Precalculus Examples

$\frac{x + 1}{{(x - 2)}^{2} (x + 4)}$

Step 1

Decompose the fraction and multiply through by the common denominator.

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Step 1.1

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 $A$ .

$\frac{A}{{(x - 2)}^{2}}$

Step 1.2

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

Step 1.3

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

Step 1.4

Multiply each fraction in the equation by the denominator of the original expression. In this case, the denominator is ${(x - 2)}^{2} (x + 4)$ .

$\frac{(x + 1) {(x - 2)}^{2} (x + 4)}{{(x - 2)}^{2} (x + 4)} = \frac{(A) {(x - 2)}^{2} (x + 4)}{{(x - 2)}^{2}} + \frac{(B) {(x - 2)}^{2} (x + 4)}{x - 2} + \frac{(C) {(x - 2)}^{2} (x + 4)}{x + 4}$

Step 1.5

Cancel the common factor of ${(x - 2)}^{2}$ .

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Step 1.5.1

Cancel the common factor.

$\frac{(x + 1) {(x - 2)}^{2} (x + 4)}{{(x - 2)}^{2} (x + 4)} = \frac{(A) {(x - 2)}^{2} (x + 4)}{{(x - 2)}^{2}} + \frac{(B) {(x - 2)}^{2} (x + 4)}{x - 2} + \frac{(C) {(x - 2)}^{2} (x + 4)}{x + 4}$

Step 1.5.2

Rewrite the expression.

$\frac{(x + 1) (x + 4)}{x + 4} = \frac{(A) {(x - 2)}^{2} (x + 4)}{{(x - 2)}^{2}} + \frac{(B) {(x - 2)}^{2} (x + 4)}{x - 2} + \frac{(C) {(x - 2)}^{2} (x + 4)}{x + 4}$

Step 1.6

Cancel the common factor of $x + 4$ .

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Step 1.6.1

Cancel the common factor.

$\frac{(x + 1) (x + 4)}{x + 4} = \frac{(A) {(x - 2)}^{2} (x + 4)}{{(x - 2)}^{2}} + \frac{(B) {(x - 2)}^{2} (x + 4)}{x - 2} + \frac{(C) {(x - 2)}^{2} (x + 4)}{x + 4}$

Step 1.6.2

Divide $x + 1$ by $1$ .

$x + 1 = \frac{(A) {(x - 2)}^{2} (x + 4)}{{(x - 2)}^{2}} + \frac{(B) {(x - 2)}^{2} (x + 4)}{x - 2} + \frac{(C) {(x - 2)}^{2} (x + 4)}{x + 4}$

Step 1.7

Simplify each term.

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Step 1.7.1

Cancel the common factor of ${(x - 2)}^{2}$ .

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Step 1.7.1.1

Cancel the common factor.

$x + 1 = \frac{A {(x - 2)}^{2} (x + 4)}{{(x - 2)}^{2}} + \frac{(B) {(x - 2)}^{2} (x + 4)}{x - 2} + \frac{(C) {(x - 2)}^{2} (x + 4)}{x + 4}$

Step 1.7.1.2

Divide $(A) (x + 4)$ by $1$ .

$x + 1 = (A) (x + 4) + \frac{(B) {(x - 2)}^{2} (x + 4)}{x - 2} + \frac{(C) {(x - 2)}^{2} (x + 4)}{x + 4}$

Step 1.7.2

Apply the distributive property.

$x + 1 = A x + A \cdot 4 + \frac{(B) {(x - 2)}^{2} (x + 4)}{x - 2} + \frac{(C) {(x - 2)}^{2} (x + 4)}{x + 4}$

Step 1.7.3

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

$x + 1 = A x + 4 \cdot A + \frac{(B) {(x - 2)}^{2} (x + 4)}{x - 2} + \frac{(C) {(x - 2)}^{2} (x + 4)}{x + 4}$

Step 1.7.4

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

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Step 1.7.4.1

Factor $x - 2$ out of $(B) {(x - 2)}^{2} (x + 4)$ .

$x + 1 = A x + 4 A + \frac{(x - 2) (B (x - 2) (x + 4))}{x - 2} + \frac{(C) {(x - 2)}^{2} (x + 4)}{x + 4}$

Step 1.7.4.2

Cancel the common factors.

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Step 1.7.4.2.1

Multiply by $1$ .

$x + 1 = A x + 4 A + \frac{(x - 2) (B (x - 2) (x + 4))}{(x - 2) \cdot 1} + \frac{(C) {(x - 2)}^{2} (x + 4)}{x + 4}$

Step 1.7.4.2.2

Cancel the common factor.

$x + 1 = A x + 4 A + \frac{(x - 2) (B (x - 2) (x + 4))}{(x - 2) \cdot 1} + \frac{(C) {(x - 2)}^{2} (x + 4)}{x + 4}$

Step 1.7.4.2.3

Rewrite the expression.

$x + 1 = A x + 4 A + \frac{B (x - 2) (x + 4)}{1} + \frac{(C) {(x - 2)}^{2} (x + 4)}{x + 4}$

Step 1.7.4.2.4

Divide $B (x - 2) (x + 4)$ by $1$ .

$x + 1 = A x + 4 A + B (x - 2) (x + 4) + \frac{(C) {(x - 2)}^{2} (x + 4)}{x + 4}$

Step 1.7.5

Apply the distributive property.

$x + 1 = A x + 4 A + (B x + B \cdot - 2) (x + 4) + \frac{(C) {(x - 2)}^{2} (x + 4)}{x + 4}$

Step 1.7.6

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

$x + 1 = A x + 4 A + (B x - 2 \cdot B) (x + 4) + \frac{(C) {(x - 2)}^{2} (x + 4)}{x + 4}$

Step 1.7.7

Expand $(B x - 2 B) (x + 4)$ using the FOIL Method.

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Step 1.7.7.1

Apply the distributive property.

$x + 1 = A x + 4 A + B x (x + 4) - 2 B (x + 4) + \frac{(C) {(x - 2)}^{2} (x + 4)}{x + 4}$

Step 1.7.7.2

Apply the distributive property.

$x + 1 = A x + 4 A + B x \cdot x + B x \cdot 4 - 2 B (x + 4) + \frac{(C) {(x - 2)}^{2} (x + 4)}{x + 4}$

Step 1.7.7.3

Apply the distributive property.

$x + 1 = A x + 4 A + B x \cdot x + B x \cdot 4 - 2 B x - 2 B \cdot 4 + \frac{(C) {(x - 2)}^{2} (x + 4)}{x + 4}$

Step 1.7.8

Simplify and combine like terms.

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Step 1.7.8.1

Simplify each term.

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Step 1.7.8.1.1

Multiply $x$ by $x$ by adding the exponents.

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Step 1.7.8.1.1.1

Move $x$ .

$x + 1 = A x + 4 A + B (x \cdot x) + B x \cdot 4 - 2 B x - 2 B \cdot 4 + \frac{(C) {(x - 2)}^{2} (x + 4)}{x + 4}$

Step 1.7.8.1.1.2

Multiply $x$ by $x$ .

$x + 1 = A x + 4 A + B x^{2} + B x \cdot 4 - 2 B x - 2 B \cdot 4 + \frac{(C) {(x - 2)}^{2} (x + 4)}{x + 4}$

Step 1.7.8.1.2

Move $4$ to the left of $B x$ .

$x + 1 = A x + 4 A + B x^{2} + 4 \cdot (B x) - 2 B x - 2 B \cdot 4 + \frac{(C) {(x - 2)}^{2} (x + 4)}{x + 4}$

Step 1.7.8.1.3

Multiply $4$ by $- 2$ .

$x + 1 = A x + 4 A + B x^{2} + 4 B x - 2 B x - 8 B + \frac{(C) {(x - 2)}^{2} (x + 4)}{x + 4}$

Step 1.7.8.2

Subtract $2 B x$ from $4 B x$ .

$x + 1 = A x + 4 A + B x^{2} + 2 B x - 8 B + \frac{(C) {(x - 2)}^{2} (x + 4)}{x + 4}$

Step 1.7.9

Cancel the common factor of $x + 4$ .

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Step 1.7.9.1

Cancel the common factor.

$x + 1 = A x + 4 A + B x^{2} + 2 B x - 8 B + \frac{(C) {(x - 2)}^{2} (x + 4)}{x + 4}$

Step 1.7.9.2

Divide $(C) {(x - 2)}^{2}$ by $1$ .

$x + 1 = A x + 4 A + B x^{2} + 2 B x - 8 B + (C) {(x - 2)}^{2}$

Step 1.7.10

Rewrite ${(x - 2)}^{2}$ as $(x - 2) (x - 2)$ .

$x + 1 = A x + 4 A + B x^{2} + 2 B x - 8 B + C ((x - 2) (x - 2))$

Step 1.7.11

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

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Step 1.7.11.1

Apply the distributive property.

$x + 1 = A x + 4 A + B x^{2} + 2 B x - 8 B + C (x (x - 2) - 2 (x - 2))$

Step 1.7.11.2

Apply the distributive property.

$x + 1 = A x + 4 A + B x^{2} + 2 B x - 8 B + C (x \cdot x + x \cdot - 2 - 2 (x - 2))$

Step 1.7.11.3

Apply the distributive property.

$x + 1 = A x + 4 A + B x^{2} + 2 B x - 8 B + C (x \cdot x + x \cdot - 2 - 2 x - 2 \cdot - 2)$

Step 1.7.12

Simplify and combine like terms.

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Step 1.7.12.1

Simplify each term.

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Step 1.7.12.1.1

Multiply $x$ by $x$ .

$x + 1 = A x + 4 A + B x^{2} + 2 B x - 8 B + C (x^{2} + x \cdot - 2 - 2 x - 2 \cdot - 2)$

Step 1.7.12.1.2

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

$x + 1 = A x + 4 A + B x^{2} + 2 B x - 8 B + C (x^{2} - 2 \cdot x - 2 x - 2 \cdot - 2)$

Step 1.7.12.1.3

Multiply $- 2$ by $- 2$ .

$x + 1 = A x + 4 A + B x^{2} + 2 B x - 8 B + C (x^{2} - 2 x - 2 x + 4)$

Step 1.7.12.2

Subtract $2 x$ from $- 2 x$ .

$x + 1 = A x + 4 A + B x^{2} + 2 B x - 8 B + C (x^{2} - 4 x + 4)$

Step 1.7.13

Apply the distributive property.

$x + 1 = A x + 4 A + B x^{2} + 2 B x - 8 B + C x^{2} + C (- 4 x) + C \cdot 4$

Step 1.7.14

Simplify.

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Step 1.7.14.1

Rewrite using the commutative property of multiplication.

$x + 1 = A x + 4 A + B x^{2} + 2 B x - 8 B + C x^{2} - 4 C x + C \cdot 4$

Step 1.7.14.2

Move $4$ to the left of $C$ .

$x + 1 = A x + 4 A + B x^{2} + 2 B x - 8 B + C x^{2} - 4 C x + 4 \cdot C$

$x + 1 = A x + 4 A + B x^{2} + 2 B x - 8 B + C x^{2} - 4 C x + 4 C$

Step 1.8

Simplify the expression.

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Step 1.8.1

Move $B$ .

$x + 1 = A x + 4 A + B x^{2} + 2 x B - 8 B + C x^{2} - 4 C x + 4 C$

Step 1.8.2

Move $- 8 B$ .

$x + 1 = A x + 4 A + B x^{2} + 2 x B + C x^{2} - 4 C x - 8 B + 4 C$

Step 1.8.3

Move $4 A$ .

$x + 1 = A x + B x^{2} + 2 x B + C x^{2} - 4 C x + 4 A - 8 B + 4 C$

Step 1.8.4

Move $2 x B$ .

$x + 1 = A x + B x^{2} + C x^{2} + 2 x B - 4 C x + 4 A - 8 B + 4 C$

Step 1.8.5

Move $A x$ .

$x + 1 = B x^{2} + C x^{2} + A x + 2 x B - 4 C x + 4 A - 8 B + 4 C$

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.

$0 = B + C$

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.

$1 = A + 2 B - 4 C$

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.

$1 = 4 A - 8 B + 4 C$

Step 2.4

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

$0 = B + C$

$1 = A + 2 B - 4 C$

$1 = 4 A - 8 B + 4 C$

$0 = B + C$

$1 = A + 2 B - 4 C$

$1 = 4 A - 8 B + 4 C$

Step 3

Solve the system of equations.

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Step 3.1

Solve for $B$ in $0 = B + C$ .

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Step 3.1.1

Rewrite the equation as $B + C = 0$ .

$B + C = 0$

$1 = A + 2 B - 4 C$

$1 = 4 A - 8 B + 4 C$

Step 3.1.2

Subtract $C$ from both sides of the equation.

$B = - C$

$1 = A + 2 B - 4 C$

$1 = 4 A - 8 B + 4 C$

$B = - C$

$1 = A + 2 B - 4 C$

$1 = 4 A - 8 B + 4 C$

Step 3.2

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

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Step 3.2.1

Replace all occurrences of $B$ in $1 = A + 2 B - 4 C$ with $- C$ .

$1 = A + 2 (- C) - 4 C$

$B = - C$

$1 = 4 A - 8 B + 4 C$

Step 3.2.2

Simplify the right side.

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Step 3.2.2.1

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

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Step 3.2.2.1.1

Multiply $- 1$ by $2$ .

$1 = A - 2 C - 4 C$

$B = - C$

$1 = 4 A - 8 B + 4 C$

Step 3.2.2.1.2

Subtract $4 C$ from $- 2 C$ .

$1 = A - 6 C$

$B = - C$

$1 = 4 A - 8 B + 4 C$

$1 = A - 6 C$

$B = - C$

$1 = 4 A - 8 B + 4 C$

$1 = A - 6 C$

$B = - C$

$1 = 4 A - 8 B + 4 C$

Step 3.2.3

Replace all occurrences of $B$ in $1 = 4 A - 8 B + 4 C$ with $- C$ .

$1 = 4 A - 8 (- C) + 4 C$

$1 = A - 6 C$

$B = - C$

Step 3.2.4

Simplify the right side.

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Step 3.2.4.1

Simplify $4 A - 8 (- C) + 4 C$ .

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Step 3.2.4.1.1

Multiply $- 1$ by $- 8$ .

$1 = 4 A + 8 C + 4 C$

$1 = A - 6 C$

$B = - C$

Step 3.2.4.1.2

Add $8 C$ and $4 C$ .

$1 = 4 A + 12 C$

$1 = A - 6 C$

$B = - C$

$1 = 4 A + 12 C$

$1 = A - 6 C$

$B = - C$

$1 = 4 A + 12 C$

$1 = A - 6 C$

$B = - C$

$1 = 4 A + 12 C$

$1 = A - 6 C$

$B = - C$

Step 3.3

Solve for $A$ in $1 = A - 6 C$ .

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Step 3.3.1

Rewrite the equation as $A - 6 C = 1$ .

$A - 6 C = 1$

$1 = 4 A + 12 C$

$B = - C$

Step 3.3.2

Add $6 C$ to both sides of the equation.

$A = 1 + 6 C$

$1 = 4 A + 12 C$

$B = - C$

$A = 1 + 6 C$

$1 = 4 A + 12 C$

$B = - C$

Step 3.4

Replace all occurrences of $A$ with $1 + 6 C$ in each equation.

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Step 3.4.1

Replace all occurrences of $A$ in $1 = 4 A + 12 C$ with $1 + 6 C$ .

$1 = 4 (1 + 6 C) + 12 C$

$A = 1 + 6 C$

$B = - C$

Step 3.4.2

Simplify the right side.

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Step 3.4.2.1

Simplify $4 (1 + 6 C) + 12 C$ .

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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.

$1 = 4 \cdot 1 + 4 (6 C) + 12 C$

$A = 1 + 6 C$

$B = - C$

Step 3.4.2.1.1.2

Multiply $4$ by $1$ .

$1 = 4 + 4 (6 C) + 12 C$

$A = 1 + 6 C$

$B = - C$

Step 3.4.2.1.1.3

Multiply $6$ by $4$ .

$1 = 4 + 24 C + 12 C$

$A = 1 + 6 C$

$B = - C$

$1 = 4 + 24 C + 12 C$

$A = 1 + 6 C$

$B = - C$

Step 3.4.2.1.2

Add $24 C$ and $12 C$ .

$1 = 4 + 36 C$

$A = 1 + 6 C$

$B = - C$

$1 = 4 + 36 C$

$A = 1 + 6 C$

$B = - C$

$1 = 4 + 36 C$

$A = 1 + 6 C$

$B = - C$

$1 = 4 + 36 C$

$A = 1 + 6 C$

$B = - C$

Step 3.5

Solve for $C$ in $1 = 4 + 36 C$ .

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Step 3.5.1

Rewrite the equation as $4 + 36 C = 1$ .

$4 + 36 C = 1$

$A = 1 + 6 C$

$B = - C$

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

Subtract $4$ from both sides of the equation.

$36 C = 1 - 4$

$A = 1 + 6 C$

$B = - C$

Step 3.5.2.2

Subtract $4$ from $1$ .

$36 C = - 3$

$A = 1 + 6 C$

$B = - C$

$36 C = - 3$

$A = 1 + 6 C$

$B = - C$

Step 3.5.3

Divide each term in $36 C = - 3$ by $36$ and simplify.

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Step 3.5.3.1

Divide each term in $36 C = - 3$ by $36$ .

$\frac{36 C}{36} = \frac{- 3}{36}$

$A = 1 + 6 C$

$B = - C$

Step 3.5.3.2

Simplify the left side.

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Step 3.5.3.2.1

Cancel the common factor of $36$ .

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Step 3.5.3.2.1.1

Cancel the common factor.

$\frac{36 C}{36} = \frac{- 3}{36}$

$A = 1 + 6 C$

$B = - C$

Step 3.5.3.2.1.2

Divide $C$ by $1$ .

$C = \frac{- 3}{36}$

$A = 1 + 6 C$

$B = - C$

$C = \frac{- 3}{36}$

$A = 1 + 6 C$

$B = - C$

$C = \frac{- 3}{36}$

$A = 1 + 6 C$

$B = - C$

Step 3.5.3.3

Simplify the right side.

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Step 3.5.3.3.1

Cancel the common factor of $- 3$ and $36$ .

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Step 3.5.3.3.1.1

Factor $3$ out of $- 3$ .

$C = \frac{3 (- 1)}{36}$

$A = 1 + 6 C$

$B = - C$

Step 3.5.3.3.1.2

Cancel the common factors.

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Step 3.5.3.3.1.2.1

Factor $3$ out of $36$ .

$C = \frac{3 \cdot - 1}{3 \cdot 12}$

$A = 1 + 6 C$

$B = - C$

Step 3.5.3.3.1.2.2

Cancel the common factor.

$C = \frac{3 \cdot - 1}{3 \cdot 12}$

$A = 1 + 6 C$

$B = - C$

Step 3.5.3.3.1.2.3

Rewrite the expression.

$C = \frac{- 1}{12}$

$A = 1 + 6 C$

$B = - C$

$C = \frac{- 1}{12}$

$A = 1 + 6 C$

$B = - C$

$C = \frac{- 1}{12}$

$A = 1 + 6 C$

$B = - C$

Step 3.5.3.3.2

Move the negative in front of the fraction.

$C = - \frac{1}{12}$

$A = 1 + 6 C$

$B = - C$

$C = - \frac{1}{12}$

$A = 1 + 6 C$

$B = - C$

$C = - \frac{1}{12}$

$A = 1 + 6 C$

$B = - C$

$C = - \frac{1}{12}$

$A = 1 + 6 C$

$B = - C$

Step 3.6

Replace all occurrences of $C$ with $- \frac{1}{12}$ in each equation.

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Step 3.6.1

Replace all occurrences of $C$ in $A = 1 + 6 C$ with $- \frac{1}{12}$ .

$A = 1 + 6 (- \frac{1}{12})$

$C = - \frac{1}{12}$

$B = - C$

Step 3.6.2

Simplify the right side.

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Step 3.6.2.1

Simplify $1 + 6 (- \frac{1}{12})$ .

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Step 3.6.2.1.1

Simplify each term.

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Step 3.6.2.1.1.1

Cancel the common factor of $6$ .

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Step 3.6.2.1.1.1.1

Move the leading negative in $- \frac{1}{12}$ into the numerator.

$A = 1 + 6 (\frac{- 1}{12})$

$C = - \frac{1}{12}$

$B = - C$

Step 3.6.2.1.1.1.2

Factor $6$ out of $12$ .

$A = 1 + 6 (\frac{- 1}{6 (2)})$

$C = - \frac{1}{12}$

$B = - C$

Step 3.6.2.1.1.1.3

Cancel the common factor.

$A = 1 + 6 (\frac{- 1}{6 \cdot 2})$

$C = - \frac{1}{12}$

$B = - C$

Step 3.6.2.1.1.1.4

Rewrite the expression.

$A = 1 + \frac{- 1}{2}$

$C = - \frac{1}{12}$

$B = - C$

$A = 1 + \frac{- 1}{2}$

$C = - \frac{1}{12}$

$B = - C$

Step 3.6.2.1.1.2

Move the negative in front of the fraction.

$A = 1 - \frac{1}{2}$

$C = - \frac{1}{12}$

$B = - C$

$A = 1 - \frac{1}{2}$

$C = - \frac{1}{12}$

$B = - C$

Step 3.6.2.1.2

Simplify the expression.

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Step 3.6.2.1.2.1

Write $1$ as a fraction with a common denominator.

$A = \frac{2}{2} - \frac{1}{2}$

$C = - \frac{1}{12}$

$B = - C$

Step 3.6.2.1.2.2

Combine the numerators over the common denominator.

$A = \frac{2 - 1}{2}$

$C = - \frac{1}{12}$

$B = - C$

Step 3.6.2.1.2.3

Subtract $1$ from $2$ .

$A = \frac{1}{2}$

$C = - \frac{1}{12}$

$B = - C$

$A = \frac{1}{2}$

$C = - \frac{1}{12}$

$B = - C$

$A = \frac{1}{2}$

$C = - \frac{1}{12}$

$B = - C$

$A = \frac{1}{2}$

$C = - \frac{1}{12}$

$B = - C$

Step 3.6.3

Replace all occurrences of $C$ in $B = - C$ with $- \frac{1}{12}$ .

$B = - (- \frac{1}{12})$

$A = \frac{1}{2}$

$C = - \frac{1}{12}$

Step 3.6.4

Simplify the right side.

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Step 3.6.4.1

Multiply $- (- \frac{1}{12})$ .

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Step 3.6.4.1.1

Multiply $- 1$ by $- 1$ .

$B = 1 (\frac{1}{12})$

$A = \frac{1}{2}$

$C = - \frac{1}{12}$

Step 3.6.4.1.2

Multiply $\frac{1}{12}$ by $1$ .

$B = \frac{1}{12}$

$A = \frac{1}{2}$

$C = - \frac{1}{12}$

$B = \frac{1}{12}$

$A = \frac{1}{2}$

$C = - \frac{1}{12}$

$B = \frac{1}{12}$

$A = \frac{1}{2}$

$C = - \frac{1}{12}$

$B = \frac{1}{12}$

$A = \frac{1}{2}$

$C = - \frac{1}{12}$

Step 3.7

List all of the solutions.

$B = \frac{1}{12}, A = \frac{1}{2}, C = - \frac{1}{12}$

Step 4

Replace each of the partial fraction coefficients in $\frac{A}{{(x - 2)}^{2}} + \frac{B}{x - 2} + \frac{C}{x + 4}$ with the values found for $A$ , $B$ , and $C$ .

$\frac{\frac{1}{2}}{{(x - 2)}^{2}} + \frac{\frac{1}{12}}{x - 2} + \frac{- \frac{1}{12}}{x + 4}$