Find the Holes in the Graph (5x^2)/(3x^2+7x)

$\frac{5 x^{2}}{3 x^{2} + 7 x}$

Step 1

Factor $x$ out of $3 x^{2} + 7 x$ .

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

$\frac{5 x^{2}}{x (3 x) + 7 x}$

Factor $x$ out of $7 x$ .

$\frac{5 x^{2}}{x (3 x) + x \cdot 7}$

Factor $x$ out of $x (3 x) + x \cdot 7$ .

$\frac{5 x^{2}}{x (3 x + 7)}$

Step 2

Cancel the common factor of $x^{2}$ and $x$ .

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

$\frac{x (5 x)}{x (3 x + 7)}$

Cancel the common factors.

Cancel the common factor.

$\frac{x (5 x)}{x (3 x + 7)}$

Rewrite the expression.

$\frac{5 x}{3 x + 7}$

Step 3

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

$x$

Step 4

To find the coordinates of the holes, set each factor that was cancelled equal to $0$ , solve, and substitute back in to $\frac{5 x}{3 x + 7}$ .

Set $x$ equal to $0$ .

$x = 0$

Substitute $0$ for $x$ in $\frac{5 x}{3 x + 7}$ and simplify.

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

$\frac{5 \cdot 0}{3 \cdot 0 + 7}$

Simplify.

Cancel the common factor of $0$ and $3 \cdot 0 + 7$ .

Factor $7$ out of $5 \cdot 0$ .

$\frac{7 (5 \cdot 0)}{3 \cdot 0 + 7}$

Cancel the common factors.

Factor $7$ out of $3 \cdot 0$ .

$\frac{7 (5 \cdot 0)}{7 (3 \cdot 0) + 7}$

Factor $7$ out of $7$ .

$\frac{7 (5 \cdot 0)}{7 (3 \cdot 0) + 7 (1)}$

Factor $7$ out of $7 (3 \cdot 0) + 7 (1)$ .

$\frac{7 (5 \cdot 0)}{7 (3 \cdot 0 + 1)}$

Cancel the common factor.

$\frac{7 (5 \cdot 0)}{7 (3 \cdot 0 + 1)}$

Rewrite the expression.

$\frac{5 \cdot 0}{3 \cdot 0 + 1}$

Multiply $5$ by $0$ .

$\frac{0}{3 \cdot 0 + 1}$

Simplify the denominator.

Multiply $3$ by $0$ .

$\frac{0}{0 + 1}$

Add $0$ and $1$ .

$\frac{0}{1}$

Divide $0$ by $1$ .

$0$

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

$(0, 0)$

Step 5

Find the Holes in the Graph (5x^2)/(3x^2+7x)

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