Find the Holes in the Graph (3x^2-12x)/(x^2-2x-3)

$\frac{3 x^{2} - 12 x}{x^{2} - 2 x - 3}$

Step 1

Factor $3 x$ out of $3 x^{2} - 12 x$ .

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

$\frac{3 x (x) - 12 x}{x^{2} - 2 x - 3}$

Factor $3 x$ out of $- 12 x$ .

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

Factor $3 x$ out of $3 x (x) + 3 x (- 4)$ .

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

Step 2

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.

$\frac{3 x (x - 4)}{(x - 3) (x + 1)}$

Step 3

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

No holes

Step 4

Find the Holes in the Graph (3x^2-12x)/(x^2-2x-3)

Related Questions

Support LCMGCF.com by disabling your adblocker.