Precalculus Examples

$\frac{x}{(x^{2} + 9) (x + 1)}$

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 is 2nd order, $2$ terms are required in the numerator. The number of terms required in the numerator is always equal to the order of the factor in the denominator.

$\frac{A x + B}{x^{2} + 9}$

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 $C$ .

$\frac{A x + B}{x^{2} + 9} + \frac{C}{x + 1}$

Step 1.3

Multiply each fraction in the equation by the denominator of the original expression. In this case, the denominator is $(x^{2} + 9) (x + 1)$ .

$\frac{x (x^{2} + 9) (x + 1)}{(x^{2} + 9) (x + 1)} = \frac{(A x + B) (x^{2} + 9) (x + 1)}{x^{2} + 9} + \frac{(C) (x^{2} + 9) (x + 1)}{x + 1}$

Step 1.4

Cancel the common factor of $x^{2} + 9$ .

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Step 1.4.1

Cancel the common factor.

$\frac{x (x^{2} + 9) (x + 1)}{(x^{2} + 9) (x + 1)} = \frac{(A x + B) (x^{2} + 9) (x + 1)}{x^{2} + 9} + \frac{(C) (x^{2} + 9) (x + 1)}{x + 1}$

Step 1.4.2

Rewrite the expression.

$\frac{x (x + 1)}{x + 1} = \frac{(A x + B) (x^{2} + 9) (x + 1)}{x^{2} + 9} + \frac{(C) (x^{2} + 9) (x + 1)}{x + 1}$

Step 1.5

Cancel the common factor of $x + 1$ .

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Step 1.5.1

Cancel the common factor.

$\frac{x (x + 1)}{x + 1} = \frac{(A x + B) (x^{2} + 9) (x + 1)}{x^{2} + 9} + \frac{(C) (x^{2} + 9) (x + 1)}{x + 1}$

Step 1.5.2

Divide $x$ by $1$ .

$x = \frac{(A x + B) (x^{2} + 9) (x + 1)}{x^{2} + 9} + \frac{(C) (x^{2} + 9) (x + 1)}{x + 1}$

Step 1.6

Simplify each term.

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Step 1.6.1

Cancel the common factor of $x^{2} + 9$ .

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Step 1.6.1.1

Cancel the common factor.

$x = \frac{(A x + B) (x^{2} + 9) (x + 1)}{x^{2} + 9} + \frac{(C) (x^{2} + 9) (x + 1)}{x + 1}$

Step 1.6.1.2

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

$x = (A x + B) (x + 1) + \frac{(C) (x^{2} + 9) (x + 1)}{x + 1}$

Step 1.6.2

Expand $(A x + B) (x + 1)$ using the FOIL Method.

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Step 1.6.2.1

Apply the distributive property.

$x = A x (x + 1) + B (x + 1) + \frac{(C) (x^{2} + 9) (x + 1)}{x + 1}$

Step 1.6.2.2

Apply the distributive property.

$x = A x \cdot x + A x \cdot 1 + B (x + 1) + \frac{(C) (x^{2} + 9) (x + 1)}{x + 1}$

Step 1.6.2.3

Apply the distributive property.

$x = A x \cdot x + A x \cdot 1 + B x + B \cdot 1 + \frac{(C) (x^{2} + 9) (x + 1)}{x + 1}$

Step 1.6.3

Simplify each term.

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Step 1.6.3.1

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

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Step 1.6.3.1.1

Move $x$ .

$x = A (x \cdot x) + A x \cdot 1 + B x + B \cdot 1 + \frac{(C) (x^{2} + 9) (x + 1)}{x + 1}$

Step 1.6.3.1.2

Multiply $x$ by $x$ .

$x = A x^{2} + A x \cdot 1 + B x + B \cdot 1 + \frac{(C) (x^{2} + 9) (x + 1)}{x + 1}$

Step 1.6.3.2

Multiply $A$ by $1$ .

$x = A x^{2} + A x + B x + B \cdot 1 + \frac{(C) (x^{2} + 9) (x + 1)}{x + 1}$

Step 1.6.3.3

Multiply $B$ by $1$ .

$x = A x^{2} + A x + B x + B + \frac{(C) (x^{2} + 9) (x + 1)}{x + 1}$

Step 1.6.4

Cancel the common factor of $x + 1$ .

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Step 1.6.4.1

Cancel the common factor.

$x = A x^{2} + A x + B x + B + \frac{(C) (x^{2} + 9) (x + 1)}{x + 1}$

Step 1.6.4.2

Divide $(C) (x^{2} + 9)$ by $1$ .

$x = A x^{2} + A x + B x + B + (C) (x^{2} + 9)$

Step 1.6.5

Apply the distributive property.

$x = A x^{2} + A x + B x + B + C x^{2} + C \cdot 9$

Step 1.6.6

Move $9$ to the left of $C$ .

$x = A x^{2} + A x + B x + B + C x^{2} + 9 C$

Step 1.7

Simplify the expression.

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Step 1.7.1

Reorder $B$ and $x$ .

$x = A x^{2} + A x + B x + B + C x^{2} + 9 C$

Step 1.7.2

Move $B$ .

$x = A x^{2} + A x + B x + C x^{2} + B + 9 C$

Step 1.7.3

Move $B x$ .

$x = A x^{2} + A x + C x^{2} + B x + B + 9 C$

Step 1.7.4

Move $A x$ .

$x = A x^{2} + C x^{2} + A x + B x + B + 9 C$

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.

$0 = A + C$

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.

$1 = A + B$

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.

$0 = B + 9 C$

Step 2.4

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

$0 = A + C$

$1 = A + B$

$0 = B + 9 C$

$0 = A + C$

$1 = A + B$

$0 = B + 9 C$

Step 3

Solve the system of equations.

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Step 3.1

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

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Step 3.1.1

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

$A + C = 0$

$1 = A + B$

$0 = B + 9 C$

Step 3.1.2

Subtract $C$ from both sides of the equation.

$A = - C$

$1 = A + B$

$0 = B + 9 C$

$A = - C$

$1 = A + B$

$0 = B + 9 C$

Step 3.2

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

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Step 3.2.1

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

$1 = (- C) + B$

$A = - C$

$0 = B + 9 C$

Step 3.2.2

Simplify the right side.

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Step 3.2.2.1

Remove parentheses.

$1 = - C + B$

$A = - C$

$0 = B + 9 C$

$1 = - C + B$

$A = - C$

$0 = B + 9 C$

$1 = - C + B$

$A = - C$

$0 = B + 9 C$

Step 3.3

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

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Step 3.3.1

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

$- C + B = 1$

$A = - C$

$0 = B + 9 C$

Step 3.3.2

Add $C$ to both sides of the equation.

$B = 1 + C$

$A = - C$

$0 = B + 9 C$

$B = 1 + C$

$A = - C$

$0 = B + 9 C$

Step 3.4

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

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Step 3.4.1

Replace all occurrences of $B$ in $0 = B + 9 C$ with $1 + C$ .

$0 = (1 + C) + 9 C$

$B = 1 + C$

$A = - C$

Step 3.4.2

Simplify the right side.

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Step 3.4.2.1

Add $C$ and $9 C$ .

$0 = 1 + 10 C$

$B = 1 + C$

$A = - C$

$0 = 1 + 10 C$

$B = 1 + C$

$A = - C$

$0 = 1 + 10 C$

$B = 1 + C$

$A = - C$

Step 3.5

Solve for $C$ in $0 = 1 + 10 C$ .

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Step 3.5.1

Rewrite the equation as $1 + 10 C = 0$ .

$1 + 10 C = 0$

$B = 1 + C$

$A = - C$

Step 3.5.2

Subtract $1$ from both sides of the equation.

$10 C = - 1$

$B = 1 + C$

$A = - C$

Step 3.5.3

Divide each term in $10 C = - 1$ by $10$ and simplify.

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Step 3.5.3.1

Divide each term in $10 C = - 1$ by $10$ .

$\frac{10 C}{10} = \frac{- 1}{10}$

$B = 1 + C$

$A = - C$

Step 3.5.3.2

Simplify the left side.

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Step 3.5.3.2.1

Cancel the common factor of $10$ .

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Step 3.5.3.2.1.1

Cancel the common factor.

$\frac{10 C}{10} = \frac{- 1}{10}$

$B = 1 + C$

$A = - C$

Step 3.5.3.2.1.2

Divide $C$ by $1$ .

$C = \frac{- 1}{10}$

$B = 1 + C$

$A = - C$

$C = \frac{- 1}{10}$

$B = 1 + C$

$A = - C$

$C = \frac{- 1}{10}$

$B = 1 + C$

$A = - C$

Step 3.5.3.3

Simplify the right side.

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Step 3.5.3.3.1

Move the negative in front of the fraction.

$C = - \frac{1}{10}$

$B = 1 + C$

$A = - C$

$C = - \frac{1}{10}$

$B = 1 + C$

$A = - C$

$C = - \frac{1}{10}$

$B = 1 + C$

$A = - C$

$C = - \frac{1}{10}$

$B = 1 + C$

$A = - C$

Step 3.6

Replace all occurrences of $C$ with $- \frac{1}{10}$ in each equation.

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Step 3.6.1

Replace all occurrences of $C$ in $B = 1 + C$ with $- \frac{1}{10}$ .

$B = 1 - \frac{1}{10}$

$C = - \frac{1}{10}$

$A = - C$

Step 3.6.2

Simplify $B = 1 - \frac{1}{10}$ .

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Step 3.6.2.1

Simplify the left side.

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Step 3.6.2.1.1

Remove parentheses.

$B = 1 - \frac{1}{10}$

$C = - \frac{1}{10}$

$A = - C$

$B = 1 - \frac{1}{10}$

$C = - \frac{1}{10}$

$A = - C$

Step 3.6.2.2

Simplify the right side.

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Step 3.6.2.2.1

Simplify $1 - \frac{1}{10}$ .

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Step 3.6.2.2.1.1

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

$B = \frac{10}{10} - \frac{1}{10}$

$C = - \frac{1}{10}$

$A = - C$

Step 3.6.2.2.1.2

Combine the numerators over the common denominator.

$B = \frac{10 - 1}{10}$

$C = - \frac{1}{10}$

$A = - C$

Step 3.6.2.2.1.3

Subtract $1$ from $10$ .

$B = \frac{9}{10}$

$C = - \frac{1}{10}$

$A = - C$

$B = \frac{9}{10}$

$C = - \frac{1}{10}$

$A = - C$

$B = \frac{9}{10}$

$C = - \frac{1}{10}$

$A = - C$

$B = \frac{9}{10}$

$C = - \frac{1}{10}$

$A = - C$

Step 3.6.3

Replace all occurrences of $C$ in $A = - C$ with $- \frac{1}{10}$ .

$A = - (- \frac{1}{10})$

$B = \frac{9}{10}$

$C = - \frac{1}{10}$

Step 3.6.4

Simplify the right side.

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Step 3.6.4.1

Multiply $- (- \frac{1}{10})$ .

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Step 3.6.4.1.1

Multiply $- 1$ by $- 1$ .

$A = 1 (\frac{1}{10})$

$B = \frac{9}{10}$

$C = - \frac{1}{10}$

Step 3.6.4.1.2

Multiply $\frac{1}{10}$ by $1$ .

$A = \frac{1}{10}$

$B = \frac{9}{10}$

$C = - \frac{1}{10}$

$A = \frac{1}{10}$

$B = \frac{9}{10}$

$C = - \frac{1}{10}$

$A = \frac{1}{10}$

$B = \frac{9}{10}$

$C = - \frac{1}{10}$

$A = \frac{1}{10}$

$B = \frac{9}{10}$

$C = - \frac{1}{10}$

Step 3.7

List all of the solutions.

$A = \frac{1}{10}, B = \frac{9}{10}, C = - \frac{1}{10}$

Step 4

Replace each of the partial fraction coefficients in $\frac{A x + B}{x^{2} + 9} + \frac{C}{x + 1}$ with the values found for $A$ , $B$ , and $C$ .

$\frac{\frac{1}{10 x} + \frac{9}{10}}{x^{2} + 9} + \frac{- \frac{1}{10}}{x + 1}$

Step 5

Combine $\frac{1}{10}$ and $x$ .

$\frac{\frac{x}{10} + \frac{9}{10}}{x^{2} + 9} + \frac{- \frac{1}{10}}{x + 1}$