Precalculus Examples

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

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 $x^{2} (x - 1)$ .

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

Step 1.3

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

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Step 1.3.1

Cancel the common factor.

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

Step 1.3.2

Rewrite the expression.

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

Step 1.4

Cancel the common factor of $x - 1$ .

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Step 1.4.1

Cancel the common factor.

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

Step 1.4.2

Divide $x - 4$ by $1$ .

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

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.

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

Step 1.5.1.2

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

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

Step 1.5.2

Apply the distributive property.

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

Step 1.5.3

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

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

Step 1.5.4

Rewrite $- 1 A$ as $- A$ .

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

Step 1.5.5

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

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Step 1.5.5.1

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

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

Step 1.5.5.2

Cancel the common factors.

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Step 1.5.5.2.1

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

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

Step 1.5.5.2.2

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

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

Step 1.5.5.2.3

Cancel the common factor.

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

Step 1.5.5.2.4

Rewrite the expression.

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

Step 1.5.5.2.5

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

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

Step 1.5.6

Apply the distributive property.

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

Step 1.5.7

Multiply $x$ by $x$ .

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

Step 1.5.8

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

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

Step 1.5.9

Rewrite $- 1 x$ as $- x$ .

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

Step 1.5.10

Apply the distributive property.

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

Step 1.5.11

Rewrite using the commutative property of multiplication.

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

Step 1.5.12

Cancel the common factor of $x - 1$ .

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Step 1.5.12.1

Cancel the common factor.

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

Step 1.5.12.2

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

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

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

Step 1.6

Simplify the expression.

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Step 1.6.1

Move $B$ .

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

Step 1.6.2

Move $- A$ .

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

Step 1.6.3

Move $- x B$ .

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

Step 1.6.4

Move $A x$ .

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

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.

$0 = B + C$

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.

$1 = A - 1 B$

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.

$- 4 = - 1 A$

Step 2.4

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

$0 = B + C$

$1 = A - 1 B$

$- 4 = - 1 A$

$0 = B + C$

$1 = A - 1 B$

$- 4 = - 1 A$

Step 3

Solve the system of equations.

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Step 3.1

Solve for $A$ in $- 4 = - 1 A$ .

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Step 3.1.1

Rewrite the equation as $- 1 A = - 4$ .

$- 1 A = - 4$

$0 = B + C$

$1 = A - 1 B$

Step 3.1.2

Divide each term in $- 1 A = - 4$ by $- 1$ and simplify.

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Step 3.1.2.1

Divide each term in $- 1 A = - 4$ by $- 1$ .

$\frac{- 1 A}{- 1} = \frac{- 4}{- 1}$

$0 = B + C$

$1 = A - 1 B$

Step 3.1.2.2

Simplify the left side.

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Step 3.1.2.2.1

Dividing two negative values results in a positive value.

$\frac{A}{1} = \frac{- 4}{- 1}$

$0 = B + C$

$1 = A - 1 B$

Step 3.1.2.2.2

Divide $A$ by $1$ .

$A = \frac{- 4}{- 1}$

$0 = B + C$

$1 = A - 1 B$

$A = \frac{- 4}{- 1}$

$0 = B + C$

$1 = A - 1 B$

Step 3.1.2.3

Simplify the right side.

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Step 3.1.2.3.1

Divide $- 4$ by $- 1$ .

$A = 4$

$0 = B + C$

$1 = A - 1 B$

$A = 4$

$0 = B + C$

$1 = A - 1 B$

$A = 4$

$0 = B + C$

$1 = A - 1 B$

$A = 4$

$0 = B + C$

$1 = A - 1 B$

Step 3.2

Replace all occurrences of $A$ with $4$ in each equation.

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Step 3.2.1

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

$1 = (4) - 1 B$

$A = 4$

$0 = B + C$

Step 3.2.2

Simplify the right side.

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Step 3.2.2.1

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

$1 = 4 - B$

$A = 4$

$0 = B + C$

$1 = 4 - B$

$A = 4$

$0 = B + C$

$1 = 4 - B$

$A = 4$

$0 = B + C$

Step 3.3

Solve for $B$ in $1 = 4 - B$ .

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Step 3.3.1

Rewrite the equation as $4 - B = 1$ .

$4 - B = 1$

$A = 4$

$0 = B + C$

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

Subtract $4$ from both sides of the equation.

$- B = 1 - 4$

$A = 4$

$0 = B + C$

Step 3.3.2.2

Subtract $4$ from $1$ .

$- B = - 3$

$A = 4$

$0 = B + C$

$- B = - 3$

$A = 4$

$0 = B + C$

Step 3.3.3

Divide each term in $- B = - 3$ by $- 1$ and simplify.

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Step 3.3.3.1

Divide each term in $- B = - 3$ by $- 1$ .

$\frac{- B}{- 1} = \frac{- 3}{- 1}$

$A = 4$

$0 = B + C$

Step 3.3.3.2

Simplify the left side.

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Step 3.3.3.2.1

Dividing two negative values results in a positive value.

$\frac{B}{1} = \frac{- 3}{- 1}$

$A = 4$

$0 = B + C$

Step 3.3.3.2.2

Divide $B$ by $1$ .

$B = \frac{- 3}{- 1}$

$A = 4$

$0 = B + C$

$B = \frac{- 3}{- 1}$

$A = 4$

$0 = B + C$

Step 3.3.3.3

Simplify the right side.

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Step 3.3.3.3.1

Divide $- 3$ by $- 1$ .

$B = 3$

$A = 4$

$0 = B + C$

$B = 3$

$A = 4$

$0 = B + C$

$B = 3$

$A = 4$

$0 = B + C$

$B = 3$

$A = 4$

$0 = B + C$

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 $0 = B + C$ with $3$ .

$0 = (3) + C$

$B = 3$

$A = 4$

Step 3.4.2

Simplify the right side.

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Step 3.4.2.1

Remove parentheses.

$0 = 3 + C$

$B = 3$

$A = 4$

$0 = 3 + C$

$B = 3$

$A = 4$

$0 = 3 + C$

$B = 3$

$A = 4$

Step 3.5

Solve for $C$ in $0 = 3 + C$ .

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Step 3.5.1

Rewrite the equation as $3 + C = 0$ .

$3 + C = 0$

$B = 3$

$A = 4$

Step 3.5.2

Subtract $3$ from both sides of the equation.

$C = - 3$

$B = 3$

$A = 4$

$C = - 3$

$B = 3$

$A = 4$

Step 3.6

Solve the system of equations.

$C = - 3 B = 3 A = 4$

Step 3.7

List all of the solutions.

$C = - 3, B = 3, A = 4$

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 $C$ .

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