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# Help

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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?

Apr 9, 2020

#1
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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)

Apr 9, 2020
#2
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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,

$$\displaystyle A(x-3)^{2}+B(x+5)(x-3)+C(x+5)\equiv1\;\;\;\textbf{ for all values of x}.$$

$$\displaystyle \text{If }\quad x=3, \quad 8C=1, \quad \text{so}\quad C=1/8.$$

$$\displaystyle \text{If}\quad x=-5 \quad 64A=1, \quad \text{so}\quad A=1/64.$$

$$\displaystyle \text{If}\quad x = 0 \quad 9A-15B+5C=1, \quad \text{so}\quad B = -1/64.$$

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Apr 10, 2020