Find the Holes in the Graph (3x+12)/(-x^2-3x+4)

$\frac{3 x + 12}{- x^{2} - 3 x + 4}$

Factor $3$ out of $3 x + 12$ .

Factor $3$ out of $3 x$ .

$\frac{3 (x) + 12}{- x^{2} - 3 x + 4}$

Factor $3$ out of $12$ .

$\frac{3 x + 3 \cdot 4}{- x^{2} - 3 x + 4}$

Factor $3$ out of $3 x + 3 \cdot 4$ .

$\frac{3 (x + 4)}{- x^{2} - 3 x + 4}$

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 $a \cdot c = - 1 \cdot 4 = - 4$ and whose sum is $b = - 3$ .

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

$\frac{3 (x + 4)}{- x^{2} - 3 (x) + 4}$

Rewrite $- 3$ as $1$ plus $- 4$

$\frac{3 (x + 4)}{- x^{2} + (1 - 4) x + 4}$

Apply the distributive property.

$\frac{3 (x + 4)}{- x^{2} + 1 x - 4 x + 4}$

Multiply $x$ by $1$ .

$\frac{3 (x + 4)}{- x^{2} + x - 4 x + 4}$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$\frac{3 (x + 4)}{(- x^{2} + x) - 4 x + 4}$

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

$\frac{3 (x + 4)}{x (- x + 1) + 4 (- x + 1)}$

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

$\frac{3 (x + 4)}{(- x + 1) (x + 4)}$

Cancel the common factor of $x + 4$ .

Cancel the common factor.

$\frac{3 (x + 4)}{(- x + 1) (x + 4)}$

Rewrite the expression.

$\frac{3}{- x + 1}$

To find the holes in the graph, look at the denominator factors that were cancelled.

$x + 4$

To find the coordinates of the holes, set each factor that was cancelled equal to $0$ , solve, and substitute back in to $\frac{3}{- x + 1}$ .

Set $x + 4$ equal to $0$ .

$x + 4 = 0$

Subtract $4$ from both sides of the equation.

$x = - 4$

Substitute $- 4$ for $x$ in $\frac{3}{- x + 1}$ and simplify.

Substitute $- 4$ for $x$ to find the $y$ coordinate of the hole.

$\frac{3}{4 + 1}$

Add $4$ and $1$ .

$\frac{3}{5}$

The holes in the graph are the points where any of the cancelled factors are equal to $0$ .

$(- 4, \frac{3}{5})$

Find the Holes in the Graph (3x+12)/(-x^2-3x+4)

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