Precalculus Examples

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

Step 1

Decompose the fraction and multiply through by the common denominator.

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Step 1.1

Factor $x$ out of $x^{2} - x$ .

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Step 1.1.1

Factor $x$ out of $x^{2}$ .

$\frac{- 5 x + 4}{x \cdot x - x}$

Step 1.1.2

Factor $x$ out of $- x$ .

$\frac{- 5 x + 4}{x \cdot x + x \cdot - 1}$

Step 1.1.3

Factor $x$ out of $x \cdot x + x \cdot - 1$ .

$\frac{- 5 x + 4}{x (x - 1)}$

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 $B$ .

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

Step 1.3

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

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

Step 1.4

Cancel the common factor of $x$ .

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Step 1.4.1

Cancel the common factor.

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

Step 1.4.2

Rewrite the expression.

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

Step 1.5

Cancel the common factor of $x - 1$ .

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Step 1.5.1

Cancel the common factor.

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

Step 1.5.2

Divide $- 5 x + 4$ by $1$ .

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

Step 1.6

Simplify each term.

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Step 1.6.1

Cancel the common factor of $x$ .

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Step 1.6.1.1

Cancel the common factor.

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

Step 1.6.1.2

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

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

Step 1.6.2

Apply the distributive property.

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

Step 1.6.3

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

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

Step 1.6.4

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

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

Step 1.6.5

Cancel the common factor of $x - 1$ .

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Step 1.6.5.1

Cancel the common factor.

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

Step 1.6.5.2

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

$- 5 x + 4 = A x - A + (B) (x)$

$- 5 x + 4 = A x - A + B x$

Step 1.7

Move $- A$ .

$- 5 x + 4 = A x + B x - A$

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 $x$ from each side of the equation. For the equation to be equal, the equivalent coefficients on each side of the equation must be equal.

$- 5 = A + B$

Step 2.2

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 2.3

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

$- 5 = A + B$

$4 = - 1 A$

$- 5 = A + B$

$4 = - 1 A$

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 $- 1 A = 4$ .

$- 1 A = 4$

$- 5 = A + B$

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 $- 1$ .

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

$- 5 = A + B$

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.

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

$- 5 = A + B$

Step 3.1.2.2.2

Divide $A$ by $1$ .

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

$- 5 = A + B$

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

$- 5 = A + B$

Step 3.1.2.3

Simplify the right side.

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Step 3.1.2.3.1

Divide $4$ by $- 1$ .

$A = - 4$

$- 5 = A + B$

$A = - 4$

$- 5 = A + B$

$A = - 4$

$- 5 = A + B$

$A = - 4$

$- 5 = A + B$

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 $- 5 = A + B$ with $- 4$ .

$- 5 = (- 4) + B$

$A = - 4$

Step 3.2.2

Simplify the right side.

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Step 3.2.2.1

Remove parentheses.

$- 5 = - 4 + B$

$A = - 4$

$- 5 = - 4 + B$

$A = - 4$

$- 5 = - 4 + B$

$A = - 4$

Step 3.3

Solve for $B$ in $- 5 = - 4 + B$ .

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Step 3.3.1

Rewrite the equation as $- 4 + B = - 5$ .

$- 4 + B = - 5$

$A = - 4$

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 $4$ to both sides of the equation.

$B = - 5 + 4$

$A = - 4$

Step 3.3.2.2

Add $- 5$ and $4$ .

$B = - 1$

$A = - 4$

$B = - 1$

$A = - 4$

$B = - 1$

$A = - 4$

Step 3.4

Solve the system of equations.

$B = - 1 A = - 4$

Step 3.5

List all of the solutions.

$B = - 1, A = - 4$

Step 4

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

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