If frac{x^3}{(2 x-1)(x+2)(x-3)} = A+frac{B}{2 | vedclass

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

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$\frac{1}{2}$

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(A) Given the expression: $\frac{x^3}{(2 x-1)(x+2)(x-3)} = A+\frac{B}{2 x-1}+\frac{C}{x+2}+\frac{D}{x-3}$.First,expand the denominator: $(2x-1)(x+2)(x-3) = (2x-1)(x^2-x-6) = 2x^3 - 2x^2 - 12x - x^2 + x + 6 = 2x^3 - 3x^2 - 11x + 6$.Since the degree of the numerator $(3)$ is equal to the degree of the denominator $(3)$,we perform long division to find the constant $A$.Dividing $x^3$ by $2x^3 - 3x^2 - 11x + 6$:$\frac{x^3}{2x^3 - 3x^2 - 11x + 6} = \frac{1}{2} \left( \frac{2x^3}{2x^3 - 3x^2 - 11x + 6} \right) = \frac{1}{2} \left( \frac{2x^3 - 3x^2 - 11x + 6 + 3x^2 + 11x - 6}{2x^3 - 3x^2 - 11x + 6} \right) = \frac{1}{2} + \frac{\frac{3}{2}x^2 + \frac{11}{2}x - 3}{(2x-1)(x+2)(x-3)}$.Comparing this with the given form $A+\frac{B}{2 x-1}+\frac{C}{x+2}+\frac{D}{x-3}$,we get $A = \frac{1}{2}$.

If $\frac{x^3}{(2 x-1)(x+2)(x-3)} = A+\frac{B}{2 x-1}+\frac{C}{x+2}+\frac{D}{x-3}$,then $A$ is equal to

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