



 
Hi Selvamani, Both the numerator $x^3+3x^29x2$ and the denominator $x^3x6$ are zero when $x = 2$ and hence you ca use the factor theorem on both of them. For example, since the denominator is zero when $x = 2$ there is a polynomial $p(x)$ so that $x^3+3x^29x2 = (x  2) p(x).$ You can find $p(x)$ by dividing $x^3+3x^29x2$ by $x  2.$ Similarly for the denominator there is a polynomial $q(x)$ so that $x^3x6 = (x  2) q(x).$ Thus \[\frac{x^3+3x^29x2}{x^3x6} = \frac{(x  2) p(x)}{(x  2) q(x)}\] and since $x \neq 2$ the $(x  2)$ factors can be cancelled to get \[\frac{x^3+3x^29x2}{x^3x6} = \frac{p(x)}{q(x)}.\] Penny  


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