Question 277257
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I think you are trying to say, at least in part, that the factors of *[tex \Large x\ +\ 4] are *[tex \Large x\ +\ 2] and *[tex \Large x\ +\ 2].  Tell you what -- I'll give you *[tex \Large x\ +\ 4] dollars and you give me *[tex \Large (x\ +\ 2)(x\ +\ 2)\ =\ x^2\ +\ 4x\ + 4] dollars in return so long as *[tex \Large x\ \geq\ 1].  So if *[tex \Large x] is 1, I give you 5 and you give me 9.  If *[tex \Large x] is  10, I give you 14 and you give me 144...


Ok, now that we have debunked that theory, let's look at what we have:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{\frac{x+4}{x-2}\ -\ \frac{x-3}{x+1}}{x\ +\ 1}]


The first thing we need to do is combine the fractions in the numerator.  Since there are no common factors between the two denominators, the LCD is simply the product of the two denominators, namely:  *[tex \Large (x\ -\ 2)(x\ +\ 1)].  Applying the LCD we get:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{\frac{(x+4)(x+1)-\left((x-3)(x-2)\right)}{(x-2)(x+1)}}{x\ +\ 1}]


Expand the three sets of binomials in the numerator using FOIL:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{\frac{(x^2\ +\ 5x\ +\ 4)-\left((x^2\ -\ 5x\ + 6)\right)}{(x^2\ -\ x\ -2)}}{x\ +\ 1}]


Remove the parentheses in the numerator of the numerator and collect like terms:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{\frac{x^2\ +\ 5x\ +\ 4\ -\ x^2\ +\ 5x\ -\ 6}{x^2\ -\ x\ -\ 2}}{x\ +\ 1}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{\frac{10x\ -\ 2}{x^2\ -\ x\ -\ 2}}{x\ +\ 1}]


Finally, take the *[tex \Large x\ +\ 1] denominator -- invert and multiply:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{10x\ -\ 2}{x^2\ -\ x\ -\ 2}\,\cdot\,\frac{1}{x\ +\ 1}]



*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{10x\ -\ 2}{x^3\ -\ 3x\ -\ 2}]


And that is the best you can do with this one.


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