SOLUTION: Let f(x)=(2x^2-7x-1)/(x-2)(x^2+3). (i)Express f(x) in partial fractions. (ii)Hence obtain the expression of f(x) in ascending powers of x, up to and including the term x^2. Ple

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Question 854282: Let f(x)=(2x^2-7x-1)/(x-2)(x^2+3).
(i)Express f(x) in partial fractions.
(ii)Hence obtain the expression of f(x) in ascending powers of x, up to and including the term x^2.
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Found 2 solutions by rothauserc, Vladdroid:
Answer by rothauserc(4718) (Show Source):
You can put this solution on YOUR website!
we are given
(2x^2-7x-1)/(x-2)(x^2+3)
i) we can write
(2x^2-7x-1)/(x-2)(x^2+3) = A / (x-2) + (Bx+C) / (x^2+3)
multiply through by common denominator (x-2)(x^2+3)
2x^2-7x-1 = A(x^2+3) + (Bx+C)(x-2)
2x^2-7x-1 = Ax^2 +A3 + Bx^2 + (Cx-B2x)+ A3-2C
group the x terms and the constant terms
2x^2-7x-1 = (A+B)x^2 + (C-B2)x + (A3-2C)
the only 0 in the original denominator is x = 2
8 -14 -1 = (A+B)4 + (C-B2)2 + (A3-C2)
-7 = 4A+4B+2C-4B+A3-2C
-7 = 7A
A = -1
reconsider
2x^2-7x-1 = (A+B)x^2 + (C-B2)x + (A3-2C)
therefore
A+B = 2 and B=3
C-2B = -7
C = -1
then the decomposition is
(2x^2-7x-1)/(x-2)(x^2+3) = -1 / (x-2) + (3x-1) / (x^2+3)
ii) f(x) = -1*(x-2)^-1 + (3x-1)*(x^2+3)^-1

Answer by Vladdroid(91) (Show Source):
You can put this solution on YOUR website!
given:
(2x^2-7x-1)/(x-2)(x^2+3)
i) we can write
(2x^2-7x-1)/(x-2)(x^2+3) = A / (x-2) + (Bx+C) / (x^2+3)
multiply through by common denominator (x-2)(x^2+3)
2x^2-7x-1 = A(x^2+3) + (Bx+C)(x-2)
2x^2-7x-1 = Ax^2 +A3 + Bx^2 + (Cx-B2x)+ A3-2C
group the x terms and the constant terms
2x^2-7x-1 = (A+B)x^2 + (C-B2)x + (A3-2C)
the only 0 in the original denominator is x = 2
8 -14 -1 = (A+B)4 + (C-B2)2 + (A3-C2)
-7 = 4A+4B+2C-4B+A3-2C
-7 = 7A
A = -1
reconsider
2x^2-7x-1 = (A+B)x^2 + (C-B2)x + (A3-2C)
therefore
A+B = 2 and B=3
C-2B = -7
C = -1
then the decomposition is
(2x^2-7x-1)/(x-2)(x^2+3) = -1 / (x-2) + (3x-1) / (x^2+3)
ii) f(x) = -1*(x-2)^-1 + (3x-1)*(x^2+3)^-1