Question 246820
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Start with *[tex \LARGE x^5\ -\ 4x^3]:


Take out a *[tex \LARGE x^3], leaving you with *[tex \LARGE x^3(x^2\ -\ 4)], then factor the difference of two squares, leaving you with *[tex \LARGE x^3(x\ -\ 2)(x\ +\ 2)]


Now look at *[tex \LARGE x^3\ +\ 4x^2\ +\ 4x ]:


Take out an *[tex \LARGE x], leaving you with *[tex \LARGE x(x^2\ +\ 4x\ +\ 4)] and then factor the trinomial leaving you with *[tex \LARGE x(x\ +\ 2)(x\ +\ 2)]


The least common multiple requires:


3 factors of *[tex \LARGE x]  (because the factor that has the most factors of *[tex \LARGE x] has 3 -- one has three and the other has one)


2 factors of *[tex \LARGE x\ +\ 2]  (because the factor that has the most factors of *[tex \LARGE x\ +\ 2] has 2 -- one has one and the other has two)


and


1 factor of *[tex \LARGE x\ -\ 2]  (because the factor that has the most factors of *[tex \LARGE x\ -\ 2] has 1 -- one has one and the other has none)


Hence the LCM is:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^3(x\ +\ 2)^2(x\ -\ 2)]


Once you multiply it out it will be a very handsome little 6th degree polynomial.  Well, handsome being a relative term anyway.  It will certainly be prettier than the 8th degree horror you would have had if you had simply multiplied *[tex \LARGE x^5\ -\ 4x^3] by *[tex \LARGE x^3\ +\ 4x^2\ +\ 4x]


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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 7\ -\ x\ \neq\ x\ -\ 7]


Multiply *[tex \LARGE 7\ -\ x] by -1:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ -1(7\ -\ x\)\ =\ (-1)\,\cdot\,7\ -\ (-1)\,\cdot\,x\ =\ -7\ -\ (-x)\ =\ -7\ +\ x\ =\ x\ -\ 7]


See?  In general, *[tex \LARGE a\ -\ b\ =\ -(b\ -\ a)]


The only specific case where *[tex \LARGE a\ -\ b\ =\ b\ -\ a] is where *[tex \LARGE a\ =\ b].


So, for your problem:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{x^2}{x\ -\ 7}\ +\ \frac{49}{7\ -\ x}]


Multiply either of the fractions by -1 over -1:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{x^2}{x\ -\ 7}\ +\ \frac{49}{7\ -\ x}\left(\frac{-1}{-1}\right)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{x^2}{x\ -\ 7}\ -\ \frac{49}{x\ -\ 7}]


Now you have the denominators equal, so:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{x^2\ -\ 49}{x\ -\ 7}]


Now factor the difference of two squares numerator:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{(x\ +\ 7)(x\ -\ 7)}{x\ -\ 7}]


And then eliminate the common factor from the numerator and denominator by applying the multiplicative identity:  *[tex \Large \frac{a}{a}\ =\ 1]. (Please eliminate the word "cancel" from your mathematical vocabulary.  The misuse of that idea has screwed up more math students than could ever be counted)


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ +\ 7]


But remember to specify that


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ \neq\ 7]


because, while 14 is a perfectly good value for *[tex \LARGE x\ +\ 7], *[tex \LARGE x] taking on a value of 7 would cause you to have a zero denominator in both of your original fractions.  If you are not yet comfortable with this idea, get a graphing calculator or graphing program for your computer, graph the function:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ f(x)\ =\ \frac{x^2\ -\ 49}{x\ -\ 7}]


And see what the calculator or computer tells you when you try to evaluate


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ f(7)]


[TILT!!]


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