If frac{x^4}{(x-1)(x-2)(x-3)} = x+k+frac{A}{x | vedclass

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(C) Given,$\frac{x^4}{(x-1)(x-2)(x-3)} = x+k+\frac{A}{x-1}+\frac{B}{x-2}+\frac{C}{x-3}$.First,perform polynomial division: $\frac{x^4}{x^3-6x^2+11x-6}

Vedclass · Accepted Answer

$63$

Vedclass · Answer

(C) Given,$\frac{x^4}{(x-1)(x-2)(x-3)} = x+k+\frac{A}{x-1}+\frac{B}{x-2}+\frac{C}{x-3}$.First,perform polynomial division: $\frac{x^4}{x^3-6x^2+11x-6} = x+6 + \frac{31x^2-72x+36}{(x-1)(x-2)(x-3)}$.Comparing,we get $k=6$.Now,decompose the remainder: $\frac{31x^2-72x+36}{(x-1)(x-2)(x-3)} = \frac{A}{x-1}+\frac{B}{x-2}+\frac{C}{x-3}$.Using partial fractions:For $A$: $A = \frac{31(1)^2-72(1)+36}{(1-2)(1-3)} = \frac{31-72+36}{2} = \frac{-5}{2}$.For $B$: $B = \frac{31(2)^2-72(2)+36}{(2-1)(2-3)} = \frac{124-144+36}{-1} = \frac{16}{-1} = -16$.For $C$: $C = \frac{31(3)^2-72(3)+36}{(3-1)(3-2)} = \frac{279-216+36}{2} = \frac{99}{2}$.Thus,$k+A-B+C = 6 + (-\frac{5}{2}) - (-16) + \frac{99}{2} = 6 - 2.5 + 16 + 49.5 = 69$.

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

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