Let \(f(x) = \frac{14 - 3x + 2x^2}{(2 + x)(3 + x^2)}\).
(a) Express \(f(x)\) in partial fractions.
(b) Hence obtain the expansion of \(f(x)\) in ascending powers of \(x\), up to and including the term in \(x^2\).
Solution
(a) Express \(f(x)\) in the form \(\frac{A}{2+x} + \frac{B+Cx}{3+x^2}\).
Equating coefficients, solve for \(A, B,\) and \(C\):
\(A = 4, \; B = 1, \; C = -2\).
(b) Expand \(\left(1 + \frac{1}{2}x\right)^{-1}\) and \(\left(1 + \frac{1}{3}x^2\right)^{-1}\) up to \(x^2\):
\(\left(1 + \frac{1}{2}x\right)^{-1} = 1 - \frac{1}{2}x + \left(\frac{1}{2}x\right)^2\)
\(\left(1 + \frac{1}{3}x^2\right)^{-1} = 1 - \frac{1}{3}x^2\)
Substitute back into the partial fractions and simplify:
\(2\left(1 - \frac{1}{2}x + \left(\frac{1}{2}x\right)^2\right) + \frac{1}{3}(1 - 2x)\left(1 - \frac{1}{3}x^2\right)\)
Multiply out and collect terms up to \(x^2\):
\(\frac{7}{3} - \frac{5}{3}x + \frac{7}{18}x^2\)
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