Algebra Examples

$f (x) = \frac{x^{2} + 3 x}{x^{2} - x}$

Step 1

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

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Step 1.1

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

$f (x) = \frac{x \cdot x + 3 x}{x^{2} - x}$

Step 1.2

Factor $x$ out of $3 x$ .

$f (x) = \frac{x \cdot x + x \cdot 3}{x^{2} - x}$

Step 1.3

Factor $x$ out of $x \cdot x + x \cdot 3$ .

$f (x) = \frac{x (x + 3)}{x^{2} - x}$

Step 2

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

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Step 2.1

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

$f (x) = \frac{x (x + 3)}{x \cdot x - x}$

Step 2.2

Factor $x$ out of $- x$ .

$f (x) = \frac{x (x + 3)}{x \cdot x + x \cdot - 1}$

Step 2.3

Factor $x$ out of $x \cdot x + x \cdot - 1$ .

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

Step 3

Cancel the common factor of $x$ .

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Step 3.1

Cancel the common factor.

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

Step 3.2

Rewrite the expression.

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

Step 4

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

$x$

Step 5

To find the coordinates of the holes, set each factor that was cancelled equal to $0$ , solve, and substitute back in to $\frac{x + 3}{x - 1}$ .

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Step 5.1

Set $x$ equal to $0$ .

$x = 0$

Step 5.2

Substitute $0$ for $x$ in $\frac{x + 3}{x - 1}$ and simplify.

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Step 5.2.1

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

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

Step 5.2.2

Simplify.

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Step 5.2.2.1

Add $0$ and $3$ .

$\frac{3}{0 - 1}$

Step 5.2.2.2

Subtract $1$ from $0$ .

$\frac{3}{- 1}$

Step 5.2.2.3

Divide $3$ by $- 1$ .

$- 3$

Step 5.3

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

$(0, - 3)$

Step 6