Question 312238
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First factor the denominator.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 4n^2\ +\ 23n\ -6\ =\ (4n\ -\ 1)(n\ +\ 6)]


Since there are two linear factors, you only need two numerator variables:


You are looking for the values of *[tex \Large A] and *[tex \Large B] such that 


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{17n\ -\ 23}{4n^2\ +\ 23n\ -6}\ =\ \frac{A}{4n\ -\ 1}\ +\ \frac{B}{n\ +\ 6}]


Add the two terms in the right hand side using the LCD which is the original denominator:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{17n\ -\ 23}{4n^2\ +\ 23n\ -6}\ =\ \frac{A(n\ +\ 6)\ +\ B(4n\ -\ 1)}{4n^2\ +\ 23n\ -6}]


Collect like terms:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{17n\ -\ 23}{4n^2\ +\ 23n\ -6}\ =\ \frac{(A\ +\ 4B)n\ +\ (6A\ -\ B)}{4n^2\ +\ 23n\ -6}]


In order for the LHS to be equal to the RHS, the coefficients must be equal, so:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ A\ +\ 4B\ =\ 17]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 6A\ -\ B\ =\ -23]


Solve the system:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ -6A\ +\ -24B\ =\ -102]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 6A\ -\ B\ =\ -23]


===================

*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 0A\ -\ 25B\ =\ -125]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ B\ =\ 5]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ A\ =\ -3]


And then substitute:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{17n\ -\ 23}{4n^2\ +\ 23n\ -6}\ =\ \frac{-3}{4n\ -\ 1}\ +\ \frac{5}{n\ +\ 6}]





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