Find the Holes in the Graph f(x)=(-x^2+2x+8)/(2x^2-16x+32)

$f (x) = \frac{- x^{2} + 2 x + 8}{2 x^{2} - 16 x + 32}$

Step 1

Factor by grouping.

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = - 1 \cdot 8 = - 8$ and whose sum is $b = 2$ .

Factor $2$ out of $2 x$ .

$f (x) = \frac{- x^{2} + 2 (x) + 8}{2 x^{2} - 16 x + 32}$

Rewrite $2$ as $- 2$ plus $4$

$f (x) = \frac{- x^{2} + (- 2 + 4) x + 8}{2 x^{2} - 16 x + 32}$

Apply the distributive property.

$f (x) = \frac{- x^{2} - 2 x + 4 x + 8}{2 x^{2} - 16 x + 32}$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$f (x) = \frac{(- x^{2} - 2 x) + 4 x + 8}{2 x^{2} - 16 x + 32}$

Factor out the greatest common factor (GCF) from each group.

$f (x) = \frac{x (- x - 2) - 4 (- x - 2)}{2 x^{2} - 16 x + 32}$

Factor the polynomial by factoring out the greatest common factor, $- x - 2$ .

$f (x) = \frac{(- x - 2) (x - 4)}{2 x^{2} - 16 x + 32}$

Step 2

Factor $2 x^{2} - 16 x + 32$ .

Factor $2$ out of $2 x^{2} - 16 x + 32$ .

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

$f (x) = \frac{(- x - 2) (x - 4)}{2 (x^{2}) - 16 x + 32}$

Factor $2$ out of $- 16 x$ .

$f (x) = \frac{(- x - 2) (x - 4)}{2 (x^{2}) + 2 (- 8 x) + 32}$

Factor $2$ out of $32$ .

$f (x) = \frac{(- x - 2) (x - 4)}{2 x^{2} + 2 (- 8 x) + 2 \cdot 16}$

Factor $2$ out of $2 x^{2} + 2 (- 8 x)$ .

$f (x) = \frac{(- x - 2) (x - 4)}{2 (x^{2} - 8 x) + 2 \cdot 16}$

Factor $2$ out of $2 (x^{2} - 8 x) + 2 \cdot 16$ .

$f (x) = \frac{(- x - 2) (x - 4)}{2 (x^{2} - 8 x + 16)}$

Factor using the perfect square rule.

Rewrite $16$ as $4^{2}$ .

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

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

$8 x = 2 \cdot x \cdot 4$

Rewrite the polynomial.

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

Factor using the perfect square trinomial rule $a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ , where $a = x$ and $b = 4$ .

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

Step 3

Cancel the common factor of $x - 4$ and ${(x - 4)}^{2}$ .

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

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

Cancel the common factors.

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

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

Cancel the common factor.

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

Rewrite the expression.

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

Step 4

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

No holes

Step 5

Find the Holes in the Graph f(x)=(-x^2+2x+8)/(2x^2-16x+32)

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