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Find the partial decomposition of x^3-4x^2+9x-5/(x^2-2x+3)^2

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Answer: To find the partial fraction decomposition of

f(x) = (x^3-4x^2+9x-5)/(x^2-2x+3)^2

We can start by factoring the denominator into irreducible quadratic factors:

x^2 - 2x + 3 = (x - 1 + 2i)(x - 1 - 2i)

where i is the imaginary unit.

Since the denominator has repeated irreducible quadratic factors, we can write the partial fraction decomposition as:

f(x) = A/(x - 1 + 2i) + B/(x - 1 - 2i) + C/(x - 1)^2

where A, B, and C are constants that we need to solve for.

To find A, we can multiply both sides by (x - 1 + 2i)^2 and then substitute x = 1 - 2i, which gives:

A = lim_(x -> 1 - 2i) [(x - 1 + 2i)^2 f(x)]

= (1 - 1 + 2i + 2i)^2 [(1 - 2i)^3 - 4(1 - 2i)^2 + 9(1 - 2i) - 5]/[(1 - 2i - 1 + 2i)^2(1 - 1 + 2i)^2]

= -2 - 3i

Similarly, to find B, we can multiply both sides by (x - 1 - 2i)^2 and then substitute x = 1 + 2i, which gives:

B = lim_(x -> 1 + 2i) [(x - 1 - 2i)^2 f(x)]

= (1 - 1 - 2i + 2i)^2 [(1 + 2i)^3 - 4(1 + 2i)^2 + 9(1 + 2i) - 5]/[(1 + 2i - 1 - 2i)^2(1 - 1 - 2i)^2]

= -2 + 3i

To find C, we can multiply both sides by (x - 1)^2 and then substitute x = 1, which gives:

C = lim_(x -> 1) [(x - 1)^2 f(x)]

= (1 - 1)^2 [(1)^3 - 4(1)^2 + 9(1) - 5]/[(1 - 1)^2(1 - 2i)^2(1 + 2i)^2]

= 1

Therefore, the partial fraction decomposition of f(x) is:

f(x) = (-2 - 3i)/(x - 1 + 2i) + (-2 + 3i)/(x - 1 - 2i) + 1/(x - 1)^2

Note that there are different ways to write the partial fraction decomposition, depending on how you choose to combine the complex constants.

User Matthew Kraus
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