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\[\dfrac{x^2 - 19}{x^3 - 2x^2 -5x + 6} = \dfrac{A}{x-1} +  \dfrac{B}{x+2} + \dfrac{C}{x-3}\]

 

Find constants A, B, and C so that the above expression holds for all real numbers other than 1, -2, and 3.

 Jan 18, 2021
 #1
avatar+116126 
+1

Note  that    x^3  -2x^2  - 5x  + 6     factors as  ( x - 1)(x + 2)(x -3)

 

Multiplying through  by  this factorization  we  get that

 

x^2   -19   =  A( X + 2)(x - 3)    +  B(x - 1)(x - 3)  +  C( x -1)(x + 2)      simpliify

 

x^2 -19 =  A(x^2 - x - 6)   + B(x^2 - 4x + 3)   + C(x^2 + x - 2)

 

1x^2  + 0x  - 19 =   (A + B + C)x^2   -  (A + 4B  - C) x +  (-6A + 3B - 2C)

 

Equating  coefficients and  constant terms we have this system

 

A + B + C    = 1           (1)

-A - 4B + C  =   0         (2)

-6A + 3B - 2C  =  -19   (3)

 

Add (1) and (2)  and we get that

 

-3B + 2C  = 1        (4)

 

Multiply  (2) through by   -6

 

6A + 24B - 6C  =  0       add this to   (3)

 

27B  -8C  =  -19      (5)

 

Multiply  (4)   by  9

 

-27B + 18C    =  9     add this to (5)

 

10C  = -10

C  = -1

 

And 

 -3B + 2(-1)   =  1

-3B  = 3

B = -1

 

And 

 

A + B + C   =1

 

A - 1 - 1  =  1

A  -2  = 1

A  = 3

 

So    ({A, B ,C}  =  {  3,  -1,  -1  }   

 

 

cool cool cool

 Jan 18, 2021
 #2
avatar+186 
0

\(\displaystyle \frac{x^{2}-19}{x^{3}-2x^{2}-5x+6} \equiv \frac{A}{x-1}+\frac{B}{x+2}+\frac{C}{x-3} \\ \text{so} \\ x^{2}-19 \equiv A(x+2)(x-3)+B(x-3)(x-1)+C(x-1)(x+2).\)

 

\(\displaystyle \text{Let }x=1,\qquad -18=A(3)(-2), \qquad A=3. \\ \text{Let }x=-2, \quad \, -15=B(-5)(-3), \qquad B= -1.\\ \text{Let }x=3,\qquad-10=C(2)(5),\qquad C=-1.\)

.
 Jan 19, 2021

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