Thread: Find the LCD of the Following Fractions

$\displaystyle \frac{2a}{(3a^2 +12a + 12)b}$ $\displaystyle \ \ \frac{-7b}{a(4b^2 - 8b + 4)}$ $\displaystyle \ \ \frac{3}{ab^3 + 2b^3}$

The answer given by the book: $\displaystyle 12ab^3(a+2)^2(b-1)^2 $

When factoring the denominators I got the following: (3a + 6)(a + 2)b, (2b - 2)(2b - 2)a, (a + 2)b^3 edit: whoops, I wrote the 6 as a b the first time.

I found the different factors with the highest powers to be: (3a + b)(a + 2)(2b - 2)(a)(b^3)

I don't understand how to get from the last step I performed to the answer given in the book. I suspect I'm doing something wrong. Any help would be greatly appreciated. Thank you.

Originally Posted by fogownz

$\displaystyle \frac{2a}{(3a^2 +12a + 12)b}$ $\displaystyle \ \ \frac{-7b}{a(4b^2 - 8b + 4)}$ $\displaystyle \ \ \frac{3}{ab^3 + 2b^3}$
The answer given by the book: $\displaystyle 12ab^3(a+2)^2(b-1)^2 $

Your factoring is off.

The first denominator is $\displaystyle 3(a+2)^2$.

The second denominator is $\displaystyle 4a(b-1)^2$.

The third denominator is $\displaystyle b^3(a+2)$.