Find the Holes in the Graph (x^2-5x+4)/(3x^2+6x-72)

$\frac{x^{2} - 5 x + 4}{3 x^{2} + 6 x - 72}$

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 $- 5$ .

$- 4, - 1$

Write the factored form using these integers.

$\frac{(x - 4) (x - 1)}{3 x^{2} + 6 x - 72}$

Factor $3 x^{2} + 6 x - 72$ .

Factor $3$ out of $3 x^{2} + 6 x - 72$ .

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

$\frac{(x - 4) (x - 1)}{3 (x^{2}) + 6 x - 72}$

Factor $3$ out of $6 x$ .

$\frac{(x - 4) (x - 1)}{3 (x^{2}) + 3 (2 x) - 72}$

Factor $3$ out of $- 72$ .

$\frac{(x - 4) (x - 1)}{3 x^{2} + 3 (2 x) + 3 \cdot - 24}$

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

$\frac{(x - 4) (x - 1)}{3 (x^{2} + 2 x) + 3 \cdot - 24}$

Factor $3$ out of $3 (x^{2} + 2 x) + 3 \cdot - 24$ .

$\frac{(x - 4) (x - 1)}{3 (x^{2} + 2 x - 24)}$

Factor.

Factor $x^{2} + 2 x - 24$ 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 $- 24$ and whose sum is $2$ .

$- 4, 6$

Write the factored form using these integers.

$\frac{(x - 4) (x - 1)}{3 ((x - 4) (x + 6))}$

Remove unnecessary parentheses.

$\frac{(x - 4) (x - 1)}{3 (x - 4) (x + 6)}$

Cancel the common factor of $x - 4$ .

Cancel the common factor.

$\frac{(x - 4) (x - 1)}{3 (x - 4) (x + 6)}$

Rewrite the expression.

$\frac{x - 1}{3 (x + 6)}$

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 $\frac{x - 1}{3 (x + 6)}$ .

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

$x - 4 = 0$

Add $4$ to both sides of the equation.

$x = 4$

Substitute $4$ for $x$ in $\frac{x - 1}{3 (x + 6)}$ and simplify.

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

$\frac{4 - 1}{3 (4 + 6)}$

Simplify.

Subtract $1$ from $4$ .

$\frac{3}{3 (4 + 6)}$

Add $4$ and $6$ .

$\frac{3}{3 \cdot 10}$

Multiply $3$ by $10$ .

$\frac{3}{30}$

Cancel the common factor of $3$ and $30$ .

Factor $3$ out of $3$ .

$\frac{3 (1)}{30}$

Cancel the common factors.

Factor $3$ out of $30$ .

$\frac{3 \cdot 1}{3 \cdot 10}$

Cancel the common factor.

$\frac{3 \cdot 1}{3 \cdot 10}$

Rewrite the expression.

$\frac{1}{10}$

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

$(4, \frac{1}{10})$

Find the Holes in the Graph (x^2-5x+4)/(3x^2+6x-72)

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