Precalculus Examples

$\frac{2}{x^{3} - 1}$

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

Rewrite $1$ as $1^{3}$ .

$\frac{2}{x^{3} - 1^{3}}$

Step 1.1.2

Since both terms are perfect cubes, factor using the difference of cubes formula, $a^{3} - b^{3} = (a - b) (a^{2} + a b + b^{2})$ where $a = x$ and $b = 1$ .

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

Step 1.1.3

Simplify.

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Step 1.1.3.1

Multiply $x$ by $1$ .

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

Step 1.1.3.2

One to any power is one.

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

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 $A$ .

$\frac{A}{x - 1}$

Step 1.3

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.

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

Step 1.4

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

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

Step 1.5

Cancel the common factor of $x - 1$ .

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Step 1.5.1

Cancel the common factor.

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

Step 1.5.2

Rewrite the expression.

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

Step 1.6

Cancel the common factor of $x^{2} + x + 1$ .

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Step 1.6.1

Cancel the common factor.

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

Step 1.6.2

Divide $2$ by $1$ .

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

Step 1.7

Simplify each term.

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Step 1.7.1

Cancel the common factor of $x - 1$ .

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Step 1.7.1.1

Cancel the common factor.

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

Step 1.7.1.2

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

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

Step 1.7.2

Apply the distributive property.

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

Step 1.7.3

Multiply $A$ by $1$ .

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

Step 1.7.4

Cancel the common factor of $x^{2} + x + 1$ .

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Step 1.7.4.1

Cancel the common factor.

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

Step 1.7.4.2

Divide $(B x + C) (x - 1)$ by $1$ .

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

Step 1.7.5

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

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Step 1.7.5.1

Apply the distributive property.

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

Step 1.7.5.2

Apply the distributive property.

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

Step 1.7.5.3

Apply the distributive property.

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

Step 1.7.6

Simplify each term.

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Step 1.7.6.1

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

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Step 1.7.6.1.1

Move $x$ .

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

Step 1.7.6.1.2

Multiply $x$ by $x$ .

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

Step 1.7.6.2

Move $- 1$ to the left of $B x$ .

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

Step 1.7.6.3

Rewrite $- 1 (B x)$ as $- (B x)$ .

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

Step 1.7.6.4

Move $- 1$ to the left of $C$ .

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

Step 1.7.6.5

Rewrite $- 1 C$ as $- C$ .

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

Step 1.8

Simplify the expression.

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Step 1.8.1

Reorder $B$ and $x^{2}$ .

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

Step 1.8.2

Move $B$ .

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

Step 1.8.3

Move $A$ .

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

Step 1.8.4

Move $A x$ .

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

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.

$0 = A + B$

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.

$0 = A - 1 B + C$

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.

$2 = A - 1 C$

Step 2.4

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

$0 = A + B$

$0 = A - 1 B + C$

$2 = A - 1 C$

$0 = A + B$

$0 = A - 1 B + C$

$2 = A - 1 C$

Step 3

Solve the system of equations.

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Step 3.1

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

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Step 3.1.1

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

$A + B = 0$

$0 = A - 1 B + C$

$2 = A - 1 C$

Step 3.1.2

Subtract $B$ from both sides of the equation.

$A = - B$

$0 = A - 1 B + C$

$2 = A - 1 C$

$A = - B$

$0 = A - 1 B + C$

$2 = A - 1 C$

Step 3.2

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

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Step 3.2.1

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

$0 = (- B) - 1 B + C$

$A = - B$

$2 = A - 1 C$

Step 3.2.2

Simplify the right side.

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Step 3.2.2.1

Simplify $(- B) - 1 B + C$ .

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Step 3.2.2.1.1

Rewrite $- 1 B$ as $- B$ .

$0 = - B - B + C$

$A = - B$

$2 = A - 1 C$

Step 3.2.2.1.2

Subtract $B$ from $- B$ .

$0 = - 2 B + C$

$A = - B$

$2 = A - 1 C$

$0 = - 2 B + C$

$A = - B$

$2 = A - 1 C$

$0 = - 2 B + C$

$A = - B$

$2 = A - 1 C$

Step 3.2.3

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

$2 = (- B) - 1 C$

$0 = - 2 B + C$

$A = - B$

Step 3.2.4

Simplify the right side.

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Step 3.2.4.1

Rewrite $- 1 C$ as $- C$ .

$2 = - B - C$

$0 = - 2 B + C$

$A = - B$

$2 = - B - C$

$0 = - 2 B + C$

$A = - B$

$2 = - B - C$

$0 = - 2 B + C$

$A = - B$

Step 3.3

Solve for $C$ in $0 = - 2 B + C$ .

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Step 3.3.1

Rewrite the equation as $- 2 B + C = 0$ .

$- 2 B + C = 0$

$2 = - B - C$

$A = - B$

Step 3.3.2

Add $2 B$ to both sides of the equation.

$C = 2 B$

$2 = - B - C$

$A = - B$

$C = 2 B$

$2 = - B - C$

$A = - B$

Step 3.4

Replace all occurrences of $C$ with $2 B$ in each equation.

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Step 3.4.1

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

$2 = - B - (2 B)$

$C = 2 B$

$A = - B$

Step 3.4.2

Simplify the right side.

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Step 3.4.2.1

Simplify $- B - (2 B)$ .

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Step 3.4.2.1.1

Multiply $2$ by $- 1$ .

$2 = - B - 2 B$

$C = 2 B$

$A = - B$

Step 3.4.2.1.2

Subtract $2 B$ from $- B$ .

$2 = - 3 B$

$C = 2 B$

$A = - B$

$2 = - 3 B$

$C = 2 B$

$A = - B$

$2 = - 3 B$

$C = 2 B$

$A = - B$

$2 = - 3 B$

$C = 2 B$

$A = - B$

Step 3.5

Solve for $B$ in $2 = - 3 B$ .

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Step 3.5.1

Rewrite the equation as $- 3 B = 2$ .

$- 3 B = 2$

$C = 2 B$

$A = - B$

Step 3.5.2

Divide each term in $- 3 B = 2$ by $- 3$ and simplify.

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Step 3.5.2.1

Divide each term in $- 3 B = 2$ by $- 3$ .

$\frac{- 3 B}{- 3} = \frac{2}{- 3}$

$C = 2 B$

$A = - B$

Step 3.5.2.2

Simplify the left side.

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Step 3.5.2.2.1

Cancel the common factor of $- 3$ .

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Step 3.5.2.2.1.1

Cancel the common factor.

$\frac{- 3 B}{- 3} = \frac{2}{- 3}$

$C = 2 B$

$A = - B$

Step 3.5.2.2.1.2

Divide $B$ by $1$ .

$B = \frac{2}{- 3}$

$C = 2 B$

$A = - B$

$B = \frac{2}{- 3}$

$C = 2 B$

$A = - B$

$B = \frac{2}{- 3}$

$C = 2 B$

$A = - B$

Step 3.5.2.3

Simplify the right side.

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Step 3.5.2.3.1

Move the negative in front of the fraction.

$B = - \frac{2}{3}$

$C = 2 B$

$A = - B$

$B = - \frac{2}{3}$

$C = 2 B$

$A = - B$

$B = - \frac{2}{3}$

$C = 2 B$

$A = - B$

$B = - \frac{2}{3}$

$C = 2 B$

$A = - B$

Step 3.6

Replace all occurrences of $B$ with $- \frac{2}{3}$ in each equation.

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Step 3.6.1

Replace all occurrences of $B$ in $C = 2 B$ with $- \frac{2}{3}$ .

$C = 2 (- \frac{2}{3})$

$B = - \frac{2}{3}$

$A = - B$

Step 3.6.2

Simplify the right side.

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Step 3.6.2.1

Simplify $2 (- \frac{2}{3})$ .

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Step 3.6.2.1.1

Multiply $2 (- \frac{2}{3})$ .

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Step 3.6.2.1.1.1

Multiply $- 1$ by $2$ .

$C = - 2 (\frac{2}{3})$

$B = - \frac{2}{3}$

$A = - B$

Step 3.6.2.1.1.2

Combine $- 2$ and $\frac{2}{3}$ .

$C = \frac{- 2 \cdot 2}{3}$

$B = - \frac{2}{3}$

$A = - B$

Step 3.6.2.1.1.3

Multiply $- 2$ by $2$ .

$C = \frac{- 4}{3}$

$B = - \frac{2}{3}$

$A = - B$

$C = \frac{- 4}{3}$

$B = - \frac{2}{3}$

$A = - B$

Step 3.6.2.1.2

Move the negative in front of the fraction.

$C = - \frac{4}{3}$

$B = - \frac{2}{3}$

$A = - B$

$C = - \frac{4}{3}$

$B = - \frac{2}{3}$

$A = - B$

$C = - \frac{4}{3}$

$B = - \frac{2}{3}$

$A = - B$

Step 3.6.3

Replace all occurrences of $B$ in $A = - B$ with $- \frac{2}{3}$ .

$A = - (- \frac{2}{3})$

$C = - \frac{4}{3}$

$B = - \frac{2}{3}$

Step 3.6.4

Simplify the right side.

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Step 3.6.4.1

Multiply $- (- \frac{2}{3})$ .

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Step 3.6.4.1.1

Multiply $- 1$ by $- 1$ .

$A = 1 (\frac{2}{3})$

$C = - \frac{4}{3}$

$B = - \frac{2}{3}$

Step 3.6.4.1.2

Multiply $\frac{2}{3}$ by $1$ .

$A = \frac{2}{3}$

$C = - \frac{4}{3}$

$B = - \frac{2}{3}$

$A = \frac{2}{3}$

$C = - \frac{4}{3}$

$B = - \frac{2}{3}$

$A = \frac{2}{3}$

$C = - \frac{4}{3}$

$B = - \frac{2}{3}$

$A = \frac{2}{3}$

$C = - \frac{4}{3}$

$B = - \frac{2}{3}$

Step 3.7

List all of the solutions.

$A = \frac{2}{3}, C = - \frac{4}{3}, B = - \frac{2}{3}$

Step 4

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

$\frac{\frac{2}{3}}{x - 1} + \frac{- \frac{2}{3 x} - \frac{4}{3}}{x^{2} + x + 1}$

Step 5

Simplify each term.

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Step 5.1

Combine $x$ and $\frac{2}{3}$ .

$\frac{\frac{2}{3}}{x - 1} + \frac{- \frac{x \cdot 2}{3} - \frac{4}{3}}{x^{2} + x + 1}$

Step 5.2

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

$\frac{\frac{2}{3}}{x - 1} + \frac{- \frac{2 x}{3} - \frac{4}{3}}{x^{2} + x + 1}$