. I have been taught to factor each denominator and then multiply each factor the greatest number of times that they occur. $$ (2x+1), x(x), x\\ LCD = (2x+1) \cdot x = 2x^2 + x $$ I have also been taught that you do not include anything that has been factored out (in this case, the second $x$ in $x(x)$). Thus, the greatest number of times $x$ occurs is once; similarly with $2x+1$, so the $LCD$ is the product of these two expressions.

But apparently this is not correct, and somehow, I have encountered multiple different solutions, including: $$ 2x^2 + x^3 \\ 2x^3 + x $$ What is correct? And please explain (simply) how I can calculate the $LCD$ for future problems like this.

|cite|improve this question
$\endgroup$
  • 1
    $\begingroup$ Welcome to MSE. The LCD is $x^2(2x+1)=2x^3+x^2$ $\endgroup$ – J. W. Tanner Mar 26 at 22:21
  • $\begingroup$ How do you calculate the LCD of 2 and 9? $\endgroup$ – dcolazin Mar 26 at 22:24
  • $\begingroup$ or of $2, 9, $ and $3$ (to be more analogous)? $\endgroup$ – J. W. Tanner Mar 26 at 22:30
  • $\begingroup$ "But apparently this is not correct, and somehow, I have encountered multiple different solutions, " Where did you encounter these? $\endgroup$ – fleablood Mar 26 at 22:49
  • $\begingroup$ " I have encountered multiple different solutions, including: " And I've encountered people saying the earth is flat and Donald Trump isn't scum. The LCD of $2x+1, x\cdot x, x$ is $x\cdot x \cdot(2x+1) = x^2(2x+1)$ exactly as you thought for exactly the reason you said it was. Don't listen to idiots. $\endgroup$ – fleablood Mar 26 at 23:07
0
$\begingroup$

Factoring each denominator and then multiplying each factor the greatest number of times that it occurs works. In this case, the factor $x$ occurs twice in the middle term, so the answer for least common denominator is $(2x+1)x^2=2x^3+x^2.$

The LCD has to be a multiple of all of the denominators, so it has to include all of the factors in the denominators for the greatest numbers of times they occur.

|cite|improve this answer
$\endgroup$
  • $\begingroup$ @Gerard Way: To get all the fractions to have the LCD, you multiply the numerators and denominators by whatever it takes to make each denominator into the LCD. When doing that, you multiply each numerator and denominator by all of the other denominators, leaving out factors of those that are already included. I wonder if that's where you got the idea "you do not include anything that has been factored out" $\endgroup$ – J. W. Tanner Mar 26 at 22:33

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.