Find the Holes in the Graph (x^2-3x-70)/(3x^2-30x)

$\frac{x^{2} - 3 x - 70}{3 x^{2} - 30 x}$

Factor $x^{2} - 3 x - 70$ 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 $- 70$ and whose sum is $- 3$ .

$- 10, 7$

Write the factored form using these integers.

$\frac{(x - 10) (x + 7)}{3 x^{2} - 30 x}$

Factor $3 x$ out of $3 x^{2} - 30 x$ .

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

$\frac{(x - 10) (x + 7)}{3 x (x) - 30 x}$

Factor $3 x$ out of $- 30 x$ .

$\frac{(x - 10) (x + 7)}{3 x (x) + 3 x (- 10)}$

Factor $3 x$ out of $3 x (x) + 3 x (- 10)$ .

$\frac{(x - 10) (x + 7)}{3 x (x - 10)}$

Cancel the common factor of $x - 10$ .

Cancel the common factor.

$\frac{(x - 10) (x + 7)}{3 x (x - 10)}$

Rewrite the expression.

$\frac{x + 7}{3 x}$

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

$x - 10$

To find the coordinates of the holes, set each factor that was cancelled equal to $0$ , solve, and substitute back in to $\frac{x + 7}{3 x}$ .

Set $x - 10$ equal to $0$ .

$x - 10 = 0$

Add $10$ to both sides of the equation.

$x = 10$

Substitute $10$ for $x$ in $\frac{x + 7}{3 x}$ and simplify.

Substitute $10$ for $x$ to find the $y$ coordinate of the hole.

$\frac{10 + 7}{3 \cdot 10}$

Simplify.

Add $10$ and $7$ .

$\frac{17}{3 \cdot 10}$

Multiply $3$ by $10$ .

$\frac{17}{30}$

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

$(10, \frac{17}{30})$

Find the Holes in the Graph (x^2-3x-70)/(3x^2-30x)

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