Find the Holes in the Graph f(x)=(3x^2-28x-20)/(2x+16)

$f (x) = \frac{3 x^{2} - 28 x - 20}{2 x + 16}$

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 $a \cdot c = 3 \cdot - 20 = - 60$ and whose sum is $b = - 28$ .

Factor $- 28$ out of $- 28 x$ .

$f (x) = \frac{3 x^{2} - 28 x - 20}{2 x + 16}$

Rewrite $- 28$ as $2$ plus $- 30$

$f (x) = \frac{3 x^{2} + (2 - 30) x - 20}{2 x + 16}$

Apply the distributive property.

$f (x) = \frac{3 x^{2} + 2 x - 30 x - 20}{2 x + 16}$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$f (x) = \frac{(3 x^{2} + 2 x) - 30 x - 20}{2 x + 16}$

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

$f (x) = \frac{x (3 x + 2) - 10 (3 x + 2)}{2 x + 16}$

Factor the polynomial by factoring out the greatest common factor, $3 x + 2$ .

$f (x) = \frac{(3 x + 2) (x - 10)}{2 x + 16}$

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

Factor $2$ out of $2 x$ .

$f (x) = \frac{(3 x + 2) (x - 10)}{2 (x) + 16}$

Factor $2$ out of $16$ .

$f (x) = \frac{(3 x + 2) (x - 10)}{2 x + 2 \cdot 8}$

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

$f (x) = \frac{(3 x + 2) (x - 10)}{2 (x + 8)}$

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)=(3x^2-28x-20)/(2x+16)

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