Precalculus Examples

$\frac{1}{4 x^{2} - 9}$

Step 1

Decompose the fraction and multiply through by the common denominator.

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Step 1.1

Factor the fraction.

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Step 1.1.1

Rewrite $4 x^{2}$ as ${(2 x)}^{2}$ .

$\frac{1}{{(2 x)}^{2} - 9}$

Step 1.1.2

Rewrite $9$ as $3^{2}$ .

$\frac{1}{{(2 x)}^{2} - 3^{2}}$

Step 1.1.3

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = 2 x$ and $b = 3$ .

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

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 $A$ .

$\frac{A}{2 x + 3}$

Step 1.3

$\frac{A}{2 x + 3} + \frac{B}{2 x - 3}$

Step 1.4

Multiply each fraction in the equation by the denominator of the original expression. In this case, the denominator is $(2 x + 3) (2 x - 3)$ .

$\frac{1 (2 x + 3) (2 x - 3)}{(2 x + 3) (2 x - 3)} = \frac{(A) (2 x + 3) (2 x - 3)}{2 x + 3} + \frac{(B) (2 x + 3) (2 x - 3)}{2 x - 3}$

Step 1.5

Cancel the common factor of $2 x + 3$ .

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Step 1.5.1

Cancel the common factor.

$\frac{1 (2 x + 3) (2 x - 3)}{(2 x + 3) (2 x - 3)} = \frac{(A) (2 x + 3) (2 x - 3)}{2 x + 3} + \frac{(B) (2 x + 3) (2 x - 3)}{2 x - 3}$

Step 1.5.2

Rewrite the expression.

$\frac{1 (2 x - 3)}{2 x - 3} = \frac{(A) (2 x + 3) (2 x - 3)}{2 x + 3} + \frac{(B) (2 x + 3) (2 x - 3)}{2 x - 3}$

Step 1.6

Cancel the common factor of $2 x - 3$ .

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Step 1.6.1

Cancel the common factor.

$\frac{1 (2 x - 3)}{2 x - 3} = \frac{(A) (2 x + 3) (2 x - 3)}{2 x + 3} + \frac{(B) (2 x + 3) (2 x - 3)}{2 x - 3}$

Step 1.6.2

Rewrite the expression.

$1 = \frac{(A) (2 x + 3) (2 x - 3)}{2 x + 3} + \frac{(B) (2 x + 3) (2 x - 3)}{2 x - 3}$

Step 1.7

Simplify each term.

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Step 1.7.1

Cancel the common factor of $2 x + 3$ .

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Step 1.7.1.1

Cancel the common factor.

$1 = \frac{A (2 x + 3) (2 x - 3)}{2 x + 3} + \frac{(B) (2 x + 3) (2 x - 3)}{2 x - 3}$

Step 1.7.1.2

Divide $(A) (2 x - 3)$ by $1$ .

$1 = (A) (2 x - 3) + \frac{(B) (2 x + 3) (2 x - 3)}{2 x - 3}$

Step 1.7.2

Apply the distributive property.

$1 = A (2 x) + A \cdot - 3 + \frac{(B) (2 x + 3) (2 x - 3)}{2 x - 3}$

Step 1.7.3

Rewrite using the commutative property of multiplication.

$1 = 2 A x + A \cdot - 3 + \frac{(B) (2 x + 3) (2 x - 3)}{2 x - 3}$

Step 1.7.4

Move $- 3$ to the left of $A$ .

$1 = 2 A x - 3 \cdot A + \frac{(B) (2 x + 3) (2 x - 3)}{2 x - 3}$

Step 1.7.5

Cancel the common factor of $2 x - 3$ .

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Step 1.7.5.1

Cancel the common factor.

$1 = 2 A x - 3 A + \frac{(B) (2 x + 3) (2 x - 3)}{2 x - 3}$

Step 1.7.5.2

Divide $(B) (2 x + 3)$ by $1$ .

$1 = 2 A x - 3 A + (B) (2 x + 3)$

Step 1.7.6

Apply the distributive property.

$1 = 2 A x - 3 A + B (2 x) + B \cdot 3$

Step 1.7.7

Rewrite using the commutative property of multiplication.

$1 = 2 A x - 3 A + 2 B x + B \cdot 3$

Step 1.7.8

Move $3$ to the left of $B$ .

$1 = 2 A x - 3 A + 2 B x + 3 B$

Step 1.8

Reorder.

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Step 1.8.1

Move $A$ .

$1 = 2 x A - 3 A + 2 B x + 3 B$

Step 1.8.2

Move $B$ .

$1 = 2 x A - 3 A + 2 x B + 3 B$

Step 1.8.3

Move $- 3 A$ .

$1 = 2 x A + 2 x B - 3 A + 3 B$

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.

$0 = 2 A + 2 B$

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.

$1 = - 3 A + 3 B$

Step 2.3

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

$0 = 2 A + 2 B$

$1 = - 3 A + 3 B$

$0 = 2 A + 2 B$

$1 = - 3 A + 3 B$

Step 3

Solve the system of equations.

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Step 3.1

Solve for $A$ in $0 = 2 A + 2 B$ .

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Step 3.1.1

Rewrite the equation as $2 A + 2 B = 0$ .

$2 A + 2 B = 0$

$1 = - 3 A + 3 B$

Step 3.1.2

Subtract $2 B$ from both sides of the equation.

$2 A = - 2 B$

$1 = - 3 A + 3 B$

Step 3.1.3

Divide each term in $2 A = - 2 B$ by $2$ and simplify.

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Step 3.1.3.1

Divide each term in $2 A = - 2 B$ by $2$ .

$\frac{2 A}{2} = \frac{- 2 B}{2}$

$1 = - 3 A + 3 B$

Step 3.1.3.2

Simplify the left side.

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Step 3.1.3.2.1

Cancel the common factor of $2$ .

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Step 3.1.3.2.1.1

Cancel the common factor.

$\frac{2 A}{2} = \frac{- 2 B}{2}$

$1 = - 3 A + 3 B$

Step 3.1.3.2.1.2

Divide $A$ by $1$ .

$A = \frac{- 2 B}{2}$

$1 = - 3 A + 3 B$

$A = \frac{- 2 B}{2}$

$1 = - 3 A + 3 B$

$A = \frac{- 2 B}{2}$

$1 = - 3 A + 3 B$

Step 3.1.3.3

Simplify the right side.

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Step 3.1.3.3.1

Cancel the common factor of $- 2$ and $2$ .

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Step 3.1.3.3.1.1

Factor $2$ out of $- 2 B$ .

$A = \frac{2 (- B)}{2}$

$1 = - 3 A + 3 B$

Step 3.1.3.3.1.2

Cancel the common factors.

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Step 3.1.3.3.1.2.1

Factor $2$ out of $2$ .

$A = \frac{2 (- B)}{2 (1)}$

$1 = - 3 A + 3 B$

Step 3.1.3.3.1.2.2

Cancel the common factor.

$A = \frac{2 (- B)}{2 \cdot 1}$

$1 = - 3 A + 3 B$

Step 3.1.3.3.1.2.3

Rewrite the expression.

$A = \frac{- B}{1}$

$1 = - 3 A + 3 B$

Step 3.1.3.3.1.2.4

Divide $- B$ by $1$ .

$A = - B$

$1 = - 3 A + 3 B$

$A = - B$

$1 = - 3 A + 3 B$

$A = - B$

$1 = - 3 A + 3 B$

$A = - B$

$1 = - 3 A + 3 B$

$A = - B$

$1 = - 3 A + 3 B$

$A = - B$

$1 = - 3 A + 3 B$

Step 3.2

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

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Step 3.2.1

Replace all occurrences of $A$ in $1 = - 3 A + 3 B$ with $- B$ .

$1 = - 3 (- B) + 3 B$

$A = - B$

Step 3.2.2

Simplify the right side.

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Step 3.2.2.1

Simplify $- 3 (- B) + 3 B$ .

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Step 3.2.2.1.1

Multiply $- 1$ by $- 3$ .

$1 = 3 B + 3 B$

$A = - B$

Step 3.2.2.1.2

Add $3 B$ and $3 B$ .

$1 = 6 B$

$A = - B$

$1 = 6 B$

$A = - B$

$1 = 6 B$

$A = - B$

$1 = 6 B$

$A = - B$

Step 3.3

Solve for $B$ in $1 = 6 B$ .

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Step 3.3.1

Rewrite the equation as $6 B = 1$ .

$6 B = 1$

$A = - B$

Step 3.3.2

Divide each term in $6 B = 1$ by $6$ and simplify.

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Step 3.3.2.1

Divide each term in $6 B = 1$ by $6$ .

$\frac{6 B}{6} = \frac{1}{6}$

$A = - B$

Step 3.3.2.2

Simplify the left side.

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Step 3.3.2.2.1

Cancel the common factor of $6$ .

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Step 3.3.2.2.1.1

Cancel the common factor.

$\frac{6 B}{6} = \frac{1}{6}$

$A = - B$

Step 3.3.2.2.1.2

Divide $B$ by $1$ .

$B = \frac{1}{6}$

$A = - B$

$B = \frac{1}{6}$

$A = - B$

$B = \frac{1}{6}$

$A = - B$

$B = \frac{1}{6}$

$A = - B$

$B = \frac{1}{6}$

$A = - B$

Step 3.4

Replace all occurrences of $B$ with $\frac{1}{6}$ in each equation.

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Step 3.4.1

Replace all occurrences of $B$ in $A = - B$ with $\frac{1}{6}$ .

$A = - (\frac{1}{6})$

$B = \frac{1}{6}$

Step 3.4.2

Simplify the right side.

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Step 3.4.2.1

Multiply $- 1$ by $\frac{1}{6}$ .

$A = - \frac{1}{6}$

$B = \frac{1}{6}$

$A = - \frac{1}{6}$

$B = \frac{1}{6}$

$A = - \frac{1}{6}$

$B = \frac{1}{6}$

Step 3.5

List all of the solutions.

$A = - \frac{1}{6}, B = \frac{1}{6}$

Step 4

Replace each of the partial fraction coefficients in $\frac{A}{2 x + 3} + \frac{B}{2 x - 3}$ with the values found for $A$ and $B$ .

$\frac{- \frac{1}{6}}{2 x + 3} + \frac{\frac{1}{6}}{2 x - 3}$