Find the Holes in the Graph f(x)=(x^2-3x)/(x-5)

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

Factor $x$ out of $x^{2} - 3 x$ .

Factor $x$ out of $x^{2}$ .

$f (x) = \frac{x \cdot x - 3 x}{x - 5}$

Factor $x$ out of $- 3 x$ .

$f (x) = \frac{x \cdot x + x \cdot - 3}{x - 5}$

Factor $x$ out of $x \cdot x + x \cdot - 3$ .

$f (x) = \frac{x (x - 3)}{x - 5}$

Since no factors can be removed from the denominator, there are no holes in the graph.

No holes