Question 244077
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Problem 1:  Factor the denominator in the right hand rational expression:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ -\ 12x\ +\ 36\ =\ (x\ -\ 6)^2]


So the left hand fraction has one factor of *[tex \LARGE x\ -\ 6] and the right hand fraction has 2 such factors.  The LCD is the product of all the factors with duplicates removed.  Hence, the LCD is just:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (x\ -\ 6)^2\ =\ x^2\ -\ 12x\ +\ 36]


If I had to continue to manipulate these rational expressions, I would leave the LCD expressed as the squared binomial.


Problem 2:


The first denominator has one factor of 7, one factor of *[tex \Large a], and five factors of *[tex \Large b]


The second denominator has one factor of 7, three factors of *[tex \Large a], and three factors of *[tex \Large b].


At most you have:  one factor of 7, three factors of *[tex \Large a], and five factors of *[tex \Large b].


Hence, your LCD is:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 7a^3b^5]



John
*[tex \LARGE e^{i\pi} + 1 = 0]
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