Precalculus Examples

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

Step 1

Decompose the fraction and multiply through by the common denominator.

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Step 1.1

Factor $x^{2} - x - 2$ using the AC method.

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Step 1.1.1

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 2$ and whose sum is $- 1$ .

$- 2, 1$

Step 1.1.2

Write the factored form using these integers.

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

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 $A$ .

$\frac{A}{x - 2}$

Step 1.3

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

Step 1.4

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

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

Step 1.5

Cancel the common factor of $x - 2$ .

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Step 1.5.1

Cancel the common factor.

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

Step 1.5.2

Rewrite the expression.

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

Step 1.6

Cancel the common factor of $x + 1$ .

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Step 1.6.1

Cancel the common factor.

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

Step 1.6.2

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

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

Step 1.7

Simplify each term.

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Step 1.7.1

Cancel the common factor of $x - 2$ .

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Step 1.7.1.1

Cancel the common factor.

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

Step 1.7.1.2

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

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

Step 1.7.2

Apply the distributive property.

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

Step 1.7.3

Multiply $A$ by $1$ .

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

Step 1.7.4

Cancel the common factor of $x + 1$ .

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Step 1.7.4.1

Cancel the common factor.

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

Step 1.7.4.2

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

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

Step 1.7.5

Apply the distributive property.

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

Step 1.7.6

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

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

Step 1.8

Move $A$ .

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

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.

$- 4 = A - 2 B$

Step 2.3

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

$5 = A + B$

$- 4 = A - 2 B$

$5 = A + B$

$- 4 = A - 2 B$

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 $A + B = 5$ .

$A + B = 5$

$- 4 = A - 2 B$

Step 3.1.2

Subtract $B$ from both sides of the equation.

$A = 5 - B$

$- 4 = A - 2 B$

$A = 5 - B$

$- 4 = A - 2 B$

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 $- 4 = A - 2 B$ with $5 - B$ .

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

$A = 5 - B$

Step 3.2.2

Simplify the right side.

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Step 3.2.2.1

Subtract $2 B$ from $- B$ .

$- 4 = 5 - 3 B$

$A = 5 - B$

$- 4 = 5 - 3 B$

$A = 5 - B$

$- 4 = 5 - 3 B$

$A = 5 - B$

Step 3.3

Solve for $B$ in $- 4 = 5 - 3 B$ .

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Step 3.3.1

Rewrite the equation as $5 - 3 B = - 4$ .

$5 - 3 B = - 4$

$A = 5 - B$

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 $5$ from both sides of the equation.

$- 3 B = - 4 - 5$

$A = 5 - B$

Step 3.3.2.2

Subtract $5$ from $- 4$ .

$- 3 B = - 9$

$A = 5 - B$

$- 3 B = - 9$

$A = 5 - B$

Step 3.3.3

Divide each term in $- 3 B = - 9$ by $- 3$ and simplify.

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Step 3.3.3.1

Divide each term in $- 3 B = - 9$ by $- 3$ .

$\frac{- 3 B}{- 3} = \frac{- 9}{- 3}$

$A = 5 - B$

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.

$\frac{- 3 B}{- 3} = \frac{- 9}{- 3}$

$A = 5 - B$

Step 3.3.3.2.1.2

Divide $B$ by $1$ .

$B = \frac{- 9}{- 3}$

$A = 5 - B$

$B = \frac{- 9}{- 3}$

$A = 5 - B$

$B = \frac{- 9}{- 3}$

$A = 5 - B$

Step 3.3.3.3

Simplify the right side.

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Step 3.3.3.3.1

Divide $- 9$ by $- 3$ .

$B = 3$

$A = 5 - B$

$B = 3$

$A = 5 - B$

$B = 3$

$A = 5 - B$

$B = 3$

$A = 5 - B$

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 $A = 5 - B$ with $3$ .

$A = 5 - (3)$

$B = 3$

Step 3.4.2

Simplify the right side.

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Step 3.4.2.1

Simplify $5 - (3)$ .

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Step 3.4.2.1.1

Multiply $- 1$ by $3$ .

$A = 5 - 3$

$B = 3$

Step 3.4.2.1.2

Subtract $3$ from $5$ .

$A = 2$

$B = 3$

$A = 2$

$B = 3$

$A = 2$

$B = 3$

$A = 2$

$B = 3$

Step 3.5

List all of the solutions.

$A = 2, B = 3$

Step 4

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

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

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