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

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(A) 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, expand the denominator:$(2x-1)(x+

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$\frac{1}{2}$

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(A) 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, 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 polynomial long division to find the constant $A$:$\frac{x^3}{2x^3 - 3x^2 - 11x + 6} = \frac{\frac{1}{2}(2x^3 - 3x^2 - 11x + 6) + (\frac{3}{2}x^2 + \frac{11}{2}x - 3)}{2x^3 - 3x^2 - 11x + 6}$.This simplifies to:$\frac{1}{2} + \frac{\frac{3}{2}x^2 + \frac{11}{2}x - 3}{(2x-1)(x+2)(x-3)}$.Comparing this with the given form $A + \frac{B}{2x-1} + \frac{C}{x+2} + \frac{D}{x-3}$, we identify $A = \frac{1}{2}$.

Element	Co-factor
$A$. $-1$	$(1)$ $-2$
$B$. $1$	$(2)$ $32$
$C$. $3$	$(3)$ $4$
$D$. $6$	$(4)$ $6$
	$(5)$ $-6$

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

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