Question 247216
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Like this:


You write:  f(x) = (5x+6)/(2x-6)  [The parentheses are essential]


I read that as


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ f(x)\ =\ \frac{5x\ +\ 6}{2x\ -\ 6}]


The domain of a rational function is all real numbers EXCEPT any number that would make any denominator equal to 0.  So what number or numbers would make *[tex \LARGE 2x\ -\ 6\ = 0]?


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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{-7xy^2}{28x^2}]


There is one factor of -1 in the numerator and none in the denominator, so the -1 stays put.


There is one factor of 7 in the numerator and one factor of 7 in the denominator because 28 equals 7 times 4.  So the 7 that is common to both numerator and denominator goes away, leaving you with a 4 in the denominator.


There is one factor of *[tex \LARGE x] in the numerator and two factors of *[tex \LARGE x] in the denominator.  The one factor that is common goes away and you are left with one factor of *[tex \LARGE x] in the denominator.


There are 2 factors of *[tex \LARGE y] in the numerator and none in the denominator, so the 2 that are in the numerator stay put.


In summary, you have a factor of -1 and two factors of *[tex \LARGE y], (which is to say, one factor of *[tex \LARGE y^2]) in the numerator.  You have one factor of 4 and one factor of *[tex \LARGE x] in the denominator.  Hence:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{-7xy^2}{28x^2}\ =\ \frac{-y^2}{4x}]


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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{x\ -\ 2}{8x^2\ -\ 32}]


The two terms in the denominator expression have a factor of 8 in common, so:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{x\ -\ 2}{8\left(x^2\ -\ 4\right)}]


The binomial expression in the denominator is the difference of two squares.  Use the difference of two squares factorization:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ a^2\ -\ b^2\ =\ (a\ +\ b)(a\ -\ b)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{x\ -\ 2}{8\left(x\ +\ 2\right)\left(x\ -\ 2\right)}]


Now you can see that the numerator and denominator have a factor of *[tex \LARGE x\ -\ 2] in common.  Eliminate that factor from both numerator and denominator:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{1}{8\left(x\ +\ 2\right)}]





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