Find the Holes in the Graph f(x)=(3x^2+12x)/(x^2+5x+4)

$f (x) = \frac{3 x^{2} + 12 x}{x^{2} + 5 x + 4}$

Step 1

Factor $3 x$ out of $3 x^{2} + 12 x$ .

Factor $3 x$ out of $3 x^{2}$ .

$f (x) = \frac{3 x (x) + 12 x}{x^{2} + 5 x + 4}$

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

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

Factor $3 x$ out of $3 x (x) + 3 x (4)$ .

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

Step 2

Factor $x^{2} + 5 x + 4$ using the AC method.

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $4$ and whose sum is $5$ .

$1, 4$

Write the factored form using these integers.

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

Step 3

Cancel the common factor of $x + 4$ .

Cancel the common factor.

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

Rewrite the expression.

$f (x) = \frac{3 x}{x + 1}$

Step 4

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

$x + 4$

Step 5

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

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

$x + 4 = 0$

Subtract $4$ from both sides of the equation.

$x = - 4$

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

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

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

Simplify.

Multiply $3$ by $- 4$ .

$\frac{- 12}{- 4 + 1}$

Add $- 4$ and $1$ .

$\frac{- 12}{- 3}$

Divide $- 12$ by $- 3$ .

$4$

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

$(- 4, 4)$

Step 6

Find the Holes in the Graph f(x)=(3x^2+12x)/(x^2+5x+4)

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