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I realize this is basic, but this little doubt has been around with me for quite a while:

I have this:

$$\frac{(2n+3)n+1}{(2n+1)(2n+3)}$$

I need it to end up with this shape:

$$\frac{n+1}{2n+3}$$

At first, I thought "well, I simply remove the $(2n+3)$ from the numerator and from the denominator and done!", but I would like to know if there are any other steps in the middle I am "jumping" by doing so.

Basically, I would like to do it as "slowly" as possible (because my professor will want to during the tests...). Is there any small step I am missing?

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  • $\begingroup$ @WillHunting: Ah! I see! After seeing the answers below, I realized that.. you're right. Thank you :) $\endgroup$ – Zol Tun Kul Nov 27 '12 at 2:19
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$\quad\dfrac{(2n+3)(n+1)}{(2n+1)(2n+3)}\quad$ can be simplified to $\quad\dfrac{n+1}{2n+1}\quad$ by canceling out the common factor of

$(2n + 3)$ in the numerator and denominator.

But then you need to check for what happens at $\;n = -\dfrac{3}{2}$.

At $n = -\dfrac{3}{2}$, the original fraction is undefined (division by zero). You lose that information when simplifying.


For $\quad\dfrac{(2n+3)n+1}{(2n+1)(2n+3)},\quad$ multiply out the numerator: $(2n+3)n+1 = 2n^2 + 3n +1$. You can then factor the numerator to obtain $(2n+1)(n+1)$. This gives you:

$$\frac{(2n+1)(n+1)}{(2n+1)(2n+3)} = \frac{n+1}{2n+3}.$$

But in this case, you need to consider that $2n + 1 = 0 \rightarrow n = -\dfrac12$, where the original fraction is undefined.

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  • $\begingroup$ +1 for the $\frac{-3}{2}$ undefinition thing. Didn't think about it (not sure if it will really destroy my proof, but it is nice to keep in mind). $\endgroup$ – Zol Tun Kul Nov 27 '12 at 2:36
  • $\begingroup$ It is good to keep in mind! You need to rule out any values of $n$, in this case, for which the denominator evaluates to zero. $\endgroup$ – Namaste Nov 27 '12 at 2:38
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This will take you there $$(2n+3)n + 1 = 2n^2 + 3n + 1 = (2n +1)(n+1). $$

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