Precalculus Examples

$\frac{8 x + 17}{x^{2} + 3 x - 4}$

Step 1

Decompose the fraction and multiply through by the common denominator.

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Step 1.1

Factor $x^{2} + 3 x - 4$ 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 $- 4$ and whose sum is $3$ .

$- 1, 4$

Step 1.1.2

Write the factored form using these integers.

$\frac{8 x + 17}{(x - 1) (x + 4)}$

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 $A$ .

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

Step 1.3

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

Step 1.4

Multiply each fraction in the equation by the denominator of the original expression. In this case, the denominator is $(x - 1) (x + 4)$ .

$\frac{(8 x + 17) (x - 1) (x + 4)}{(x - 1) (x + 4)} = \frac{(A) (x - 1) (x + 4)}{x - 1} + \frac{(B) (x - 1) (x + 4)}{x + 4}$

Step 1.5

Cancel the common factor of $x - 1$ .

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Step 1.5.1

Cancel the common factor.

$\frac{(8 x + 17) (x - 1) (x + 4)}{(x - 1) (x + 4)} = \frac{(A) (x - 1) (x + 4)}{x - 1} + \frac{(B) (x - 1) (x + 4)}{x + 4}$

Step 1.5.2

Rewrite the expression.

$\frac{(8 x + 17) (x + 4)}{x + 4} = \frac{(A) (x - 1) (x + 4)}{x - 1} + \frac{(B) (x - 1) (x + 4)}{x + 4}$

Step 1.6

Cancel the common factor of $x + 4$ .

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Step 1.6.1

Cancel the common factor.

$\frac{(8 x + 17) (x + 4)}{x + 4} = \frac{(A) (x - 1) (x + 4)}{x - 1} + \frac{(B) (x - 1) (x + 4)}{x + 4}$

Step 1.6.2

Divide $8 x + 17$ by $1$ .

$8 x + 17 = \frac{(A) (x - 1) (x + 4)}{x - 1} + \frac{(B) (x - 1) (x + 4)}{x + 4}$

Step 1.7

Simplify each term.

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Step 1.7.1

Cancel the common factor of $x - 1$ .

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Step 1.7.1.1

Cancel the common factor.

$8 x + 17 = \frac{A (x - 1) (x + 4)}{x - 1} + \frac{(B) (x - 1) (x + 4)}{x + 4}$

Step 1.7.1.2

Divide $(A) (x + 4)$ by $1$ .

$8 x + 17 = (A) (x + 4) + \frac{(B) (x - 1) (x + 4)}{x + 4}$

Step 1.7.2

Apply the distributive property.

$8 x + 17 = A x + A \cdot 4 + \frac{(B) (x - 1) (x + 4)}{x + 4}$

Step 1.7.3

Move $4$ to the left of $A$ .

$8 x + 17 = A x + 4 \cdot A + \frac{(B) (x - 1) (x + 4)}{x + 4}$

Step 1.7.4

Cancel the common factor of $x + 4$ .

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Step 1.7.4.1

Cancel the common factor.

$8 x + 17 = A x + 4 A + \frac{(B) (x - 1) (x + 4)}{x + 4}$

Step 1.7.4.2

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

$8 x + 17 = A x + 4 A + (B) (x - 1)$

Step 1.7.5

Apply the distributive property.

$8 x + 17 = A x + 4 A + B x + B \cdot - 1$

Step 1.7.6

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

$8 x + 17 = A x + 4 A + B x - 1 \cdot B$

Step 1.7.7

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

$8 x + 17 = A x + 4 A + B x - B$

Step 1.8

Move $4 A$ .

$8 x + 17 = A x + B x + 4 A - B$

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.

$8 = A + B$

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.

$17 = 4 A - 1 B$

Step 2.3

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

$8 = A + B$

$17 = 4 A - 1 B$

$8 = A + B$

$17 = 4 A - 1 B$

Step 3

Solve the system of equations.

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Step 3.1

Solve for $A$ in $8 = A + B$ .

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Step 3.1.1

Rewrite the equation as $A + B = 8$ .

$A + B = 8$

$17 = 4 A - 1 B$

Step 3.1.2

Subtract $B$ from both sides of the equation.

$A = 8 - B$

$17 = 4 A - 1 B$

$A = 8 - B$

$17 = 4 A - 1 B$

Step 3.2

Replace all occurrences of $A$ with $8 - B$ in each equation.

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Step 3.2.1

Replace all occurrences of $A$ in $17 = 4 A - 1 B$ with $8 - B$ .

$17 = 4 (8 - B) - 1 B$

$A = 8 - B$

Step 3.2.2

Simplify the right side.

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Step 3.2.2.1

Simplify $4 (8 - B) - 1 B$ .

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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.

$17 = 4 \cdot 8 + 4 (- B) - 1 B$

$A = 8 - B$

Step 3.2.2.1.1.2

Multiply $4$ by $8$ .

$17 = 32 + 4 (- B) - 1 B$

$A = 8 - B$

Step 3.2.2.1.1.3

Multiply $- 1$ by $4$ .

$17 = 32 - 4 B - 1 B$

$A = 8 - B$

Step 3.2.2.1.1.4

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

$17 = 32 - 4 B - B$

$A = 8 - B$

$17 = 32 - 4 B - B$

$A = 8 - B$

Step 3.2.2.1.2

Subtract $B$ from $- 4 B$ .

$17 = 32 - 5 B$

$A = 8 - B$

$17 = 32 - 5 B$

$A = 8 - B$

$17 = 32 - 5 B$

$A = 8 - B$

$17 = 32 - 5 B$

$A = 8 - B$

Step 3.3

Solve for $B$ in $17 = 32 - 5 B$ .

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Step 3.3.1

Rewrite the equation as $32 - 5 B = 17$ .

$32 - 5 B = 17$

$A = 8 - B$

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 $32$ from both sides of the equation.

$- 5 B = 17 - 32$

$A = 8 - B$

Step 3.3.2.2

Subtract $32$ from $17$ .

$- 5 B = - 15$

$A = 8 - B$

$- 5 B = - 15$

$A = 8 - B$

Step 3.3.3

Divide each term in $- 5 B = - 15$ by $- 5$ and simplify.

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Step 3.3.3.1

Divide each term in $- 5 B = - 15$ by $- 5$ .

$\frac{- 5 B}{- 5} = \frac{- 15}{- 5}$

$A = 8 - B$

Step 3.3.3.2

Simplify the left side.

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Step 3.3.3.2.1

Cancel the common factor of $- 5$ .

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Step 3.3.3.2.1.1

Cancel the common factor.

$\frac{- 5 B}{- 5} = \frac{- 15}{- 5}$

$A = 8 - B$

Step 3.3.3.2.1.2

Divide $B$ by $1$ .

$B = \frac{- 15}{- 5}$

$A = 8 - B$

$B = \frac{- 15}{- 5}$

$A = 8 - B$

$B = \frac{- 15}{- 5}$

$A = 8 - B$

Step 3.3.3.3

Simplify the right side.

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Step 3.3.3.3.1

Divide $- 15$ by $- 5$ .

$B = 3$

$A = 8 - B$

$B = 3$

$A = 8 - B$

$B = 3$

$A = 8 - B$

$B = 3$

$A = 8 - B$

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 $A = 8 - B$ with $3$ .

$A = 8 - (3)$

$B = 3$

Step 3.4.2

Simplify the right side.

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Step 3.4.2.1

Simplify $8 - (3)$ .

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Step 3.4.2.1.1

Multiply $- 1$ by $3$ .

$A = 8 - 3$

$B = 3$

Step 3.4.2.1.2

Subtract $3$ from $8$ .

$A = 5$

$B = 3$

$A = 5$

$B = 3$

$A = 5$

$B = 3$

$A = 5$

$B = 3$

Step 3.5

List all of the solutions.

$A = 5, B = 3$

Step 4

Replace each of the partial fraction coefficients in $\frac{A}{x - 1} + \frac{B}{x + 4}$ with the values found for $A$ and $B$ .

$\frac{5}{x - 1} + \frac{3}{x + 4}$