SAT Mathematics Practice Test 7

The numerator of the given expression can be rewritten in terms of the denominator, x – 3, as follows: ${\mathit{x}}^2-2\mathit{x}-5={\mathit{x}}^2-3\mathit{x}+\mathit{x}-3-2$, which is equivalent to $\mathit{x}\left(\mathit{x}\mathit{–}3\right)+\left(\mathit{x}\mathit{–}3\right)\mathit{–}2$. So t he given expression is equivalent to $\frac{\mathit{x}\left(\mathit{x}-3\right)+\left(\mathit{x}-3\right)-2}{\mathit{x}-3}$ = $\frac{\mathit{x}\left(\mathit{x}-3\right)}{\mathit{x}-3}+\frac{\mathit{x}-3}{\mathit{x}-3}-\frac{2}{\mathit{x}-3}$ Since the given expression is defined for x ≠ 3, the expression can be rewritten as $\mathit{x}+1-\frac{2}{\mathit{x}-3}$.
Long division can also be used as an alternate approach.