Lecture : Least Common Denominator
Just as with fractions, when we add or subtract rational expression, we must have a common denominator.
When we deal with rational expressions with monomial denominators, we find the common denominator for the coefficients, and for variables we use the highest exponent.
Here is an example:
5 |
+ |
7 |
In this problem we have monomial denominators, so we look at the coefficients, 6 and 8, a common denominator for these is 24. We then look at each variable individually. In this problem we have x3 and x2, since 3 is our largest exponent we will use x3 in our LCD. We also have y5 and y7, since 7 is our larger exponent we will use y7 in the LCD.
So our LCD in this problem would be 24x3y7
If our rational expressions have polynomial denominators, then to find the LCD we must first factor each denominator. After we factor the denominators, we get the LCD by writing each factor only once. The only time a factor will appear twice in the LCD is if it appears twice in a single denominator.
Here is an example:
3 |
- |
8 |
We first factor our denominators:
3 |
- |
8 |
Now we write down each factor once to get our LCD. LCD = (x + 2)(x + 6)(x - 3). Note we do not write (x + 6) twice because it does not appear in a single denominator twice. The only time a factor will appear twice in the LCD is if it appears twice in a single denominator.
Find the least common denominator for each of the following pairs of rational expressions.
1. |
2 |
- |
1 |
2. |
3x |
- |
8 |
3. |
x - 2 |
+ |
x - 2 |
2 |
- |
1 |
In this problem we have monomial denominators, so we look at the coefficients, 4 and 10, a common denominator for these is 20. We then look at each variable individually. In this problem we have x4 and x2, since 4 is our largest exponent we will use x4 in our LCD. We also have y3 and y5, since 5 is our larger exponent we will use y5 in the LCD.
So our LCD in this problem would be 20x4y5
3x |
- |
8 |
Since we have polynomial denominators, we first factor our denominators:
3x |
- |
8 |
Now we write down each factor once to get our LCD. LCD = (2x + 1)(x - 5)(2x + 1), or (2x + 1)2(x - 5).
Note we write (2x + 1) twice because it appears in a single denominator twice. The only time a factor will appear twice in the LCD is if it appears twice in a single denominator.
x - 2 |
+ |
x - 2 |
We first factor our denominators:
x - 2 |
+ |
x - 2 |
Now we write down each factor once to get our LCD. LCD = (x + 9)(x - 9)(x + 2).
Note we do not write (x - 9) twice because it does not appear in a single denominator twice. The only time a factor will appear twice in the LCD is if it appears twice in a single denominator.