+0

# question on algebra

-2
26
1
+1347

How do I solve this?

Let $A$, $B$, and $C$ be constants so that
$\frac{x^2-21x-3}{x^3-x^2-21x+45}=\frac{A}{x+5}+\frac{B}{x-3} + \frac{C}{(x - 3)^2}$
holds for all real numbers $x$ other than $-5$ and $3.$  What is $A$?

Aug 17, 2023

#1
+128732
+1

Note that     x^3 - x^2 -21x + 45 =  (x + 5) (x - 3) (x - 3)^2

Multiply through by this factorization and we  have

x^2 - 21x -3 = A(x - 3)^2  + B( x + 5) ( x-3) + C( x + 5)

x^2 -21x - 3 =  A( x^2 - 6x + 9) + B( x^2 + 2x -15) + C(x + 5)

Equate coefficients on each  side

1 =  A + B    →    B =  1 - A      (1)

-21 = -6A + 2B + C     (2)

-3 = 9A -15B + 5C     (3)

Sub (1) into (2) , (3)

-21 = -6A + 2 ( 1 - A) + C     →  -21 = - 6A + 2 - 2A + C →         -8A + C  =  -23      (4)

-3  = 9A - 15 ( 1 - A) + 5C  →  -3  =  9A - 15 + 15A  + 5C   →      24A + 5C =   12      (5)

Multiply  (4)  through by (-5)    add to  (5)

40A - 5C = 115

24A + 5C =  12

____________

64 A      =  127

A = 127/64

Aug 17, 2023