# When is it necessary to find the least common denominator (LCD) of two rational expressions and how do you find the LCD of two rational expressions. How is factoring related to this process? Give an example.

First of all, finding the LCD and factoring an expression are not really go-with-each-other processes though sometimes one would need to use both in order to simplify an expression.

It is necessary to look for the least common denominator when one is trying to add or subtract rational expressions that do not have the same denominator.

Take for example the following:

`1/(x+2) + 1/(x-2) = (x-2)/((x-2)(x+2)) + (x+2)/((x-2)(x+2)) = ((x-2) + (x+2))/((x-2)(x+2)) = (2x)/((x-2)(x+2)).`

In this case, the denominator of the two addends are not the same. One has (x+2), the other (x-2). Hence, we had to look for the least common denominator. The LCD, as the term implies, is the smallest number that is the multiple of all denominators involved. In this case, we simply multiply (x+2) and (x-2) to get (x+2)(x-2) (or the multiplied version: `x^2 - 4` ) as the LCD. [Note that least common denominator of fractions is simply the least common multiple of the denominators.]

Let's consider two numbers, 6 and 12. Simply multiplying 6 and 12 gives us 72. This number, 72, is indeed a common multiple of 6 and 12. Meaning that 72 would be a common denominator of `x/6` and `x/12` . However, notice that it is not the least common denominator.

Method 1:

To look for the LEAST common denominator of fractions (or the least common multiple of the denominators), we first have to list the multiples of each expression. The first number common to both is the LCM.

Method 2:

For the second method, we write numbers or expressions as products of primes (or in the case of rational expressions as products of non-factorable terms). For 6, we have `2*3` and for 12 we have `2^2*3` . Then, we look at each prime numbers and note the highest number of that specific primes. For 2, we have 1 for 6 and 2 for 12 so we take `2^2` for "two 2's". In the case of three we have 1 for 12 and 1 for 6 and so we simply have `3^1 = 1` for "one 3". Then, we multiply these numbers: `2^2*3 = 12` .

The first rational expression (at the start) has (x+2) as the denominator, it is not factorable already. The second term has (x - 2) as denominator, which is also already not factorable. Hence, to get the LCD, we simply multiply them.

Now, suppose we have the following:

`1/(x^2 - 6x + 9) - 1/(x-3).`

The denominator of the first term is `x^2 - 6x + 9`  - factorable to `(x-3)(x+3)` . The second term's denominator is (x-3). Hence, the LCD is (x-3)(x+3)

Last example:

`1/(x^2 + 2x + 1) + x/(x+1)`

The denominator of the first term is  (x+1)(x+1). The denominator of the second term is (x+1). Notice that we only have one "non-factorable expression" - (x+1). The first term has two, the second 1. We have `(x+1)^2 = x^2 + 2x + 1` as the LCD. Hence:

`1/(x+1)^2 + x/(x+1) = 1/(x+1)^2 + (x*(x+1))/(x+1)^2 = (x^2 + x + 1)/(x+1)^2.`

## Note: if you are not familiar with simplifying fractions, here is an illustration:

We simply transformed the fraction. We do this by dividing the LCD with the original denominator and then multiplying by the numerator. The answer would be the new numerator. As an illustration: the LCD is bd. The original denominator of the first term is b and the numerator is a. Hence, bd/b = d. Then, d*a = ad. Hence, ad is the numerator when the denominator is already the LCD. The same is true for the second term.

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