SOLUTION: In class, we derived that \frac{1}{n(n + 1)} = \frac{1}{n} - \frac{1}{n + 1}. Fill in the blanks to make a true equation: \frac{5x}{(x - 1)(x^2 + 2)(x + 7)^3)} = \frac{A}{x

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: In class, we derived that \frac{1}{n(n + 1)} = \frac{1}{n} - \frac{1}{n + 1}. Fill in the blanks to make a true equation: \frac{5x}{(x - 1)(x^2 + 2)(x + 7)^3)} = \frac{A}{x       Log On


   

Question 1209392: In class, we derived that
\frac{1}{n(n + 1)} = \frac{1}{n} - \frac{1}{n + 1}.
Fill in the blanks to make a true equation:
\frac{5x}{(x - 1)(x^2 + 2)(x + 7)^3)} = \frac{A}{x - 1} + \frac{Bx + C}{x^2 + 2} + \frac{D}{x + 7} + \frac{E}{(x + 7)^2} + \frac{F}{(x + 7)^3}
Answer by Edwin McCravy(20059) About Me  (Show Source):
You can put this solution on YOUR website!
Since there are no blanks, perhaps you meant to find values of A thru F that
would make this equation an identity.

%285x%5E%22%22%29%2F%28%28x+-+1%29%28x%5E2+%2B+2%29%28x+%2B+7%29%5E3%29%22%22=%22%22

This is a very complicated partial fraction problem.
Go to this site: 

https://www.symbolab.com/solver/partial-fractions-calculator

input this %285x%5E%22%22%29%2F%28%28x+-+1%29%28x%5E2+%2B+2%29%28x+%2B+7%29%5E3%29

and you'll get:



If your teacher expects you to do this by hand, then your teacher should
be fired.  This is NOT teaching mathematics!

This is extremely more complicated than the very simple problem of breaking
 
1%2F%28n%28n+%2B+1%29%29 

into partial fractions and getting:

1%2Fn+-+1%2F%28n+%2B+1%29

Edwin