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

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(A) Given the expression $\frac{x^4}{(x^2+1)(x-2)}$.Perform polynomial division: $x^4 = (x^2+1)(x^2-1) + 1$.So, $\frac{x^4}{(x^2+1)(x-2)} = \frac{(x^2

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$5$

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(A) Given the expression $\frac{x^4}{(x^2+1)(x-2)}$.Perform polynomial division: $x^4 = (x^2+1)(x^2-1) + 1$.So, $\frac{x^4}{(x^2+1)(x-2)} = \frac{(x^2+1)(x^2-1)+1}{(x^2+1)(x-2)} = \frac{x^2-1}{x-2} + \frac{1}{(x^2+1)(x-2)}$.Further, $\frac{x^2-1}{x-2} = \frac{x^2-4+3}{x-2} = x+2 + \frac{3}{x-2}$.Now, decompose $\frac{1}{(x^2+1)(x-2)}$ using partial fractions:$\frac{1}{(x^2+1)(x-2)} = \frac{1}{5} \left( \frac{1}{x-2} - \frac{x+2}{x^2+1} ight) = \frac{1}{5(x-2)} - \frac{x+2}{5(x^2+1)}$.Combining these, we get:$\frac{x^4}{(x^2+1)(x-2)} = x+2 + \frac{3}{x-2} + \frac{1}{5(x-2)} - \frac{x+2}{5(x^2+1)} = x+2 + \frac{16}{5(x-2)} - \frac{x+2}{5(x^2+1)}$.Comparing with $f(x) + \frac{Ax+B}{x^2+1} + \frac{C}{x-2}$, we identify:$f(x) = x+2$, $A = -\frac{1}{5}$, $B = -\frac{2}{5}$, $C = \frac{16}{5}$.Calculate $f(14) + 2A - B = (14+2) + 2(-\frac{1}{5}) - (-\frac{2}{5}) = 16 - \frac{2}{5} + \frac{2}{5} = 16$.Since $5C = 5 	imes \frac{16}{5} = 16$, the result is $5C$.

If $\frac{x^4}{(x^2+1)(x-2)}=f(x)+\frac{Ax+B}{x^2+1}+\frac{C}{x-2}$, then $f(14)+2A-B=$ (in $C$)

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