If the equivalent partial fraction of frac{x^ | vedclass

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(C) Given the expression: $\frac{x^3}{(2x-1)(x+2)(x-3)} = A + \frac{B}{2x-1} + \frac{C}{x+2} + \frac{D}{x-3}$.First,perform polynomial division since

Vedclass · Accepted Answer

$-\frac{8}{25}$

Vedclass · Answer

(C) Given the expression: $\frac{x^3}{(2x-1)(x+2)(x-3)} = A + \frac{B}{2x-1} + \frac{C}{x+2} + \frac{D}{x-3}$.First,perform polynomial division since the degree of the numerator is equal to the degree of the denominator. The denominator is $(2x-1)(x^2-x-6) = 2x^3 - 3x^2 - 11x + 6$. Dividing $x^3$ by $2x^3 - 3x^2 - 11x + 6$ gives $A = \frac{1}{2}$.Now,equate the remainder: $\frac{x^3 - \frac{1}{2}(2x^3 - 3x^2 - 11x + 6)}{(2x-1)(x+2)(x-3)} = \frac{\frac{3}{2}x^2 + \frac{11}{2}x - 3}{(2x-1)(x+2)(x-3)} = \frac{B}{2x-1} + \frac{C}{x+2} + \frac{D}{x-3}$.To find $C$,use the cover-up method by multiplying both sides by $(x+2)$ and setting $x = -2$:$C = \left[ \frac{\frac{3}{2}x^2 + \frac{11}{2}x - 3}{(2x-1)(x-3)} \right]_{x=-2}$.$C = \frac{\frac{3}{2}(4) + \frac{11}{2}(-2) - 3}{(2(-2)-1)(-2-3)} = \frac{6 - 11 - 3}{(-5)(-5)} = \frac{-8}{25}$.

If the equivalent partial fraction of $\frac{x^3}{(2x-1)(x+2)(x-3)}$ is given by $A+\frac{B}{2x-1}+\frac{C}{x+2}+\frac{D}{x-3}$,then the value of $C$ is

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