Partial fraction sums
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Partial-fraction is the process of starting with the simplified answer and taking it back apart, of decomposing the final expression into its initial polynomial fractions.
The numerator is of degree 5; the denominator is of degree 3. So first I have to do the long division:The long division rearranges the rational expression to give me:Now I can decompose the fractional part. The denominator factors as (x2 + 1)(x – 2).The x2 + 1 is irreducible, so the decomposition will be of the form:Multiplying out and solving, I get:2x2 + x + 5 = A(x2 + 1) + (Bx + C)(x – 2)
x = 2: 8 + 2 + 5 = A(5) + (2B + C)(0), 15 = 5A, and A = 3
x = 0: 0 + 0 + 5 = 3(1) + (0 + C)(0 – 2),
5 = 3 – 2C, 2 = –2C, and C = –1
x = 1: 2 + 1 + 5 = 3(1 + 1) + (1B – 1)(1 – 2),
8 = 6 + (B – 1)(–1) = 6 – B + 1,
8 = 7 – B, 1 = – B, and B = –1Then the complete expansion is:
x²+3/x-2+ -x-1/x²+1=x²+3/x-2-x+1/x²+1A
The numerator is of degree 5; the denominator is of degree 3. So first I have to do the long division:The long division rearranges the rational expression to give me:Now I can decompose the fractional part. The denominator factors as (x2 + 1)(x – 2).The x2 + 1 is irreducible, so the decomposition will be of the form:Multiplying out and solving, I get:2x2 + x + 5 = A(x2 + 1) + (Bx + C)(x – 2)
x = 2: 8 + 2 + 5 = A(5) + (2B + C)(0), 15 = 5A, and A = 3
x = 0: 0 + 0 + 5 = 3(1) + (0 + C)(0 – 2),
5 = 3 – 2C, 2 = –2C, and C = –1
x = 1: 2 + 1 + 5 = 3(1 + 1) + (1B – 1)(1 – 2),
8 = 6 + (B – 1)(–1) = 6 – B + 1,
8 = 7 – B, 1 = – B, and B = –1Then the complete expansion is:
x²+3/x-2+ -x-1/x²+1=x²+3/x-2-x+1/x²+1A
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