Precalculus Examples

$- \frac{5 x - 4}{(x + 1) (x - 2)}$

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 $A$ .

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

Step 1.2

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

Step 1.3

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

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

Step 1.4

Cancel the common factor of $x + 1$ .

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Step 1.4.1

Cancel the common factor.

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

Step 1.4.2

Rewrite the expression.

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

Step 1.5

Cancel the common factor of $x - 2$ .

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Step 1.5.1

Cancel the common factor.

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

Step 1.5.2

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

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

Step 1.6

Apply the distributive property.

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

Step 1.7

Multiply.

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Step 1.7.1

Multiply $5$ by $- 1$ .

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

Step 1.7.2

Multiply $- 1$ by $- 4$ .

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

Step 1.8

Simplify each term.

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Step 1.8.1

Cancel the common factor of $x + 1$ .

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Step 1.8.1.1

Cancel the common factor.

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

Step 1.8.1.2

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

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

Step 1.8.2

Apply the distributive property.

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

Step 1.8.3

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

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

Step 1.8.4

Cancel the common factor of $x - 2$ .

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Step 1.8.4.1

Cancel the common factor.

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

Step 1.8.4.2

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

$- 5 x + 4 = A x - 2 A + (B) (x + 1)$

Step 1.8.5

Apply the distributive property.

$- 5 x + 4 = A x - 2 A + B x + B \cdot 1$

Step 1.8.6

Multiply $B$ by $1$ .

$- 5 x + 4 = A x - 2 A + B x + B$

Step 1.9

Move $- 2 A$ .

$- 5 x + 4 = A x + B x - 2 A + B$

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.

$4 = - 2 A + B$

Step 2.3

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

$- 5 = A + B$

$4 = - 2 A + B$

$- 5 = A + B$

$4 = - 2 A + B$

Step 3

Solve the system of equations.

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Step 3.1

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

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Step 3.1.1

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

$A + B = - 5$

$4 = - 2 A + B$

Step 3.1.2

Subtract $B$ from both sides of the equation.

$A = - 5 - B$

$4 = - 2 A + B$

$A = - 5 - B$

$4 = - 2 A + B$

Step 3.2

Replace all occurrences of $A$ with $- 5 - B$ in each equation.

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Step 3.2.1

Replace all occurrences of $A$ in $4 = - 2 A + B$ with $- 5 - B$ .

$4 = - 2 (- 5 - B) + B$

$A = - 5 - B$

Step 3.2.2

Simplify the right side.

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Step 3.2.2.1

Simplify $- 2 (- 5 - B) + B$ .

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Step 3.2.2.1.1

Simplify each term.

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Step 3.2.2.1.1.1

Apply the distributive property.

$4 = - 2 \cdot - 5 - 2 (- B) + B$

$A = - 5 - B$

Step 3.2.2.1.1.2

Multiply $- 2$ by $- 5$ .

$4 = 10 - 2 (- B) + B$

$A = - 5 - B$

Step 3.2.2.1.1.3

Multiply $- 1$ by $- 2$ .

$4 = 10 + 2 B + B$

$A = - 5 - B$

$4 = 10 + 2 B + B$

$A = - 5 - B$

Step 3.2.2.1.2

Add $2 B$ and $B$ .

$4 = 10 + 3 B$

$A = - 5 - B$

$4 = 10 + 3 B$

$A = - 5 - B$

$4 = 10 + 3 B$

$A = - 5 - B$

$4 = 10 + 3 B$

$A = - 5 - B$

Step 3.3

Solve for $B$ in $4 = 10 + 3 B$ .

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Step 3.3.1

Rewrite the equation as $10 + 3 B = 4$ .

$10 + 3 B = 4$

$A = - 5 - B$

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 $10$ from both sides of the equation.

$3 B = 4 - 10$

$A = - 5 - B$

Step 3.3.2.2

Subtract $10$ from $4$ .

$3 B = - 6$

$A = - 5 - B$

$3 B = - 6$

$A = - 5 - B$

Step 3.3.3

Divide each term in $3 B = - 6$ by $3$ and simplify.

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Step 3.3.3.1

Divide each term in $3 B = - 6$ by $3$ .

$\frac{3 B}{3} = \frac{- 6}{3}$

$A = - 5 - B$

Step 3.3.3.2

Simplify the left side.

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Step 3.3.3.2.1

Cancel the common factor of $3$ .

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Step 3.3.3.2.1.1

Cancel the common factor.

$\frac{3 B}{3} = \frac{- 6}{3}$

$A = - 5 - B$

Step 3.3.3.2.1.2

Divide $B$ by $1$ .

$B = \frac{- 6}{3}$

$A = - 5 - B$

$B = \frac{- 6}{3}$

$A = - 5 - B$

$B = \frac{- 6}{3}$

$A = - 5 - B$

Step 3.3.3.3

Simplify the right side.

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Step 3.3.3.3.1

Divide $- 6$ by $3$ .

$B = - 2$

$A = - 5 - B$

$B = - 2$

$A = - 5 - B$

$B = - 2$

$A = - 5 - B$

$B = - 2$

$A = - 5 - B$

Step 3.4

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

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Step 3.4.1

Replace all occurrences of $B$ in $A = - 5 - B$ with $- 2$ .

$A = - 5 - (- 2)$

$B = - 2$

Step 3.4.2

Simplify the right side.

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Step 3.4.2.1

Simplify $- 5 - (- 2)$ .

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Step 3.4.2.1.1

Multiply $- 1$ by $- 2$ .

$A = - 5 + 2$

$B = - 2$

Step 3.4.2.1.2

Add $- 5$ and $2$ .

$A = - 3$

$B = - 2$

$A = - 3$

$B = - 2$

$A = - 3$

$B = - 2$

$A = - 3$

$B = - 2$

Step 3.5

List all of the solutions.

$A = - 3, B = - 2$

Step 4

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

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