In a problem, I noticed the author did this:

$$\frac{1}{(a+2)+(z-2)} = \frac{1}{(a+2)}\cdot \frac{1}{1+\frac{z-2}{a+2}}$$

What he is saying is to take the entire $(a+2)$ term and multiply it by $1$ and then also to $\frac{z-2}{a+2}$ in order to get the same thing as on the left. But growing up, I learned that whenever we needed to multiply $(a+b)\cdot (c+d)$ we needed to do $(a\cdot c+ a\cdot d +b\cdot c + b\cdot d)$. Why is it ok to do what the author did instead?

  • $\begingroup$ @Arashium No. When you look at the right side of the equation, you'll notice that what you need to do to return to the left side of the equation is multiply $1$ by $(a+2)$ and $\frac{z-2}{a+2}$ by $(a+2)$. Why is this allowed? $\endgroup$ Mar 13, 2016 at 2:25
  • $\begingroup$ the author has done factorization and you are talking about expansion. $\endgroup$
    – Arashium
    Mar 13, 2016 at 2:26

1 Answer 1


The author is simply factoring the denominator, and the fact that there are sums in the parentheses is not relevant. A simpler version of the same thing: $\dfrac{1}{\phantom{\big(}A+B\phantom{\big)}}=\dfrac{1}{A\left(1+\frac{B}{A}\right)}=\dfrac{1}{\phantom{\big(}A\phantom{\big(}}\cdot\dfrac{1}{\left(1+\frac{B}{A}\right)}$

  • $\begingroup$ Ah, I see. Thank you! $\endgroup$ Mar 13, 2016 at 2:27

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .