Precalculus Examples

$\frac{15}{4 x (3 x + 5)}$

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 $B$ .

$\frac{A}{x} + \frac{B}{3 x + 5}$

Step 1.2

Multiply each fraction in the equation by the denominator of the original expression. In this case, the denominator is $4 x (3 x + 5)$ .

$\frac{15 (4 x (3 x + 5))}{4 x (3 x + 5)} = \frac{A (4 x (3 x + 5))}{x} + \frac{(B) (4 x (3 x + 5))}{3 x + 5}$

Step 1.3

Cancel the common factor of $4$ .

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Step 1.3.1

Cancel the common factor.

$\frac{15 (4 x (3 x + 5))}{4 x (3 x + 5)} = \frac{A (4 x (3 x + 5))}{x} + \frac{(B) (4 x (3 x + 5))}{3 x + 5}$

Step 1.3.2

Rewrite the expression.

$\frac{15 (x (3 x + 5))}{x (3 x + 5)} = \frac{A (4 x (3 x + 5))}{x} + \frac{(B) (4 x (3 x + 5))}{3 x + 5}$

Step 1.4

Cancel the common factor of $x$ .

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Step 1.4.1

Cancel the common factor.

$\frac{15 (x (3 x + 5))}{x (3 x + 5)} = \frac{A (4 x (3 x + 5))}{x} + \frac{(B) (4 x (3 x + 5))}{3 x + 5}$

Step 1.4.2

Rewrite the expression.

$\frac{15 (3 x + 5)}{3 x + 5} = \frac{A (4 x (3 x + 5))}{x} + \frac{(B) (4 x (3 x + 5))}{3 x + 5}$

Step 1.5

Cancel the common factor of $3 x + 5$ .

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Step 1.5.1

Cancel the common factor.

$\frac{15 (3 x + 5)}{3 x + 5} = \frac{A (4 x (3 x + 5))}{x} + \frac{(B) (4 x (3 x + 5))}{3 x + 5}$

Step 1.5.2

Divide $15$ by $1$ .

$15 = \frac{A (4 x (3 x + 5))}{x} + \frac{(B) (4 x (3 x + 5))}{3 x + 5}$

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.

$15 = \frac{A (4 x (3 x + 5))}{x} + \frac{(B) (4 x (3 x + 5))}{3 x + 5}$

Step 1.6.1.2

Divide $A (4 (3 x + 5))$ by $1$ .

$15 = A (4 (3 x + 5)) + \frac{(B) (4 x (3 x + 5))}{3 x + 5}$

Step 1.6.2

Rewrite using the commutative property of multiplication.

$15 = 4 A (3 x + 5) + \frac{(B) (4 x (3 x + 5))}{3 x + 5}$

Step 1.6.3

Apply the distributive property.

$15 = 4 A (3 x) + 4 A \cdot 5 + \frac{(B) (4 x (3 x + 5))}{3 x + 5}$

Step 1.6.4

Rewrite using the commutative property of multiplication.

$15 = 4 \cdot (3 A x) + 4 A \cdot 5 + \frac{(B) (4 x (3 x + 5))}{3 x + 5}$

Step 1.6.5

Multiply $5$ by $4$ .

$15 = 4 \cdot (3 A x) + 20 A + \frac{(B) (4 x (3 x + 5))}{3 x + 5}$

Step 1.6.6

Multiply $4$ by $3$ .

$15 = 12 A x + 20 A + \frac{(B) (4 x (3 x + 5))}{3 x + 5}$

Step 1.6.7

Cancel the common factor of $3 x + 5$ .

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Step 1.6.7.1

Cancel the common factor.

$15 = 12 A x + 20 A + \frac{B (4 x (3 x + 5))}{3 x + 5}$

Step 1.6.7.2

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

$15 = 12 A x + 20 A + (B) (4 x)$

Step 1.6.8

Rewrite using the commutative property of multiplication.

$15 = 12 A x + 20 A + 4 B x$

Step 1.7

Reorder.

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Step 1.7.1

Move $A$ .

$15 = 12 x A + 20 A + 4 B x$

Step 1.7.2

Move $B$ .

$15 = 12 x A + 20 A + 4 x B$

Step 1.7.3

Move $20 A$ .

$15 = 12 x A + 4 x B + 20 A$

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.

$0 = 12 A + 4 B$

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.

$15 = 20 A$

Step 2.3

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

$0 = 12 A + 4 B$

$15 = 20 A$

$0 = 12 A + 4 B$

$15 = 20 A$

Step 3

Solve the system of equations.

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Step 3.1

Solve for $A$ in $15 = 20 A$ .

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Step 3.1.1

Rewrite the equation as $20 A = 15$ .

$20 A = 15$

$0 = 12 A + 4 B$

Step 3.1.2

Divide each term in $20 A = 15$ by $20$ and simplify.

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Step 3.1.2.1

Divide each term in $20 A = 15$ by $20$ .

$\frac{20 A}{20} = \frac{15}{20}$

$0 = 12 A + 4 B$

Step 3.1.2.2

Simplify the left side.

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Step 3.1.2.2.1

Cancel the common factor of $20$ .

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Step 3.1.2.2.1.1

Cancel the common factor.

$\frac{20 A}{20} = \frac{15}{20}$

$0 = 12 A + 4 B$

Step 3.1.2.2.1.2

Divide $A$ by $1$ .

$A = \frac{15}{20}$

$0 = 12 A + 4 B$

$A = \frac{15}{20}$

$0 = 12 A + 4 B$

$A = \frac{15}{20}$

$0 = 12 A + 4 B$

Step 3.1.2.3

Simplify the right side.

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Step 3.1.2.3.1

Cancel the common factor of $15$ and $20$ .

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Step 3.1.2.3.1.1

Factor $5$ out of $15$ .

$A = \frac{5 (3)}{20}$

$0 = 12 A + 4 B$

Step 3.1.2.3.1.2

Cancel the common factors.

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Step 3.1.2.3.1.2.1

Factor $5$ out of $20$ .

$A = \frac{5 \cdot 3}{5 \cdot 4}$

$0 = 12 A + 4 B$

Step 3.1.2.3.1.2.2

Cancel the common factor.

$A = \frac{5 \cdot 3}{5 \cdot 4}$

$0 = 12 A + 4 B$

Step 3.1.2.3.1.2.3

Rewrite the expression.

$A = \frac{3}{4}$

$0 = 12 A + 4 B$

$A = \frac{3}{4}$

$0 = 12 A + 4 B$

$A = \frac{3}{4}$

$0 = 12 A + 4 B$

$A = \frac{3}{4}$

$0 = 12 A + 4 B$

$A = \frac{3}{4}$

$0 = 12 A + 4 B$

$A = \frac{3}{4}$

$0 = 12 A + 4 B$

Step 3.2

Replace all occurrences of $A$ with $\frac{3}{4}$ in each equation.

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Step 3.2.1

Replace all occurrences of $A$ in $0 = 12 A + 4 B$ with $\frac{3}{4}$ .

$0 = 12 (\frac{3}{4}) + 4 B$

$A = \frac{3}{4}$

Step 3.2.2

Simplify the right side.

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Step 3.2.2.1

Simplify each term.

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Step 3.2.2.1.1

Cancel the common factor of $4$ .

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Step 3.2.2.1.1.1

Factor $4$ out of $12$ .

$0 = 4 (3) (\frac{3}{4}) + 4 B$

$A = \frac{3}{4}$

Step 3.2.2.1.1.2

Cancel the common factor.

$0 = 4 \cdot (3 (\frac{3}{4})) + 4 B$

$A = \frac{3}{4}$

Step 3.2.2.1.1.3

Rewrite the expression.

$0 = 3 \cdot 3 + 4 B$

$A = \frac{3}{4}$

$0 = 3 \cdot 3 + 4 B$

$A = \frac{3}{4}$

Step 3.2.2.1.2

Multiply $3$ by $3$ .

$0 = 9 + 4 B$

$A = \frac{3}{4}$

$0 = 9 + 4 B$

$A = \frac{3}{4}$

$0 = 9 + 4 B$

$A = \frac{3}{4}$

$0 = 9 + 4 B$

$A = \frac{3}{4}$

Step 3.3

Solve for $B$ in $0 = 9 + 4 B$ .

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Step 3.3.1

Rewrite the equation as $9 + 4 B = 0$ .

$9 + 4 B = 0$

$A = \frac{3}{4}$

Step 3.3.2

Subtract $9$ from both sides of the equation.

$4 B = - 9$

$A = \frac{3}{4}$

Step 3.3.3

Divide each term in $4 B = - 9$ by $4$ and simplify.

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Step 3.3.3.1

Divide each term in $4 B = - 9$ by $4$ .

$\frac{4 B}{4} = \frac{- 9}{4}$

$A = \frac{3}{4}$

Step 3.3.3.2

Simplify the left side.

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Step 3.3.3.2.1

Cancel the common factor of $4$ .

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Step 3.3.3.2.1.1

Cancel the common factor.

$\frac{4 B}{4} = \frac{- 9}{4}$

$A = \frac{3}{4}$

Step 3.3.3.2.1.2

Divide $B$ by $1$ .

$B = \frac{- 9}{4}$

$A = \frac{3}{4}$

$B = \frac{- 9}{4}$

$A = \frac{3}{4}$

$B = \frac{- 9}{4}$

$A = \frac{3}{4}$

Step 3.3.3.3

Simplify the right side.

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Step 3.3.3.3.1

Move the negative in front of the fraction.

$B = - \frac{9}{4}$

$A = \frac{3}{4}$

$B = - \frac{9}{4}$

$A = \frac{3}{4}$

$B = - \frac{9}{4}$

$A = \frac{3}{4}$

$B = - \frac{9}{4}$

$A = \frac{3}{4}$

Step 3.4

Solve the system of equations.

$B = - \frac{9}{4} A = \frac{3}{4}$

Step 3.5

List all of the solutions.

$B = - \frac{9}{4}, A = \frac{3}{4}$

Step 4

Replace each of the partial fraction coefficients in $\frac{A}{x} + \frac{B}{3 x + 5}$ with the values found for $A$ and $B$ .

$\frac{\frac{3}{4}}{x} + \frac{- \frac{9}{4}}{3 x + 5}$