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

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

Factor $x^{3} + 5 x^{2} + 6 x$ .

Factor $x$ out of $x^{3} + 5 x^{2} + 6 x$ .

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

$\frac{5 x}{x \cdot x^{2} + 5 x^{2} + 6 x}$

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

$\frac{5 x}{x \cdot x^{2} + x (5 x) + 6 x}$

Factor $x$ out of $6 x$ .

$\frac{5 x}{x \cdot x^{2} + x (5 x) + x \cdot 6}$

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

$\frac{5 x}{x (x^{2} + 5 x) + x \cdot 6}$

Factor $x$ out of $x (x^{2} + 5 x) + x \cdot 6$ .

$\frac{5 x}{x (x^{2} + 5 x + 6)}$

Factor.

Factor $x^{2} + 5 x + 6$ 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 $6$ and whose sum is $5$ .

$2, 3$

Write the factored form using these integers.

$\frac{5 x}{x ((x + 2) (x + 3))}$

Remove unnecessary parentheses.

$\frac{5 x}{x (x + 2) (x + 3)}$

Cancel the common factor of $x$ .

Cancel the common factor.

$\frac{5 x}{x (x + 2) (x + 3)}$

Rewrite the expression.

$\frac{5}{(x + 2) (x + 3)}$

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

$x$

To find the coordinates of the holes, set each factor that was cancelled equal to $0$ , solve, and substitute back in to $\frac{5}{(x + 2) (x + 3)}$ .

Set $x$ equal to $0$ .

$x = 0$

Substitute $0$ for $x$ in $\frac{5}{(x + 2) (x + 3)}$ and simplify.

Substitute $0$ for $x$ to find the $y$ coordinate of the hole.

$\frac{5}{(0 + 2) (0 + 3)}$

Simplify.

Simplify the denominator.

Add $0$ and $2$ .

$\frac{5}{2 (0 + 3)}$

Add $0$ and $3$ .

$\frac{5}{2 \cdot 3}$

Multiply $2$ by $3$ .

$\frac{5}{6}$

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

$(0, \frac{5}{6})$

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

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