Which of the following is equivalent to $\dfrac{2x^2-5x-1}{x-3}$ ?

Summary Submit

Use long division or synthetic division. Below we show long division.

$$\begin{array}{r} 2x+1\phantom{-10} \\ x-3{\overline{\smash{\big)}\,2x^2-5x-1}} \\ \underline{-(2x^2-6x)} \hphantom{-9}\\ x-1\phantom{} \\ \underline{-(x-3)} \\ \end{array}$$ $$\text{ Remainder}=2$$
$$\frac{2x^2-5x-1}{x-3} = \boxed{2x+1 + \frac{2}{x-3}}$$

Choose an arbitary choice to test, such as $x=0$.

$$\frac{2x^2-5x-1}{x-3}$$ $$\frac{2(0)^2-5(0)-1}{0-3}$$ $$= \frac{1}{3}$$

Testing the options,

We can combine the fractions in the last two answer choices and check if they are equivalent to the initial expression. Below we show verification for the correct choice:

$$2x +1 + \frac{2}{x-3}$$ $$(2x+1) \left(\frac{x-3}{x-3}\right) + \frac{2}{x-3}$$ $$\frac{(2x+1)(x-3)}{x-3}+\frac{2}{x-3}$$ $$\frac{2x^2+x-6x-3}{x-3}+\frac{2}{x-3}$$ $$\frac{2x^2-5x-3+2}{x-3}$$ $$=\frac{2x^2-5x-1}{x-3}   \checkmark$$