The partial fractions of frac{3x^3 - 8x^2 + 1 | vedclass

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Question

(C) Let $\frac{3x^3 - 8x^2 + 10}{(x - 1)^4} = \frac{A}{x - 1} + \frac{B}{(x - 1)^2} + \frac{C}{(x - 1)^3} + \frac{D}{(x - 1)^4}$.Multiplying both side

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$\frac{3}{(x - 1)} + \frac{1}{(x - 1)^2} - \frac{7}{(x - 1)^3} + \frac{5}{(x - 1)^4}$

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(C) Let $\frac{3x^3 - 8x^2 + 10}{(x - 1)^4} = \frac{A}{x - 1} + \frac{B}{(x - 1)^2} + \frac{C}{(x - 1)^3} + \frac{D}{(x - 1)^4}$.Multiplying both sides by $(x - 1)^4$,we get:$3x^3 - 8x^2 + 10 = A(x - 1)^3 + B(x - 1)^2 + C(x - 1) + D$.Expanding the terms:$3x^3 - 8x^2 + 10 = A(x^3 - 3x^2 + 3x - 1) + B(x^2 - 2x + 1) + C(x - 1) + D$.Comparing coefficients of powers of $x$:Coefficient of $x^3$: $A = 3$.Coefficient of $x^2$: $-3A + B = -8$ $\Rightarrow -3(3) + B = -8$ $\Rightarrow B = 1$.Coefficient of $x$: $3A - 2B + C = 0$ $\Rightarrow 3(3) - 2(1) + C = 0$ $\Rightarrow 9 - 2 + C = 0$ $\Rightarrow C = -7$.Constant term: $-A + B - C + D = 10$ $\Rightarrow -3 + 1 - (-7) + D = 10$ $\Rightarrow 5 + D = 10$ $\Rightarrow D = 5$.Thus,the partial fractions are $\frac{3}{x - 1} + \frac{1}{(x - 1)^2} - \frac{7}{(x - 1)^3} + \frac{5}{(x - 1)^4}$.

The partial fractions of $\frac{3x^3 - 8x^2 + 10}{(x - 1)^4}$ are:

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