If x2−10x+13(x−1)x2−5x+6=Ax−1+Bx−2+Kx−3 then K=

given that x2−10x+13(x−1)x2−5x+6=Ax−1+Bx−2+Kx−3 ⇒x2−10x+13=A(x-2)(x-3)+B(x-1)(x-3)+k(x-1)(x-2) put x=3 ∴k=-4

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$If \frac{x^{2} - 10 x + 13}{(x - 1) (x^{2} - 5 x + 6)} = \frac{A}{x - 1} + \frac{B}{x - 2} + \frac{K}{x - 3} then K =$

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answer is -4.

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Detailed Solution

$given that \frac{x^{2} - 10 x + 13}{(x - 1) (x^{2} - 5 x + 6)} = \frac{A}{x - 1} + \frac{B}{x - 2} + \frac{K}{x - 3} \Rightarrow x^{2} - 10 x + 13 = A (x - 2) (x - 3) + B (x - 1) (x - 3) + k (x - 1) (x - 2) put x = 3 ∴ k = - 4$

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