Find the Holes in the Graph f(x)=(4x-8)/(x^2-4x+4)

$f (x) = \frac{4 x - 8}{x^{2} - 4 x + 4}$

Factor $4$ out of $4 x - 8$ .

Factor $4$ out of $4 x$ .

$f (x) = \frac{4 (x) - 8}{x^{2} - 4 x + 4}$

Factor $4$ out of $- 8$ .

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

Factor $4$ out of $4 x + 4 \cdot - 2$ .

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

Factor using the perfect square rule.

Rewrite $4$ as $2^{2}$ .

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

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$4 x = 2 \cdot x \cdot 2$

Rewrite the polynomial.

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

Factor using the perfect square trinomial rule $a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ , where $a = x$ and $b = 2$ .

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

Cancel the common factor of $x - 2$ and ${(x - 2)}^{2}$ .

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

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

Cancel the common factors.

Factor $x - 2$ out of ${(x - 2)}^{2}$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

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

No holes