Precalculus Examples

$\frac{x + 14}{x^{2} - 2 x - 8}$

Step 1

Decompose the fraction and multiply through by the common denominator.

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Step 1.1

Factor $x^{2} - 2 x - 8$ 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 $- 8$ and whose sum is $- 2$ .

$- 4, 2$

Step 1.1.2

Write the factored form using these integers.

$\frac{x + 14}{(x - 4) (x + 2)}$

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 $A$ .

$\frac{A}{x - 4}$

Step 1.3

$\frac{A}{x - 4} + \frac{B}{x + 2}$

Step 1.4

Multiply each fraction in the equation by the denominator of the original expression. In this case, the denominator is $(x - 4) (x + 2)$ .

$\frac{(x + 14) (x - 4) (x + 2)}{(x - 4) (x + 2)} = \frac{(A) (x - 4) (x + 2)}{x - 4} + \frac{(B) (x - 4) (x + 2)}{x + 2}$

Step 1.5

Cancel the common factor of $x - 4$ .

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Step 1.5.1

Cancel the common factor.

$\frac{(x + 14) (x - 4) (x + 2)}{(x - 4) (x + 2)} = \frac{(A) (x - 4) (x + 2)}{x - 4} + \frac{(B) (x - 4) (x + 2)}{x + 2}$

Step 1.5.2

Rewrite the expression.

$\frac{(x + 14) (x + 2)}{x + 2} = \frac{(A) (x - 4) (x + 2)}{x - 4} + \frac{(B) (x - 4) (x + 2)}{x + 2}$

Step 1.6

Cancel the common factor of $x + 2$ .

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Step 1.6.1

Cancel the common factor.

$\frac{(x + 14) (x + 2)}{x + 2} = \frac{(A) (x - 4) (x + 2)}{x - 4} + \frac{(B) (x - 4) (x + 2)}{x + 2}$

Step 1.6.2

Divide $x + 14$ by $1$ .

$x + 14 = \frac{(A) (x - 4) (x + 2)}{x - 4} + \frac{(B) (x - 4) (x + 2)}{x + 2}$

Step 1.7

Simplify each term.

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Step 1.7.1

Cancel the common factor of $x - 4$ .

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Step 1.7.1.1

Cancel the common factor.

$x + 14 = \frac{A (x - 4) (x + 2)}{x - 4} + \frac{(B) (x - 4) (x + 2)}{x + 2}$

Step 1.7.1.2

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

$x + 14 = (A) (x + 2) + \frac{(B) (x - 4) (x + 2)}{x + 2}$

Step 1.7.2

Apply the distributive property.

$x + 14 = A x + A \cdot 2 + \frac{(B) (x - 4) (x + 2)}{x + 2}$

Step 1.7.3

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

$x + 14 = A x + 2 \cdot A + \frac{(B) (x - 4) (x + 2)}{x + 2}$

Step 1.7.4

Cancel the common factor of $x + 2$ .

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Step 1.7.4.1

Cancel the common factor.

$x + 14 = A x + 2 A + \frac{(B) (x - 4) (x + 2)}{x + 2}$

Step 1.7.4.2

Divide $(B) (x - 4)$ by $1$ .

$x + 14 = A x + 2 A + (B) (x - 4)$

Step 1.7.5

Apply the distributive property.

$x + 14 = A x + 2 A + B x + B \cdot - 4$

Step 1.7.6

Move $- 4$ to the left of $B$ .

$x + 14 = A x + 2 A + B x - 4 B$

Step 1.8

Move $2 A$ .

$x + 14 = A x + B x + 2 A - 4 B$

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.

$1 = A + B$

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.

$14 = 2 A - 4 B$

Step 2.3

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

$1 = A + B$

$14 = 2 A - 4 B$

$1 = A + B$

$14 = 2 A - 4 B$

Step 3

Solve the system of equations.

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Step 3.1

Solve for $A$ in $1 = A + B$ .

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Step 3.1.1

Rewrite the equation as $A + B = 1$ .

$A + B = 1$

$14 = 2 A - 4 B$

Step 3.1.2

Subtract $B$ from both sides of the equation.

$A = 1 - B$

$14 = 2 A - 4 B$

$A = 1 - B$

$14 = 2 A - 4 B$

Step 3.2

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

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Step 3.2.1

Replace all occurrences of $A$ in $14 = 2 A - 4 B$ with $1 - B$ .

$14 = 2 (1 - B) - 4 B$

$A = 1 - B$

Step 3.2.2

Simplify the right side.

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Step 3.2.2.1

Simplify $2 (1 - B) - 4 B$ .

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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.

$14 = 2 \cdot 1 + 2 (- B) - 4 B$

$A = 1 - B$

Step 3.2.2.1.1.2

Multiply $2$ by $1$ .

$14 = 2 + 2 (- B) - 4 B$

$A = 1 - B$

Step 3.2.2.1.1.3

Multiply $- 1$ by $2$ .

$14 = 2 - 2 B - 4 B$

$A = 1 - B$

$14 = 2 - 2 B - 4 B$

$A = 1 - B$

Step 3.2.2.1.2

Subtract $4 B$ from $- 2 B$ .

$14 = 2 - 6 B$

$A = 1 - B$

$14 = 2 - 6 B$

$A = 1 - B$

$14 = 2 - 6 B$

$A = 1 - B$

$14 = 2 - 6 B$

$A = 1 - B$

Step 3.3

Solve for $B$ in $14 = 2 - 6 B$ .

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Step 3.3.1

Rewrite the equation as $2 - 6 B = 14$ .

$2 - 6 B = 14$

$A = 1 - B$

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 $2$ from both sides of the equation.

$- 6 B = 14 - 2$

$A = 1 - B$

Step 3.3.2.2

Subtract $2$ from $14$ .

$- 6 B = 12$

$A = 1 - B$

$- 6 B = 12$

$A = 1 - B$

Step 3.3.3

Divide each term in $- 6 B = 12$ by $- 6$ and simplify.

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Step 3.3.3.1

Divide each term in $- 6 B = 12$ by $- 6$ .

$\frac{- 6 B}{- 6} = \frac{12}{- 6}$

$A = 1 - B$

Step 3.3.3.2

Simplify the left side.

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Step 3.3.3.2.1

Cancel the common factor of $- 6$ .

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Step 3.3.3.2.1.1

Cancel the common factor.

$\frac{- 6 B}{- 6} = \frac{12}{- 6}$

$A = 1 - B$

Step 3.3.3.2.1.2

Divide $B$ by $1$ .

$B = \frac{12}{- 6}$

$A = 1 - B$

$B = \frac{12}{- 6}$

$A = 1 - B$

$B = \frac{12}{- 6}$

$A = 1 - B$

Step 3.3.3.3

Simplify the right side.

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Step 3.3.3.3.1

Divide $12$ by $- 6$ .

$B = - 2$

$A = 1 - B$

$B = - 2$

$A = 1 - B$

$B = - 2$

$A = 1 - B$

$B = - 2$

$A = 1 - B$

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 $A = 1 - B$ with $- 2$ .

$A = 1 - (- 2)$

$B = - 2$

Step 3.4.2

Simplify the right side.

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Step 3.4.2.1

Simplify $1 - (- 2)$ .

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Step 3.4.2.1.1

Multiply $- 1$ by $- 2$ .

$A = 1 + 2$

$B = - 2$

Step 3.4.2.1.2

Add $1$ 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 $\frac{A}{x - 4} + \frac{B}{x + 2}$ with the values found for $A$ and $B$ .

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