Find the Holes in the Graph f(x)=(2x^2-4x)/(x^2-2x-3)

$f (x) = \frac{2 x^{2} - 4 x}{x^{2} - 2 x - 3}$

Factor $2 x$ out of $2 x^{2} - 4 x$ .

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

$f (x) = \frac{2 x (x) - 4 x}{x^{2} - 2 x - 3}$

Factor $2 x$ out of $- 4 x$ .

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

Factor $2 x$ out of $2 x (x) + 2 x (- 2)$ .

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

Factor $x^{2} - 2 x - 3$ using the AC method.

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 3$ and whose sum is $- 2$ .

$- 3, 1$

Write the factored form using these integers.

$f (x) = \frac{2 x (x - 2)}{(x - 3) (x + 1)}$

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

No holes

Find the Holes in the Graph f(x)=(2x^2-4x)/(x^2-2x-3)

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