When |x|

Question

(C) Using partial fraction decomposition,we write: $\frac{3x^2-5x+3}{(x-1)(2x+1)(x+3)} = \frac{A}{x-1} + \frac{B}{2x+1} + \frac{C}{x+3}$.Solving for c

Vedclass · Accepted Answer

$\frac{-722}{27}$

Vedclass · Answer

(C) Using partial fraction decomposition,we write: $\frac{3x^2-5x+3}{(x-1)(2x+1)(x+3)} = \frac{A}{x-1} + \frac{B}{2x+1} + \frac{C}{x+3}$.Solving for coefficients: $3x^2-5x+3 = A(2x+1)(x+3) + B(x-1)(x+3) + C(x-1)(2x+1)$.For $x=1$: $3-5+3 = A(3)(4) \implies 1 = 12A \implies A = \frac{1}{12}$.For $x=-1/2$: $3(1/4) + 5/2 + 3 = B(-3/2)(5/2) \implies 3/4 + 10/4 + 12/4 = -\frac{15}{4}B \implies 25 = -15B \implies B = -\frac{5}{3}$.For $x=-3$: $3(9) + 15 + 3 = C(-4)(-5) \implies 45 = 20C \implies C = \frac{9}{4}$.Thus,$f(x) = -\frac{1}{12}(1-x)^{-1} - \frac{5}{3}(1+2x)^{-1} + \frac{3}{4}(1+x/3)^{-1}$.The coefficient of $x^4$ is $-\frac{1}{12}(1)^4 - \frac{5}{3}(-2)^4 + \frac{3}{4}(-1/3)^4 = -\frac{1}{12} - \frac{5}{3}(16) + \frac{3}{4}(\frac{1}{81}) = -\frac{1}{12} - \frac{80}{3} + \frac{1}{108}$.$= \frac{-9 - 2880 + 1}{108} = -\frac{2888}{108} = -\frac{722}{27}$.

When $|x| < \frac{1}{2}$,the coefficient of $x^4$ in the expansion of $\frac{3x^2-5x+3}{(x-1)(2x+1)(x+3)}$ is

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