Find the Holes in the Graph f(x)=(x^2+14x+40)/(9x-15)

$f (x) = \frac{x^{2} + 14 x + 40}{9 x - 15}$

Step 1

Factor $x^{2} + 14 x + 40$ 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 $40$ and whose sum is $14$ .

$4, 10$

Write the factored form using these integers.

$f (x) = \frac{(x + 4) (x + 10)}{9 x - 15}$

Step 2

Factor $3$ out of $9 x - 15$ .

Factor $3$ out of $9 x$ .

$f (x) = \frac{(x + 4) (x + 10)}{3 (3 x) - 15}$

Factor $3$ out of $- 15$ .

$f (x) = \frac{(x + 4) (x + 10)}{3 (3 x) + 3 (- 5)}$

Factor $3$ out of $3 (3 x) + 3 (- 5)$ .

$f (x) = \frac{(x + 4) (x + 10)}{3 (3 x - 5)}$

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 f(x)=(x^2+14x+40)/(9x-15)

Related Questions

Support LCMGCF.com by disabling your adblocker.