How do you find the LCD for #2 /(2x+6)# , #15 /(2x^2 + 12x +18)#?

1 Answer
Mar 29, 2018

#=>2(x+3)^2#

Explanation:

You find the LCD for polynomial expressions the same way you would with just numbers.

For example, if we want to find the LCD for #1/3# and #1/6#, we look for the smallest multiple of #3# and #6# that is shared. In this case, it happens to be #6#. We can always find a common denominator by taking the product: #3xx6 = 18#, but this is not always the LCD. In this case, because #6# is a multiple of #3#, we can do better!

For polynomials, we apply the same logic. We try to find the smallest multiple of both polynomial denominators.

We are given two fractions which can be simplified first by factoring:

#2/(2x+6) = 2/(2(x+3)) = color(blue)(1/(x+3))#

#15/(2x^2+12x+18)=15/(2(x^2+6x+9))=color(blue)(15/(2(x+3)^2))#

We notice that the second fraction has an extra power of #(x+3)# in its denominator and an extra multiple of #2#. This second denominator is our least common denominator, because we already have #x+3# in the first fraction. There is nothing in the first fraction that isn't in the second.

To see that the second fraction denominator is the LCD, let's substitute #a equiv (x+3)#.

So we have these two fractions:

#1/a, 15/(2a^2)#

You can always find a common denominator by multiplying the two denominators. In this case we would get a common denominator of #2a^3#.

But like the example I gave first, this isn't necessarily the LCD. In this case (for the same reasons I showed with numbers), #a^2# is a multiple of #a#, so #a^2# is sufficient. Now, the only thing that differs between the fractions is a factor of #2# in the denominator, so we need to keep that as well. So we've determined that the LCD for these fractions is #2a^2#, which is precisely #color(green)(2(x+3)^2)#.