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?
+23252 The common denominator is: (x + 5)(x-3)2
Multiplying A/(x + 5) by (x - 3)2 / (x - 3)2 gives a numerator of Ax2 - 6Ax + 9A
Multiplying B/(x - 3) by (x + 5)(x - 3) / (x + 5)(x - 3) gives a numerator of Bx2 + 2Bx - 15 B
Multiplying C/(x - 3)2 by (x + 5) / (x + 5) gives a numerator of Cx + 5C
Adding these numerators together, and rearranging, we have: (A + B)x2 + (-A + 3B + C)x + (9A - 15B + 5C)
Since the term on the left-side of the equation has no x2 term: A + B = 0.
Since the term on the left-side of the equation has no x term: -A + 3B + C = 0
Since the constant on the left-side equals 1: 9A - 15B + 5C = 1
Solving these three simulaneous equations, we get A = 1/64 B = -1/64 C = 1/8
(I used matrices)
+397 Add together the terms on the rhs.
\(\displaystyle \frac{A(x-3)^{2}+B(x+5)(x-3)+C(x+5)}{(x+5)(x-3)^{2}}.\)
This is to be identical with the fraction on the lhs. The denominators are identical and the numerators must also be identical.
That is,
\(\displaystyle A(x-3)^{2}+B(x+5)(x-3)+C(x+5)\equiv1\;\;\;\textbf{ for all values of x}.\)
\(\displaystyle \text{If }\quad x=3, \quad 8C=1, \quad \text{so}\quad C=1/8.\)
\(\displaystyle \text{If}\quad x=-5 \quad 64A=1, \quad \text{so}\quad A=1/64.\)
\(\displaystyle \text{If}\quad x = 0 \quad 9A-15B+5C=1, \quad \text{so}\quad B = -1/64.\)
