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

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(B) The given expression is $\frac{x^3}{(2x-1)(x+2)(x-3)}$. Since the degree of the numerator is equal to the degree of the denominator,we perform pol

Vedclass · Accepted Answer

$4/25$

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(B) The given expression is $\frac{x^3}{(2x-1)(x+2)(x-3)}$. Since the degree of the numerator is equal to the degree of the denominator,we perform polynomial division first.\ The denominator is $(2x-1)(x^2-x-6) = 2x^3-2x^2-12x-x^2+x+6 = 2x^3-3x^2-11x+6$.\ Dividing $x^3$ by $2x^3-3x^2-11x+6$,we get $A = 1/2$ as the quotient.\ The expression becomes $\frac{1}{2} + \frac{\frac{3}{2}x^2 + \frac{11}{2}x - 3}{(2x-1)(x+2)(x-3)} = A + \frac{B}{2x-1} + \frac{C}{x+2} + \frac{D}{x-3}$.\ Using the cover-up rule or substitution:\ For $B$: Put $x = 1/2$,$B = \frac{\frac{3}{2}(1/4) + \frac{11}{2}(1/2) - 3}{(1/2+2)(1/2-3)} = \frac{3/8 + 11/4 - 3}{(5/2)(-5/2)} = \frac{(3+22-24)/8}{-25/4} = \frac{1/8}{-25/4} = -1/50$.\ For $C$: Put $x = -2$,$C = \frac{\frac{3}{2}(4) + \frac{11}{2}(-2) - 3}{(-4-1)(-2-3)} = \frac{6-11-3}{(-5)(-5)} = \frac{-8}{25}$.\ For $D$: Put $x = 3$,$D = \frac{\frac{3}{2}(9) + \frac{11}{2}(3) - 3}{(6-1)(3+2)} = \frac{27/2 + 33/2 - 3}{(5)(5)} = \frac{30-3}{25} = 27/25$.\ Finally,$A+B+C = 1/2 - 1/50 - 8/25 = \frac{25-1-16}{50} = \frac{8}{50} = 4/25$.

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

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