If x4x-1x+2=13x-1-163x+2+x2-x+k then k=

x4x-1 x+2=13x-1-163x+2+x2-x+k x2+x-2)x4( x2-x3+3 x4+x3-2x2-x3+2x2 -x3-x2+2x3x2-2x 3x2+3x-6-5x+6x4(x-1) (x+2)=x2-x+3+6-5x(x-1) (x+1)k=3

If x4x-1x+2=13x-1-163x+2+x2-x+k then k=

x4x-1 x+2=13x-1-163x+2+x2-x+k x2+x-2)x4( x2-x3+3 x4+x3-2x2-x3+2x2 -x3-x2+2x3x2-2x 3x2+3x-6-5x+6x4(x-1) (x+2)=x2-x+3+6-5x(x-1) (x+1)k=3

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$If \frac{x^{4}}{(x - 1) (x + 2)} = \frac{1}{3 (x - 1)} - \frac{16}{3 (x + 2)} + x^{2} - x + k then k =$

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answer is D.

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Detailed Solution

$\begin{array}{rcl} \frac{x^{4}}{(x - 1) (x + 2)} & = & \frac{1}{3 (x - 1)} - \frac{16}{3 (x + 2)} + x^{2} - x + k \\ x^{2} + x - 2) x^{4} (x^{2} - x^{3} + 3 \\ \frac{x^{4} + x^{3} - 2 x^{2}}{- x^{3} + 2 x^{2}} \\ \frac{- x^{3} - x^{2} + 2 x}{3 x^{2} - 2 x} \\ \frac{3 x^{2} + 3 x - 6}{- 5 x + 6} \\ \frac{x^{4}}{(x - 1) (x + 2)} & = & x^{2} - x + 3 + \frac{6 - 5 x}{(x - 1) (x + 1)} \\ k & = & 3 \end{array}$