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The integers \(G\) and\(H\) are chosen such that \(\frac{G}{x+5}+\frac{H}{x^2-4x}=\frac{x^2-2x+10}{x^3+x^2-20x}\) for all real values of \(x\) except -5, 0, and 4. Find \(H/G\).

 Jun 29, 2018
 #1
avatar+101252 
+2

We can use partial fractions, here

 

Note that   x^3 + x^2 -20x   can be factored  as   ( x^2 - 4x) ( x + 5)

 

x^2  - 2x + 10                               G                    H  

______________        =          _____         +   ________

( x^2 - 4x)(x + 5)                        x + 5               x^2  - 4x

 

Multiply through  by  ( x + 5) ( x^2  - 4x)

 

 

x^2   - 2x + 10  =  G(x^2 - 4x)  +  H (x + 5)    simplify

 

x^2  - 2x + 10  =  Gx^2 - 4Gx  + Hx + 5H      equate coefficients

 

1 = G

-2 = H  - 4G

10  = 5H

 

It's obvious that  G  = 1   and  H  = 2

 

So

 

H / G   =   2 / 1    = 2

 

 

 

cool cool cool

 Jun 29, 2018

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