Precalculus Examples

$\frac{x^{2} + 12 x + 12}{x^{3} - 4 x}$

Step 1

Decompose the fraction and multiply through by the common denominator.

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Step 1.1

Factor the fraction.

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Step 1.1.1

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

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Step 1.1.1.1

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

$\frac{x^{2} + 12 x + 12}{x \cdot x^{2} - 4 x}$

Step 1.1.1.2

Factor $x$ out of $- 4 x$ .

$\frac{x^{2} + 12 x + 12}{x \cdot x^{2} + x \cdot - 4}$

Step 1.1.1.3

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

$\frac{x^{2} + 12 x + 12}{x (x^{2} - 4)}$

Step 1.1.2

Rewrite $4$ as $2^{2}$ .

$\frac{x^{2} + 12 x + 12}{x (x^{2} - 2^{2})}$

Step 1.1.3

Factor.

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Step 1.1.3.1

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = x$ and $b = 2$ .

$\frac{x^{2} + 12 x + 12}{x ((x + 2) (x - 2))}$

Step 1.1.3.2

Remove unnecessary parentheses.

$\frac{x^{2} + 12 x + 12}{x (x + 2) (x - 2)}$

Step 1.2

For each factor in the denominator, create a new fraction using the factor as the denominator, and an unknown value as the numerator. Since the factor in the denominator is linear, put a single variable in its place $B$ .

$\frac{A}{x} + \frac{B}{x + 2}$

Step 1.3

$\frac{A}{x} + \frac{B}{x + 2} + \frac{C}{x - 2}$

Step 1.4

Multiply each fraction in the equation by the denominator of the original expression. In this case, the denominator is $x (x + 2) (x - 2)$ .

$\frac{(x^{2} + 12 x + 12) (x (x + 2) (x - 2))}{x (x + 2) (x - 2)} = \frac{A (x (x + 2) (x - 2))}{x} + \frac{(B) (x (x + 2) (x - 2))}{x + 2} + \frac{(C) (x (x + 2) (x - 2))}{x - 2}$

Step 1.5

Cancel the common factor of $x$ .

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Step 1.5.1

Cancel the common factor.

$\frac{(x^{2} + 12 x + 12) (x (x + 2) (x - 2))}{x (x + 2) (x - 2)} = \frac{A (x (x + 2) (x - 2))}{x} + \frac{(B) (x (x + 2) (x - 2))}{x + 2} + \frac{(C) (x (x + 2) (x - 2))}{x - 2}$

Step 1.5.2

Rewrite the expression.

$\frac{(x^{2} + 12 x + 12) ((x + 2) (x - 2))}{(x + 2) (x - 2)} = \frac{A (x (x + 2) (x - 2))}{x} + \frac{(B) (x (x + 2) (x - 2))}{x + 2} + \frac{(C) (x (x + 2) (x - 2))}{x - 2}$

Step 1.6

Cancel the common factor of $x + 2$ .

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Step 1.6.1

Cancel the common factor.

$\frac{(x^{2} + 12 x + 12) ((x + 2) (x - 2))}{(x + 2) (x - 2)} = \frac{A (x (x + 2) (x - 2))}{x} + \frac{(B) (x (x + 2) (x - 2))}{x + 2} + \frac{(C) (x (x + 2) (x - 2))}{x - 2}$

Step 1.6.2

Rewrite the expression.

$\frac{(x^{2} + 12 x + 12) (x - 2)}{x - 2} = \frac{A (x (x + 2) (x - 2))}{x} + \frac{(B) (x (x + 2) (x - 2))}{x + 2} + \frac{(C) (x (x + 2) (x - 2))}{x - 2}$

Step 1.7

Cancel the common factor of $x - 2$ .

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Step 1.7.1

Cancel the common factor.

$\frac{(x^{2} + 12 x + 12) (x - 2)}{x - 2} = \frac{A (x (x + 2) (x - 2))}{x} + \frac{(B) (x (x + 2) (x - 2))}{x + 2} + \frac{(C) (x (x + 2) (x - 2))}{x - 2}$

Step 1.7.2

Divide $x^{2} + 12 x + 12$ by $1$ .

$x^{2} + 12 x + 12 = \frac{A (x (x + 2) (x - 2))}{x} + \frac{(B) (x (x + 2) (x - 2))}{x + 2} + \frac{(C) (x (x + 2) (x - 2))}{x - 2}$

Step 1.8

Simplify each term.

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Step 1.8.1

Cancel the common factor of $x$ .

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Step 1.8.1.1

Cancel the common factor.

$x^{2} + 12 x + 12 = \frac{A (x (x + 2) (x - 2))}{x} + \frac{(B) (x (x + 2) (x - 2))}{x + 2} + \frac{(C) (x (x + 2) (x - 2))}{x - 2}$

Step 1.8.1.2

Divide $A ((x + 2) (x - 2))$ by $1$ .

$x^{2} + 12 x + 12 = A ((x + 2) (x - 2)) + \frac{(B) (x (x + 2) (x - 2))}{x + 2} + \frac{(C) (x (x + 2) (x - 2))}{x - 2}$

Step 1.8.2

Expand $(x + 2) (x - 2)$ using the FOIL Method.

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Step 1.8.2.1

Apply the distributive property.

$x^{2} + 12 x + 12 = A (x (x - 2) + 2 (x - 2)) + \frac{(B) (x (x + 2) (x - 2))}{x + 2} + \frac{(C) (x (x + 2) (x - 2))}{x - 2}$

Step 1.8.2.2

Apply the distributive property.

$x^{2} + 12 x + 12 = A (x \cdot x + x \cdot - 2 + 2 (x - 2)) + \frac{(B) (x (x + 2) (x - 2))}{x + 2} + \frac{(C) (x (x + 2) (x - 2))}{x - 2}$

Step 1.8.2.3

Apply the distributive property.

$x^{2} + 12 x + 12 = A (x \cdot x + x \cdot - 2 + 2 x + 2 \cdot - 2) + \frac{(B) (x (x + 2) (x - 2))}{x + 2} + \frac{(C) (x (x + 2) (x - 2))}{x - 2}$

Step 1.8.3

Combine the opposite terms in $x \cdot x + x \cdot - 2 + 2 x + 2 \cdot - 2$ .

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Step 1.8.3.1

Reorder the factors in the terms $x \cdot - 2$ and $2 x$ .

$x^{2} + 12 x + 12 = A (x \cdot x - 2 x + 2 x + 2 \cdot - 2) + \frac{(B) (x (x + 2) (x - 2))}{x + 2} + \frac{(C) (x (x + 2) (x - 2))}{x - 2}$

Step 1.8.3.2

Add $- 2 x$ and $2 x$ .

$x^{2} + 12 x + 12 = A (x \cdot x + 0 + 2 \cdot - 2) + \frac{(B) (x (x + 2) (x - 2))}{x + 2} + \frac{(C) (x (x + 2) (x - 2))}{x - 2}$

Step 1.8.3.3

Add $x \cdot x$ and $0$ .

$x^{2} + 12 x + 12 = A (x \cdot x + 2 \cdot - 2) + \frac{(B) (x (x + 2) (x - 2))}{x + 2} + \frac{(C) (x (x + 2) (x - 2))}{x - 2}$

Step 1.8.4

Simplify each term.

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Step 1.8.4.1

Multiply $x$ by $x$ .

$x^{2} + 12 x + 12 = A (x^{2} + 2 \cdot - 2) + \frac{(B) (x (x + 2) (x - 2))}{x + 2} + \frac{(C) (x (x + 2) (x - 2))}{x - 2}$

Step 1.8.4.2

Multiply $2$ by $- 2$ .

$x^{2} + 12 x + 12 = A (x^{2} - 4) + \frac{(B) (x (x + 2) (x - 2))}{x + 2} + \frac{(C) (x (x + 2) (x - 2))}{x - 2}$

Step 1.8.5

Apply the distributive property.

$x^{2} + 12 x + 12 = A x^{2} + A \cdot - 4 + \frac{(B) (x (x + 2) (x - 2))}{x + 2} + \frac{(C) (x (x + 2) (x - 2))}{x - 2}$

Step 1.8.6

Move $- 4$ to the left of $A$ .

$x^{2} + 12 x + 12 = A x^{2} - 4 \cdot A + \frac{(B) (x (x + 2) (x - 2))}{x + 2} + \frac{(C) (x (x + 2) (x - 2))}{x - 2}$

Step 1.8.7

Cancel the common factor of $x + 2$ .

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Step 1.8.7.1

Cancel the common factor.

$x^{2} + 12 x + 12 = A x^{2} - 4 A + \frac{B (x (x + 2) (x - 2))}{x + 2} + \frac{(C) (x (x + 2) (x - 2))}{x - 2}$

Step 1.8.7.2

Divide $(B) (x (x - 2))$ by $1$ .

$x^{2} + 12 x + 12 = A x^{2} - 4 A + (B) (x (x - 2)) + \frac{(C) (x (x + 2) (x - 2))}{x - 2}$

Step 1.8.8

Apply the distributive property.

$x^{2} + 12 x + 12 = A x^{2} - 4 A + B (x \cdot x + x \cdot - 2) + \frac{(C) (x (x + 2) (x - 2))}{x - 2}$

Step 1.8.9

Multiply $x$ by $x$ .

$x^{2} + 12 x + 12 = A x^{2} - 4 A + B (x^{2} + x \cdot - 2) + \frac{(C) (x (x + 2) (x - 2))}{x - 2}$

Step 1.8.10

Move $- 2$ to the left of $x$ .

$x^{2} + 12 x + 12 = A x^{2} - 4 A + B (x^{2} - 2 \cdot x) + \frac{(C) (x (x + 2) (x - 2))}{x - 2}$

Step 1.8.11

Apply the distributive property.

$x^{2} + 12 x + 12 = A x^{2} - 4 A + B x^{2} + B (- 2 x) + \frac{(C) (x (x + 2) (x - 2))}{x - 2}$

Step 1.8.12

Rewrite using the commutative property of multiplication.

$x^{2} + 12 x + 12 = A x^{2} - 4 A + B x^{2} - 2 B x + \frac{(C) (x (x + 2) (x - 2))}{x - 2}$

Step 1.8.13

Cancel the common factor of $x - 2$ .

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Step 1.8.13.1

Cancel the common factor.

$x^{2} + 12 x + 12 = A x^{2} - 4 A + B x^{2} - 2 B x + \frac{C (x (x + 2) (x - 2))}{x - 2}$

Step 1.8.13.2

Divide $(C) (x (x + 2))$ by $1$ .

$x^{2} + 12 x + 12 = A x^{2} - 4 A + B x^{2} - 2 B x + (C) (x (x + 2))$

Step 1.8.14

Apply the distributive property.

$x^{2} + 12 x + 12 = A x^{2} - 4 A + B x^{2} - 2 B x + C (x \cdot x + x \cdot 2)$

Step 1.8.15

Multiply $x$ by $x$ .

$x^{2} + 12 x + 12 = A x^{2} - 4 A + B x^{2} - 2 B x + C (x^{2} + x \cdot 2)$

Step 1.8.16

Move $2$ to the left of $x$ .

$x^{2} + 12 x + 12 = A x^{2} - 4 A + B x^{2} - 2 B x + C (x^{2} + 2 \cdot x)$

Step 1.8.17

Apply the distributive property.

$x^{2} + 12 x + 12 = A x^{2} - 4 A + B x^{2} - 2 B x + C x^{2} + C (2 x)$

Step 1.8.18

Rewrite using the commutative property of multiplication.

$x^{2} + 12 x + 12 = A x^{2} - 4 A + B x^{2} - 2 B x + C x^{2} + 2 C x$

Step 1.9

Simplify the expression.

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Step 1.9.1

Move $B$ .

$x^{2} + 12 x + 12 = A x^{2} - 4 A + B x^{2} - 2 x B + C x^{2} + 2 C x$

Step 1.9.2

Move $- 4 A$ .

$x^{2} + 12 x + 12 = A x^{2} + B x^{2} - 2 x B + C x^{2} + 2 C x - 4 A$

Step 1.9.3

Move $- 2 x B$ .

$x^{2} + 12 x + 12 = A x^{2} + B x^{2} + C x^{2} - 2 x B + 2 C x - 4 A$

Step 2

Create equations for the partial fraction variables and use them to set up a system of equations.

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Step 2.1

Create an equation for the partial fraction variables by equating the coefficients of $x^{2}$ from each side of the equation. For the equation to be equal, the equivalent coefficients on each side of the equation must be equal.

$1 = A + B + C$

Step 2.2

Create an equation for the partial fraction variables by equating the coefficients of $x$ from each side of the equation. For the equation to be equal, the equivalent coefficients on each side of the equation must be equal.

$12 = - 2 B + 2 C$

Step 2.3

Create an equation for the partial fraction variables by equating the coefficients of the terms not containing $x$ . For the equation to be equal, the equivalent coefficients on each side of the equation must be equal.

$12 = - 4 A$

Step 2.4

Set up the system of equations to find the coefficients of the partial fractions.

$1 = A + B + C$

$12 = - 2 B + 2 C$

$12 = - 4 A$

$1 = A + B + C$

$12 = - 2 B + 2 C$

$12 = - 4 A$

Step 3

Solve the system of equations.

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Step 3.1

Solve for $A$ in $12 = - 4 A$ .

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Step 3.1.1

Rewrite the equation as $- 4 A = 12$ .

$- 4 A = 12$

$1 = A + B + C$

$12 = - 2 B + 2 C$

Step 3.1.2

Divide each term in $- 4 A = 12$ by $- 4$ and simplify.

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Step 3.1.2.1

Divide each term in $- 4 A = 12$ by $- 4$ .

$\frac{- 4 A}{- 4} = \frac{12}{- 4}$

$1 = A + B + C$

$12 = - 2 B + 2 C$

Step 3.1.2.2

Simplify the left side.

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Step 3.1.2.2.1

Cancel the common factor of $- 4$ .

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Step 3.1.2.2.1.1

Cancel the common factor.

$\frac{- 4 A}{- 4} = \frac{12}{- 4}$

$1 = A + B + C$

$12 = - 2 B + 2 C$

Step 3.1.2.2.1.2

Divide $A$ by $1$ .

$A = \frac{12}{- 4}$

$1 = A + B + C$

$12 = - 2 B + 2 C$

$A = \frac{12}{- 4}$

$1 = A + B + C$

$12 = - 2 B + 2 C$

$A = \frac{12}{- 4}$

$1 = A + B + C$

$12 = - 2 B + 2 C$

Step 3.1.2.3

Simplify the right side.

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Step 3.1.2.3.1

Divide $12$ by $- 4$ .

$A = - 3$

$1 = A + B + C$

$12 = - 2 B + 2 C$

$A = - 3$

$1 = A + B + C$

$12 = - 2 B + 2 C$

$A = - 3$

$1 = A + B + C$

$12 = - 2 B + 2 C$

$A = - 3$

$1 = A + B + C$

$12 = - 2 B + 2 C$

Step 3.2

Replace all occurrences of $A$ with $- 3$ in each equation.

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Step 3.2.1

Replace all occurrences of $A$ in $1 = A + B + C$ with $- 3$ .

$1 = (- 3) + B + C$

$A = - 3$

$12 = - 2 B + 2 C$

Step 3.2.2

Simplify the right side.

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Step 3.2.2.1

Remove parentheses.

$1 = - 3 + B + C$

$A = - 3$

$12 = - 2 B + 2 C$

$1 = - 3 + B + C$

$A = - 3$

$12 = - 2 B + 2 C$

$1 = - 3 + B + C$

$A = - 3$

$12 = - 2 B + 2 C$

Step 3.3

Solve for $B$ in $1 = - 3 + B + C$ .

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Step 3.3.1

Rewrite the equation as $- 3 + B + C = 1$ .

$- 3 + B + C = 1$

$A = - 3$

$12 = - 2 B + 2 C$

Step 3.3.2

Move all terms not containing $B$ to the right side of the equation.

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Step 3.3.2.1

Add $3$ to both sides of the equation.

$B + C = 1 + 3$

$A = - 3$

$12 = - 2 B + 2 C$

Step 3.3.2.2

Subtract $C$ from both sides of the equation.

$B = 1 + 3 - C$

$A = - 3$

$12 = - 2 B + 2 C$

Step 3.3.2.3

Add $1$ and $3$ .

$B = 4 - C$

$A = - 3$

$12 = - 2 B + 2 C$

$B = 4 - C$

$A = - 3$

$12 = - 2 B + 2 C$

$B = 4 - C$

$A = - 3$

$12 = - 2 B + 2 C$

Step 3.4

Replace all occurrences of $B$ with $4 - C$ in each equation.

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Step 3.4.1

Replace all occurrences of $B$ in $12 = - 2 B + 2 C$ with $4 - C$ .

$12 = - 2 (4 - C) + 2 C$

$B = 4 - C$

$A = - 3$

Step 3.4.2

Simplify the right side.

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Step 3.4.2.1

Simplify $- 2 (4 - C) + 2 C$ .

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Step 3.4.2.1.1

Simplify each term.

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Step 3.4.2.1.1.1

Apply the distributive property.

$12 = - 2 \cdot 4 - 2 (- C) + 2 C$

$B = 4 - C$

$A = - 3$

Step 3.4.2.1.1.2

Multiply $- 2$ by $4$ .

$12 = - 8 - 2 (- C) + 2 C$

$B = 4 - C$

$A = - 3$

Step 3.4.2.1.1.3

Multiply $- 1$ by $- 2$ .

$12 = - 8 + 2 C + 2 C$

$B = 4 - C$

$A = - 3$

$12 = - 8 + 2 C + 2 C$

$B = 4 - C$

$A = - 3$

Step 3.4.2.1.2

Add $2 C$ and $2 C$ .

$12 = - 8 + 4 C$

$B = 4 - C$

$A = - 3$

$12 = - 8 + 4 C$

$B = 4 - C$

$A = - 3$

$12 = - 8 + 4 C$

$B = 4 - C$

$A = - 3$

$12 = - 8 + 4 C$

$B = 4 - C$

$A = - 3$

Step 3.5

Solve for $C$ in $12 = - 8 + 4 C$ .

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Step 3.5.1

Rewrite the equation as $- 8 + 4 C = 12$ .

$- 8 + 4 C = 12$

$B = 4 - C$

$A = - 3$

Step 3.5.2

Move all terms not containing $C$ to the right side of the equation.

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Step 3.5.2.1

Add $8$ to both sides of the equation.

$4 C = 12 + 8$

$B = 4 - C$

$A = - 3$

Step 3.5.2.2

Add $12$ and $8$ .

$4 C = 20$

$B = 4 - C$

$A = - 3$

$4 C = 20$

$B = 4 - C$

$A = - 3$

Step 3.5.3

Divide each term in $4 C = 20$ by $4$ and simplify.

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Step 3.5.3.1

Divide each term in $4 C = 20$ by $4$ .

$\frac{4 C}{4} = \frac{20}{4}$

$B = 4 - C$

$A = - 3$

Step 3.5.3.2

Simplify the left side.

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Step 3.5.3.2.1

Cancel the common factor of $4$ .

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Step 3.5.3.2.1.1

Cancel the common factor.

$\frac{4 C}{4} = \frac{20}{4}$

$B = 4 - C$

$A = - 3$

Step 3.5.3.2.1.2

Divide $C$ by $1$ .

$C = \frac{20}{4}$

$B = 4 - C$

$A = - 3$

$C = \frac{20}{4}$

$B = 4 - C$

$A = - 3$

$C = \frac{20}{4}$

$B = 4 - C$

$A = - 3$

Step 3.5.3.3

Simplify the right side.

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Step 3.5.3.3.1

Divide $20$ by $4$ .

$C = 5$

$B = 4 - C$

$A = - 3$

$C = 5$

$B = 4 - C$

$A = - 3$

$C = 5$

$B = 4 - C$

$A = - 3$

$C = 5$

$B = 4 - C$

$A = - 3$

Step 3.6

Replace all occurrences of $C$ with $5$ in each equation.

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Step 3.6.1

Replace all occurrences of $C$ in $B = 4 - C$ with $5$ .

$B = 4 - (5)$

$C = 5$

$A = - 3$

Step 3.6.2

Simplify the right side.

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Step 3.6.2.1

Simplify $4 - (5)$ .

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Step 3.6.2.1.1

Multiply $- 1$ by $5$ .

$B = 4 - 5$

$C = 5$

$A = - 3$

Step 3.6.2.1.2

Subtract $5$ from $4$ .

$B = - 1$

$C = 5$

$A = - 3$

$B = - 1$

$C = 5$

$A = - 3$

$B = - 1$

$C = 5$

$A = - 3$

$B = - 1$

$C = 5$

$A = - 3$

Step 3.7

List all of the solutions.

$B = - 1, C = 5, A = - 3$

Step 4

Replace each of the partial fraction coefficients in $\frac{A}{x} + \frac{B}{x + 2} + \frac{C}{x - 2}$ with the values found for $A$ , $B$ , and $C$ .

$- \frac{3}{x} + \frac{- 1}{x + 2} + \frac{5}{x - 2}$