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

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(A) Given the expression: $\frac{x^3}{(2 x-1)(x+2)(x-3)} = A + \frac{B}{2 x-1} + \frac{C}{x+2} + \frac{D}{x-3}$ Multiply both sides by $(2x-1)(x+2)(x-

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$\frac{1}{2}$

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(A) Given the expression: $\frac{x^3}{(2 x-1)(x+2)(x-3)} = A + \frac{B}{2 x-1} + \frac{C}{x+2} + \frac{D}{x-3}$ Multiply both sides by $(2x-1)(x+2)(x-3)$: $x^3 = A(2 x-1)(x+2)(x-3) + B(x+2)(x-3) + C(x-3)(2 x-1) + D(2 x-1)(x+2)$ To find $A$,we observe the coefficient of $x^3$ on both sides. The denominator is $(2x-1)(x+2)(x-3) = (2x^2+3x-2)(x-3) = 2x^3 - 6x^2 + 3x^2 - 9x - 2x + 6 = 2x^3 - 3x^2 - 11x + 6$. Since the degree of the numerator and denominator is the same $(3)$,$A$ is the ratio of the leading coefficients: $A = \frac{1}{2}$ Alternatively,by substituting values: For $x=3$: $27 = D(5)(5) \Rightarrow D = \frac{27}{25}$ For $x=-2$: $-8 = C(-5)(-5) \Rightarrow C = -8/25$ For $x=1/2$: $1/8 = B(5/2)(-5/2) \Rightarrow B = -1/50$ For $x=0$: $0 = A(-1)(2)(-3) + B(2)(-3) + C(-3)(-1) + D(-1)(2)$ $0 = 6A - 6B + 3C - 2D$ $6A = 6(-1/50) - 3(-8/25) + 2(27/25) = -6/50 + 24/25 + 54/25 = -3/25 + 78/25 = 75/25 = 3$ $6A = 3 \Rightarrow A = 1/2$

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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