Hi Selvamani,

Both the numerator $x^3+3x^2-9x-2$ and the denominator $x^3-x-6$ 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 $x^3+3x^2-9x-2 = (x - 2) p(x).$ You can find $p(x)$ by dividing $x^3+3x^2-9x-2$ by $x - 2.$ Similarly for the denominator there is a polynomial $q(x)$ so that $x^3-x-6 = (x - 2) q(x).$ Thus

\[\frac{x^3+3x^2-9x-2}{x^3-x-6} = \frac{(x - 2) p(x)}{(x - 2) q(x)}\]

and since $x \neq 2$ the $(x - 2)$ factors can be cancelled to get

\[\frac{x^3+3x^2-9x-2}{x^3-x-6} = \frac{p(x)}{q(x)}.\]

Penny