Show that $\displaystyle f(x) = (x^3 + 1)/(x - 7)$ is order $\displaystyle x^2$
The 'division' between polynomials can be performed as...
$\displaystyle f(x)=\frac{x^{3}+1}{x-7} = \frac{x^{3}-7x^{2}}{x-7} + \frac{7x^{2} +1}{x-7}=$
$\displaystyle = x^{2} + \frac{7x^{2}-49x}{x-7} + \frac{49x+1}{x-7}=$
$\displaystyle = x^{2} +7x + \frac{49x-343}{x-7} + \frac{344}{x-7}= x^{2} + 7x + 49 + \frac{344}{x-7}$
... so that is...
$\displaystyle f(x)= \mathcal{O} \{x^{2}\}$
Kind regards
$\displaystyle \chi$ $\displaystyle \sigma$