Precalculus Examples

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

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 in the denominator is linear, put a single variable in its place $C$ .

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

Step 1.2

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

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

Step 1.3

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

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Step 1.3.1

Cancel the common factor.

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

Step 1.3.2

Rewrite the expression.

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

Step 1.4

Cancel the common factor of $x + 1$ .

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Step 1.4.1

Cancel the common factor.

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

Step 1.4.2

Divide $4 x^{2} + 2 x - 1$ by $1$ .

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

Step 1.5

Simplify each term.

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Step 1.5.1

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

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Step 1.5.1.1

Cancel the common factor.

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

Step 1.5.1.2

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

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

Step 1.5.2

Apply the distributive property.

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

Step 1.5.3

Multiply $A$ by $1$ .

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

Step 1.5.4

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

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Step 1.5.4.1

Factor $x$ out of $B (x^{2} (x + 1))$ .

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

Step 1.5.4.2

Cancel the common factors.

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Step 1.5.4.2.1

Raise $x$ to the power of $1$ .

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

Step 1.5.4.2.2

Factor $x$ out of $x^{1}$ .

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

Step 1.5.4.2.3

Cancel the common factor.

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

Step 1.5.4.2.4

Rewrite the expression.

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

Step 1.5.4.2.5

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

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

Step 1.5.5

Apply the distributive property.

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

Step 1.5.6

Multiply $x$ by $x$ .

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

Step 1.5.7

Multiply $x$ by $1$ .

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

Step 1.5.8

Apply the distributive property.

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

Step 1.5.9

Cancel the common factor of $x + 1$ .

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Step 1.5.9.1

Cancel the common factor.

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

Step 1.5.9.2

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

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

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

Step 1.6

Simplify the expression.

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Step 1.6.1

Reorder $B$ and $x$ .

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

Step 1.6.2

Move $A$ .

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

Step 1.6.3

Move $B x$ .

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

Step 1.6.4

Move $A x$ .

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

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.

$4 = B + C$

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.

$2 = A + B$

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.

$- 1 = A$

Step 2.4

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

$4 = B + C$

$2 = A + B$

$- 1 = A$

$4 = B + C$

$2 = A + B$

$- 1 = A$

Step 3

Solve the system of equations.

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Step 3.1

Rewrite the equation as $A = - 1$ .

$A = - 1$

$4 = B + C$

$2 = A + B$

Step 3.2

Replace all occurrences of $A$ with $- 1$ in each equation.

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Step 3.2.1

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

$2 = (- 1) + B$

$A = - 1$

$4 = B + C$

Step 3.2.2

Simplify the right side.

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Step 3.2.2.1

Remove parentheses.

$2 = - 1 + B$

$A = - 1$

$4 = B + C$

$2 = - 1 + B$

$A = - 1$

$4 = B + C$

$2 = - 1 + B$

$A = - 1$

$4 = B + C$

Step 3.3

Solve for $B$ in $2 = - 1 + B$ .

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Step 3.3.1

Rewrite the equation as $- 1 + B = 2$ .

$- 1 + B = 2$

$A = - 1$

$4 = B + C$

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 $1$ to both sides of the equation.

$B = 2 + 1$

$A = - 1$

$4 = B + C$

Step 3.3.2.2

Add $2$ and $1$ .

$B = 3$

$A = - 1$

$4 = B + C$

$B = 3$

$A = - 1$

$4 = B + C$

$B = 3$

$A = - 1$

$4 = B + C$

Step 3.4

Replace all occurrences of $B$ with $3$ in each equation.

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Step 3.4.1

Replace all occurrences of $B$ in $4 = B + C$ with $3$ .

$4 = (3) + C$

$B = 3$

$A = - 1$

Step 3.4.2

Simplify the right side.

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Step 3.4.2.1

Remove parentheses.

$4 = 3 + C$

$B = 3$

$A = - 1$

$4 = 3 + C$

$B = 3$

$A = - 1$

$4 = 3 + C$

$B = 3$

$A = - 1$

Step 3.5

Solve for $C$ in $4 = 3 + C$ .

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Step 3.5.1

Rewrite the equation as $3 + C = 4$ .

$3 + C = 4$

$B = 3$

$A = - 1$

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 $3$ from both sides of the equation.

$C = 4 - 3$

$B = 3$

$A = - 1$

Step 3.5.2.2

Subtract $3$ from $4$ .

$C = 1$

$B = 3$

$A = - 1$

$C = 1$

$B = 3$

$A = - 1$

$C = 1$

$B = 3$

$A = - 1$

Step 3.6

Solve the system of equations.

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

Step 3.7

List all of the solutions.

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

Step 4

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

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