Find the Holes in the Graph g(x)=(x+10)/(x^3+5x^2-50x)

$g (x) = \frac{x + 10}{x^{3} + 5 x^{2} - 50 x}$

Factor $x^{3} + 5 x^{2} - 50 x$ .

Factor $x$ out of $x^{3} + 5 x^{2} - 50 x$ .

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

$g (x) = \frac{x + 10}{x \cdot x^{2} + 5 x^{2} - 50 x}$

Factor $x$ out of $5 x^{2}$ .

$g (x) = \frac{x + 10}{x \cdot x^{2} + x (5 x) - 50 x}$

Factor $x$ out of $- 50 x$ .

$g (x) = \frac{x + 10}{x \cdot x^{2} + x (5 x) + x \cdot - 50}$

Factor $x$ out of $x \cdot x^{2} + x (5 x)$ .

$g (x) = \frac{x + 10}{x (x^{2} + 5 x) + x \cdot - 50}$

Factor $x$ out of $x (x^{2} + 5 x) + x \cdot - 50$ .

$g (x) = \frac{x + 10}{x (x^{2} + 5 x - 50)}$

Factor.

Factor $x^{2} + 5 x - 50$ 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 $- 50$ and whose sum is $5$ .

$- 5, 10$

Write the factored form using these integers.

$g (x) = \frac{x + 10}{x ((x - 5) (x + 10))}$

Remove unnecessary parentheses.

$g (x) = \frac{x + 10}{x (x - 5) (x + 10)}$

Cancel the common factor of $x + 10$ .

Cancel the common factor.

$g (x) = \frac{x + 10}{x (x - 5) (x + 10)}$

Rewrite the expression.

$g (x) = \frac{1}{x (x - 5)}$

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 $\frac{1}{x (x - 5)}$ .

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

$x + 10 = 0$

Subtract $10$ from both sides of the equation.

$x = - 10$

Substitute $- 10$ for $x$ in $\frac{1}{x (x - 5)}$ and simplify.

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

$\frac{1}{- 10 (- 10 - 5)}$

Simplify.

Subtract $5$ from $- 10$ .

$\frac{1}{- 10 \cdot - 15}$

Multiply $- 10$ by $- 15$ .

$\frac{1}{150}$

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

$(- 10, \frac{1}{150})$

Find the Holes in the Graph g(x)=(x+10)/(x^3+5x^2-50x)

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