Find the Holes in the Graph f(x)=(2x^2-3x+1)/(2x+16)

$f (x) = \frac{2 x^{2} - 3 x + 1}{2 x + 16}$

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 $a \cdot c = 2 \cdot 1 = 2$ and whose sum is $b = - 3$ .

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

$f (x) = \frac{2 x^{2} - 3 x + 1}{2 x + 16}$

Rewrite $- 3$ as $- 1$ plus $- 2$

$f (x) = \frac{2 x^{2} + (- 1 - 2) x + 1}{2 x + 16}$

Apply the distributive property.

$f (x) = \frac{2 x^{2} - 1 x - 2 x + 1}{2 x + 16}$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$f (x) = \frac{(2 x^{2} - 1 x) - 2 x + 1}{2 x + 16}$

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

$f (x) = \frac{x (2 x - 1) - (2 x - 1)}{2 x + 16}$

Factor the polynomial by factoring out the greatest common factor, $2 x - 1$ .

$f (x) = \frac{(2 x - 1) (x - 1)}{2 x + 16}$

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

Factor $2$ out of $2 x$ .

$f (x) = \frac{(2 x - 1) (x - 1)}{2 (x) + 16}$

Factor $2$ out of $16$ .

$f (x) = \frac{(2 x - 1) (x - 1)}{2 x + 2 \cdot 8}$

Factor $2$ out of $2 x + 2 \cdot 8$ .

$f (x) = \frac{(2 x - 1) (x - 1)}{2 (x + 8)}$

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^2-3x+1)/(2x+16)

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