Find the Holes in the Graph f(x)=(x-5)/(3x^2-17x-28)

$f (x) = \frac{x - 5}{3 x^{2} - 17 x - 28}$

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 $a \cdot c = 3 \cdot - 28 = - 84$ and whose sum is $b = - 17$ .

Factor $- 17$ out of $- 17 x$ .

$f (x) = \frac{x - 5}{3 x^{2} - 17 x - 28}$

Rewrite $- 17$ as $4$ plus $- 21$

$f (x) = \frac{x - 5}{3 x^{2} + (4 - 21) x - 28}$

Apply the distributive property.

$f (x) = \frac{x - 5}{3 x^{2} + 4 x - 21 x - 28}$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$f (x) = \frac{x - 5}{(3 x^{2} + 4 x) - 21 x - 28}$

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

$f (x) = \frac{x - 5}{x (3 x + 4) - 7 (3 x + 4)}$

Factor the polynomial by factoring out the greatest common factor, $3 x + 4$ .

$f (x) = \frac{x - 5}{(3 x + 4) (x - 7)}$

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)=(x-5)/(3x^2-17x-28)

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