If frac{3x^3-7x+1}{(x-2)^5} = frac{A}{x-2} + | vedclass

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(A) Let $u = x-2$,so $x = u+2$. Substituting this into the numerator: $3(u+2)^3 - 7(u+2) + 1 = 3(u^3 + 6u^2 + 12u + 8) - 7u - 14 + 1 = 3u^3 + 18u^2 +

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(A) Let $u = x-2$,so $x = u+2$. Substituting this into the numerator: $3(u+2)^3 - 7(u+2) + 1 = 3(u^3 + 6u^2 + 12u + 8) - 7u - 14 + 1 = 3u^3 + 18u^2 + 36u + 24 - 7u - 13 = 3u^3 + 18u^2 + 29u + 11$. Dividing by $u^5$: $\frac{3u^3 + 18u^2 + 29u + 11}{u^5} = \frac{3}{u^2} + \frac{18}{u^3} + \frac{29}{u^4} + \frac{11}{u^5}$. Comparing this with the given form: $\frac{A}{u} + \frac{B}{u^2} + \frac{C}{u^3} + \frac{D}{u^4} + \frac{E}{u^5}$,we get $A = 0$,$B = 3$,$C = 18$,$D = 29$,$E = 11$. Therefore,$A(B+C+D+E) = 0(3+18+29+11) = 0$.

If $\frac{3x^3-7x+1}{(x-2)^5} = \frac{A}{x-2} + \frac{B}{(x-2)^2} + \frac{C}{(x-2)^3} + \frac{D}{(x-2)^4} + \frac{E}{(x-2)^5}$,then $A(B+C+D+E) =$ ?

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