Precalculus Examples

$\frac{5 x^{2} - 4}{x^{2} (x + 2)}$

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 $C$ .

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

Step 1.2

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

$\frac{(5 x^{2} - 4) (x^{2} (x + 2))}{x^{2} (x + 2)} = \frac{A (x^{2} (x + 2))}{x^{2}} + \frac{B (x^{2} (x + 2))}{x} + \frac{(C) (x^{2} (x + 2))}{x + 2}$

Step 1.3

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

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Step 1.3.1

Cancel the common factor.

$\frac{(5 x^{2} - 4) (x^{2} (x + 2))}{x^{2} (x + 2)} = \frac{A (x^{2} (x + 2))}{x^{2}} + \frac{B (x^{2} (x + 2))}{x} + \frac{(C) (x^{2} (x + 2))}{x + 2}$

Step 1.3.2

Rewrite the expression.

$\frac{(5 x^{2} - 4) (x + 2)}{x + 2} = \frac{A (x^{2} (x + 2))}{x^{2}} + \frac{B (x^{2} (x + 2))}{x} + \frac{(C) (x^{2} (x + 2))}{x + 2}$

Step 1.4

Cancel the common factor of $x + 2$ .

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Step 1.4.1

Cancel the common factor.

$\frac{(5 x^{2} - 4) (x + 2)}{x + 2} = \frac{A (x^{2} (x + 2))}{x^{2}} + \frac{B (x^{2} (x + 2))}{x} + \frac{(C) (x^{2} (x + 2))}{x + 2}$

Step 1.4.2

Divide $5 x^{2} - 4$ by $1$ .

$5 x^{2} - 4 = \frac{A (x^{2} (x + 2))}{x^{2}} + \frac{B (x^{2} (x + 2))}{x} + \frac{(C) (x^{2} (x + 2))}{x + 2}$

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.

$5 x^{2} - 4 = \frac{A (x^{2} (x + 2))}{x^{2}} + \frac{B (x^{2} (x + 2))}{x} + \frac{(C) (x^{2} (x + 2))}{x + 2}$

Step 1.5.1.2

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

$5 x^{2} - 4 = A (x + 2) + \frac{B (x^{2} (x + 2))}{x} + \frac{(C) (x^{2} (x + 2))}{x + 2}$

Step 1.5.2

Apply the distributive property.

$5 x^{2} - 4 = A x + A \cdot 2 + \frac{B (x^{2} (x + 2))}{x} + \frac{(C) (x^{2} (x + 2))}{x + 2}$

Step 1.5.3

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

$5 x^{2} - 4 = A x + 2 \cdot A + \frac{B (x^{2} (x + 2))}{x} + \frac{(C) (x^{2} (x + 2))}{x + 2}$

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 $B (x^{2} (x + 2))$ .

$5 x^{2} - 4 = A x + 2 A + \frac{x (B (x (x + 2)))}{x} + \frac{(C) (x^{2} (x + 2))}{x + 2}$

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 $1$ .

$5 x^{2} - 4 = A x + 2 A + \frac{x (B (x (x + 2)))}{x} + \frac{(C) (x^{2} (x + 2))}{x + 2}$

Step 1.5.4.2.2

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

$5 x^{2} - 4 = A x + 2 A + \frac{x (B (x (x + 2)))}{x \cdot 1} + \frac{(C) (x^{2} (x + 2))}{x + 2}$

Step 1.5.4.2.3

Cancel the common factor.

$5 x^{2} - 4 = A x + 2 A + \frac{x (B (x (x + 2)))}{x \cdot 1} + \frac{(C) (x^{2} (x + 2))}{x + 2}$

Step 1.5.4.2.4

Rewrite the expression.

$5 x^{2} - 4 = A x + 2 A + \frac{B (x (x + 2))}{1} + \frac{(C) (x^{2} (x + 2))}{x + 2}$

Step 1.5.4.2.5

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

$5 x^{2} - 4 = A x + 2 A + B (x (x + 2)) + \frac{(C) (x^{2} (x + 2))}{x + 2}$

Step 1.5.5

Apply the distributive property.

$5 x^{2} - 4 = A x + 2 A + B (x \cdot x + x \cdot 2) + \frac{(C) (x^{2} (x + 2))}{x + 2}$

Step 1.5.6

Multiply $x$ by $x$ .

$5 x^{2} - 4 = A x + 2 A + B (x^{2} + x \cdot 2) + \frac{(C) (x^{2} (x + 2))}{x + 2}$

Step 1.5.7

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

$5 x^{2} - 4 = A x + 2 A + B (x^{2} + 2 \cdot x) + \frac{(C) (x^{2} (x + 2))}{x + 2}$

Step 1.5.8

Apply the distributive property.

$5 x^{2} - 4 = A x + 2 A + B x^{2} + B (2 x) + \frac{(C) (x^{2} (x + 2))}{x + 2}$

Step 1.5.9

Rewrite using the commutative property of multiplication.

$5 x^{2} - 4 = A x + 2 A + B x^{2} + 2 B x + \frac{(C) (x^{2} (x + 2))}{x + 2}$

Step 1.5.10

Cancel the common factor of $x + 2$ .

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Step 1.5.10.1

Cancel the common factor.

$5 x^{2} - 4 = A x + 2 A + B x^{2} + 2 B x + \frac{C (x^{2} (x + 2))}{x + 2}$

Step 1.5.10.2

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

$5 x^{2} - 4 = A x + 2 A + B x^{2} + 2 B x + (C) (x^{2})$

$5 x^{2} - 4 = A x + 2 A + B x^{2} + 2 B x + C x^{2}$

Step 1.6

Simplify the expression.

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Step 1.6.1

Move $B$ .

$5 x^{2} - 4 = A x + 2 A + B x^{2} + 2 x B + C x^{2}$

Step 1.6.2

Move $2 A$ .

$5 x^{2} - 4 = A x + B x^{2} + 2 x B + C x^{2} + 2 A$

Step 1.6.3

Move $2 x B$ .

$5 x^{2} - 4 = A x + B x^{2} + C x^{2} + 2 x B + 2 A$

Step 1.6.4

Move $A x$ .

$5 x^{2} - 4 = B x^{2} + C x^{2} + A x + 2 x B + 2 A$

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.

$5 = B + C$

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.

$0 = A + 2 B$

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.

$- 4 = 2 A$

Step 2.4

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

$5 = B + C$

$0 = A + 2 B$

$- 4 = 2 A$

$5 = B + C$

$0 = A + 2 B$

$- 4 = 2 A$

Step 3

Solve the system of equations.

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Step 3.1

Solve for $A$ in $- 4 = 2 A$ .

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Step 3.1.1

Rewrite the equation as $2 A = - 4$ .

$2 A = - 4$

$5 = B + C$

$0 = A + 2 B$

Step 3.1.2

Divide each term in $2 A = - 4$ by $2$ and simplify.

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Step 3.1.2.1

Divide each term in $2 A = - 4$ by $2$ .

$\frac{2 A}{2} = \frac{- 4}{2}$

$5 = B + C$

$0 = A + 2 B$

Step 3.1.2.2

Simplify the left side.

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Step 3.1.2.2.1

Cancel the common factor of $2$ .

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Step 3.1.2.2.1.1

Cancel the common factor.

$\frac{2 A}{2} = \frac{- 4}{2}$

$5 = B + C$

$0 = A + 2 B$

Step 3.1.2.2.1.2

Divide $A$ by $1$ .

$A = \frac{- 4}{2}$

$5 = B + C$

$0 = A + 2 B$

$A = \frac{- 4}{2}$

$5 = B + C$

$0 = A + 2 B$

$A = \frac{- 4}{2}$

$5 = B + C$

$0 = A + 2 B$

Step 3.1.2.3

Simplify the right side.

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Step 3.1.2.3.1

Divide $- 4$ by $2$ .

$A = - 2$

$5 = B + C$

$0 = A + 2 B$

$A = - 2$

$5 = B + C$

$0 = A + 2 B$

$A = - 2$

$5 = B + C$

$0 = A + 2 B$

$A = - 2$

$5 = B + C$

$0 = A + 2 B$

Step 3.2

Replace all occurrences of $A$ with $- 2$ in each equation.

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Step 3.2.1

Replace all occurrences of $A$ in $0 = A + 2 B$ with $- 2$ .

$0 = (- 2) + 2 B$

$A = - 2$

$5 = B + C$

Step 3.2.2

Simplify the right side.

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Step 3.2.2.1

Remove parentheses.

$0 = - 2 + 2 B$

$A = - 2$

$5 = B + C$

$0 = - 2 + 2 B$

$A = - 2$

$5 = B + C$

$0 = - 2 + 2 B$

$A = - 2$

$5 = B + C$

Step 3.3

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

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Step 3.3.1

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

$- 2 + 2 B = 0$

$A = - 2$

$5 = B + C$

Step 3.3.2

Add $2$ to both sides of the equation.

$2 B = 2$

$A = - 2$

$5 = B + C$

Step 3.3.3

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

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Step 3.3.3.1

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

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

$A = - 2$

$5 = B + C$

Step 3.3.3.2

Simplify the left side.

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Step 3.3.3.2.1

Cancel the common factor of $2$ .

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Step 3.3.3.2.1.1

Cancel the common factor.

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

$A = - 2$

$5 = B + C$

Step 3.3.3.2.1.2

Divide $B$ by $1$ .

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

$A = - 2$

$5 = B + C$

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

$A = - 2$

$5 = B + C$

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

$A = - 2$

$5 = B + C$

Step 3.3.3.3

Simplify the right side.

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Step 3.3.3.3.1

Divide $2$ by $2$ .

$B = 1$

$A = - 2$

$5 = B + C$

$B = 1$

$A = - 2$

$5 = B + C$

$B = 1$

$A = - 2$

$5 = B + C$

$B = 1$

$A = - 2$

$5 = B + C$

Step 3.4

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

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Step 3.4.1

Replace all occurrences of $B$ in $5 = B + C$ with $1$ .

$5 = (1) + C$

$B = 1$

$A = - 2$

Step 3.4.2

Simplify the right side.

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Step 3.4.2.1

Remove parentheses.

$5 = 1 + C$

$B = 1$

$A = - 2$

$5 = 1 + C$

$B = 1$

$A = - 2$

$5 = 1 + C$

$B = 1$

$A = - 2$

Step 3.5

Solve for $C$ in $5 = 1 + C$ .

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Step 3.5.1

Rewrite the equation as $1 + C = 5$ .

$1 + C = 5$

$B = 1$

$A = - 2$

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 $1$ from both sides of the equation.

$C = 5 - 1$

$B = 1$

$A = - 2$

Step 3.5.2.2

Subtract $1$ from $5$ .

$C = 4$

$B = 1$

$A = - 2$

$C = 4$

$B = 1$

$A = - 2$

$C = 4$

$B = 1$

$A = - 2$

Step 3.6

Solve the system of equations.

$C = 4 B = 1 A = - 2$

Step 3.7

List all of the solutions.

$C = 4, B = 1, A = - 2$

Step 4

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

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