Algebra Examples

$f (x) = \frac{x^{3} + 4 x^{2} - 21 x}{x^{2} + 4 x - 21}$

Step 1

Factor $x^{3} + 4 x^{2} - 21 x$ .

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Step 1.1

Factor $x$ out of $x^{3} + 4 x^{2} - 21 x$ .

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Step 1.1.1

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

$f (x) = \frac{x \cdot x^{2} + 4 x^{2} - 21 x}{x^{2} + 4 x - 21}$

Step 1.1.2

Factor $x$ out of $4 x^{2}$ .

$f (x) = \frac{x \cdot x^{2} + x (4 x) - 21 x}{x^{2} + 4 x - 21}$

Step 1.1.3

Factor $x$ out of $- 21 x$ .

$f (x) = \frac{x \cdot x^{2} + x (4 x) + x \cdot - 21}{x^{2} + 4 x - 21}$

Step 1.1.4

Factor $x$ out of $x \cdot x^{2} + x (4 x)$ .

$f (x) = \frac{x (x^{2} + 4 x) + x \cdot - 21}{x^{2} + 4 x - 21}$

Step 1.1.5

Factor $x$ out of $x (x^{2} + 4 x) + x \cdot - 21$ .

$f (x) = \frac{x (x^{2} + 4 x - 21)}{x^{2} + 4 x - 21}$

Step 1.2

Factor.

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Step 1.2.1

Factor $x^{2} + 4 x - 21$ using the AC method.

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Step 1.2.1.1

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 $- 21$ and whose sum is $4$ .

$- 3, 7$

Step 1.2.1.2

Write the factored form using these integers.

$f (x) = \frac{x ((x - 3) (x + 7))}{x^{2} + 4 x - 21}$

Step 1.2.2

Remove unnecessary parentheses.

$f (x) = \frac{x (x - 3) (x + 7)}{x^{2} + 4 x - 21}$

Step 2

Factor $x^{2} + 4 x - 21$ using the AC method.

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Step 2.1

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 $- 21$ and whose sum is $4$ .

$- 3, 7$

Step 2.2

Write the factored form using these integers.

$f (x) = \frac{x (x - 3) (x + 7)}{(x - 3) (x + 7)}$

Step 3

Cancel the common factor of $x - 3$ .

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Step 3.1

Cancel the common factor.

$f (x) = \frac{x (x - 3) (x + 7)}{(x - 3) (x + 7)}$

Step 3.2

Rewrite the expression.

$f (x) = \frac{x (x + 7)}{x + 7}$

Step 4

Cancel the common factor of $x + 7$ .

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Step 4.1

Cancel the common factor.

$f (x) = \frac{x (x + 7)}{x + 7}$

Step 4.2

Divide $x$ by $1$ .

$f (x) = x$

Step 5

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

$x - 3, x + 7$

Step 6

To find the coordinates of the holes, set each factor that was cancelled equal to $0$ , solve, and substitute back in to $x$ .

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Step 6.1

Set $x - 3$ equal to $0$ .

$x - 3 = 0$

Step 6.2

Add $3$ to both sides of the equation.

$x = 3$

Step 6.3

Substitute $3$ for $x$ to find the $y$ coordinate of the hole.

$3$

Step 6.4

Set $x + 7$ equal to $0$ .

$x + 7 = 0$

Step 6.5

Subtract $7$ from both sides of the equation.

$x = - 7$

Step 6.6

Substitute $- 7$ for $x$ to find the $y$ coordinate of the hole.

$- 7$

Step 6.7

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

$(3, 3), (- 7, - 7)$

Step 7