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# HELP! DUE In 1 HOUR!

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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?

Aug 17, 2022

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Multiplying the entire equation by $$x^3 - x^2 - 21x + 45$$$$1 = A(x-3)(x-3) + B(x+5)(x-3) + C(x-5)$$

Simplify the right-hand side: $$1 = A(x^2 - 6x + 9) + B(x^2 + 2x - 15) + C(x-5)$$

Equating terms to get the following system: $$Ax^2 + Bx^2 = 0$$

$$-6Ax + 2Bx + Cx = 0$$

$$9A - 15B - 5C = 1$$

Removing the x's from the system gives us a simpler system: $$A + B = 0$$

$$-6A + 2B + C = 0$$

$$9A - 15B - 5C = 1$$

Now, just solve for a.

:)

Aug 17, 2022