If frac{x^3 - 6x^2 + 10x - 2}{x^2 - 5x + 6} = | vedclass

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(A) Given the expression $\frac{x^3 - 6x^2 + 10x - 2}{x^2 - 5x + 6}$.First,perform polynomial long division of the numerator by the denominator.The de

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$x - 1$

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(A) Given the expression $\frac{x^3 - 6x^2 + 10x - 2}{x^2 - 5x + 6}$.First,perform polynomial long division of the numerator by the denominator.The denominator is $x^2 - 5x + 6 = (x - 2)(x - 3)$.Dividing $x^3 - 6x^2 + 10x - 2$ by $x^2 - 5x + 6$:$x^3 - 6x^2 + 10x - 2 = x(x^2 - 5x + 6) - x^2 + 4x - 2$$= x(x^2 - 5x + 6) - 1(x^2 - 5x + 6) - x + 4$$= (x - 1)(x^2 - 5x + 6) - x + 4$.Thus,$\frac{x^3 - 6x^2 + 10x - 2}{x^2 - 5x + 6} = (x - 1) + \frac{-x + 4}{(x - 2)(x - 3)}$.Using partial fractions for $\frac{-x + 4}{(x - 2)(x - 3)} = \frac{A}{x - 2} + \frac{B}{x - 3}$:$-x + 4 = A(x - 3) + B(x - 2)$.For $x = 2$: $-2 + 4 = A(2 - 3) \implies 2 = -A \implies A = -2$.For $x = 3$: $-3 + 4 = B(3 - 2) \implies 1 = B \implies B = 1$.So,the expression is $(x - 1) - \frac{2}{x - 2} + \frac{1}{x - 3}$.Comparing this with $f(x) + \frac{A}{x - 2} + \frac{B}{x - 3}$,we get $f(x) = x - 1$.

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

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