$\displaystyle \frac{6}{2(x-1)(x+1)}+\frac{1}{6(x-1)^2}$
Could someone please show me in detail how you would go about finding the LCD for a problem such as this, with multiple variables and terms in the denominator.
Many Thanks
$\displaystyle \frac{6}{2(x-1)(x+1)}+\frac{1}{6(x-1)^2}$
Could someone please show me in detail how you would go about finding the LCD for a problem such as this, with multiple variables and terms in the denominator.
Many Thanks
The denominators are very nearly factored. You should be able to do it by inspection.
First) 2, (x-1), (x+1)
Second) 2, 3, (x-1), (x-1)
Just read off the LCD by meeting the needs of all denominators.
2, (x-1), (x+1), 3, (x-1) again. ==> 2*3*(x+1)*(x-1)^2
The LCD is $\displaystyle 6(x-1)^{2}(x+1)$
Because 2 is a factor of 6 and (x-1) is a factor of (x-1)^2.
Then, you have the x+1 hanging there.
Multiply the top and bottom of whatever you need to make the denominators the LCD
$\displaystyle \frac{3(x-1)}{3(x-1)}\cdot \frac{6}{2(x-1)(x+1)}+\frac{1}{6(x-1)^{2}}\cdot \frac{(x+1)}{(x+1)}$
$\displaystyle \frac{19x-17}{6(x+1)(x-1)^{2}}$
So in doing the prime factorization for this problem it would look something like this
6 = 3 * 2
2 = 2 * 1
(x-1)
(x+1)
$\displaystyle (x-1)^2$
therefore the LCD is $\displaystyle 6(x+1)(x-1)^2 $
My only problem with this is how does the $\displaystyle (x-1)^2$ cover the $\displaystyle (x-1)$ and allow it be dropped from the LCD?
Look at my response again. There is a reason why I wrote them in that order. I'll do it again witht he thought process outloud.
Just write down the first denominator
2, (x-1), (x+1)
Now the second, one piece at a time.
2 we have already. No need to add another.
3 is new. Add that.
(x-1) we have already. No need to add another.
(x-1) again. Whoops, there are two. We DO need another.
Second) 2, 3, (x-1), (x-1)
We did not DROP anything. We just met the needs of everyone. If you include one (x-1) for the first and two more from the second, that's a total of three and NO denominator has that many. We want the LEAST possible. All three is excessive.
This has been stated in some of your other posts, but here it is again.
Find all the factors of your denominators and put them in exponential form.
$\displaystyle 2^1\cdot2^1\cdot3^1\cdot(x-1)^1\cdot(x+1)^1\cdot(x-1)^2$
Now find all the different factors using the largest exponent.
LCD=$\displaystyle 2^1\cdot3^1\cdot(x+1)^1\cdot(x-1)^2=6\cdot(x+1)\cdot(x-1)^2$