Find the Holes in the Graph f(x)=(3x^3+2x^2)/(x^2+x)

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

Step 1

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

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

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

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

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

Factor $x^{2}$ out of $x^{2} (3 x) + x^{2} \cdot 2$ .

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

Step 2

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

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

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

Raise $x$ to the power of $1$ .

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

Factor $x$ out of $x^{1}$ .

$f (x) = \frac{x^{2} (3 x + 2)}{x \cdot x + x \cdot 1}$

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

$f (x) = \frac{x^{2} (3 x + 2)}{x (x + 1)}$

Step 3

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

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

$f (x) = \frac{x (x (3 x + 2))}{x (x + 1)}$

Cancel the common factors.

Cancel the common factor.

$f (x) = \frac{x (x (3 x + 2))}{x (x + 1)}$

Rewrite the expression.

$f (x) = \frac{x (3 x + 2)}{x + 1}$

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 $\frac{x (3 x + 2)}{x + 1}$ .

Set $x$ equal to $0$ .

$x = 0$

Substitute $0$ for $x$ in $\frac{x (3 x + 2)}{x + 1}$ and simplify.

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

$\frac{0 (3 \cdot 0 + 2)}{0 + 1}$

Simplify.

Multiply $0$ by $3 \cdot 0 + 2$ .

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

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 6

Find the Holes in the Graph f(x)=(3x^3+2x^2)/(x^2+x)

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