+171 Suppose \(\frac{1}{x^3-x^2-21x+45}=\frac{A}{x+5}+\frac{B}{x-3} + \frac{C}{(x - 3)^2}\)
where A, B, and C are real constants. What is A?
+23245 x3 - x2 - 21x + 45 can be factored into (x + 5)(x -3)2
so multiply each term by (x + 5)(x -3)2
You'll end with: 1 = A(x - 3)2 + B(x - 3)(x + 5) + C(x + 5)
1 = A(x2 - 6x + 9) + B(x2 + 2x - 15) + C(x - 5)
1 = Ax2 - 6Ax + 9A + Bx2 + 2Bx - 15B + Cx - 5C
0x2 + 0x + 1 = Ax2 + Bx2 - 6Ax + 2Bx + Cx + 9A - 15B - 5C
So: Ax2 + Bx2 = 0x2 ---> (A + B)x2 = 0x2 ---> A + B = 0
-6Ax + 2Bx + Cx = 0x ---> (-6A + 2B + C)x = 0x ---> -6A + 2B + C = 0
9A - 15B - 5C = 1
Now, solve these three equations simultaneously: A + B = 0
-6A + 2B + C = 0
9A - 15B - 5C = 1
+23245 To finish:
First, eliminate C from the last two equations:
-6A + 2B + C = 0 ---> x 5 ---> -30A + 10B + 5C = 0
9A - 15B - 5C = 1 ---> 9A - 15B - 5C = 1
(add down) -11A - 5B = 1
Combining this result with the first equation:
A + B = 0 ---> x 11 ---> 11A + 11B = 0
-11A - 5B = 1 ---> -11A - 5B = 1
(add down) 6B = 1 ---> B = 1/6
A + B = 0 ----> A + (1/6) = 0 ---> A = -1/6
-6A + 2B + C = 0 ---> -6(-1/6) + 2(1/6) + C = 0 ---> 1 + 1/3 + C = 0 ---> C = -4/3
