If frac{x^4}{(x-1)(x-2)(x-3)}=Ax+B cdot frac{ | vedclass

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(D) Given the expression $\frac{x^4}{(x-1)(x-2)(x-3)}$. First,perform polynomial division since the degree of the numerator is greater than the denomi

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$32$

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(D) Given the expression $\frac{x^4}{(x-1)(x-2)(x-3)}$. First,perform polynomial division since the degree of the numerator is greater than the denominator. $(x-1)(x-2)(x-3) = x^3 - 6x^2 + 11x - 6$. Dividing $x^4$ by $x^3 - 6x^2 + 11x - 6$ gives $x+6$ with a remainder of $25x^2 - 60x + 36$. So,$\frac{x^4}{(x-1)(x-2)(x-3)} = x+6 + \frac{25x^2 - 60x + 36}{(x-1)(x-2)(x-3)}$. Using partial fractions for the remainder: $\frac{25x^2 - 60x + 36}{(x-1)(x-2)(x-3)} = \frac{B}{x-1} + \frac{C}{x-2} + \frac{D}{x-3}$. For $x=1$: $B = \frac{25-60+36}{(1-2)(1-3)} = \frac{1}{2}$. For $x=2$: $C = \frac{25(4)-60(2)+36}{(2-1)(2-3)} = \frac{100-120+36}{-1} = -16$. For $x=3$: $D = \frac{25(9)-60(3)+36}{(3-1)(3-2)} = \frac{225-180+36}{2} = \frac{81}{2} = 40.5$. Comparing with $Ax+B \cdot \frac{1}{x-1}+C \cdot \frac{1}{x-2}+D \cdot \frac{1}{x-3}+E$,we have $A=1$,$E=6$,$B=0.5$,$C=-16$,$D=40.5$. Sum $A+B+C+D+E = 1 + 0.5 - 16 + 40.5 + 6 = 32$.

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

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