Precalculus Examples

$\frac{9}{x^{3} - 8}$

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 $8$ as $2^{3}$ .

$\frac{9}{x^{3} - 2^{3}}$

Step 1.1.2

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

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

Step 1.1.3

Simplify.

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Step 1.1.3.1

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

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

Step 1.1.3.2

Raise $2$ to the power of $2$ .

$\frac{9}{(x - 2) (x^{2} + 2 x + 4)}$

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 $A$ .

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

Step 1.3

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 is 2nd order, $2$ terms are required in the numerator. The number of terms required in the numerator is always equal to the order of the factor in the denominator.

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

Step 1.4

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

$\frac{9 (x - 2) (x^{2} + 2 x + 4)}{(x - 2) (x^{2} + 2 x + 4)} = \frac{(A) (x - 2) (x^{2} + 2 x + 4)}{x - 2} + \frac{(B x + C) (x - 2) (x^{2} + 2 x + 4)}{x^{2} + 2 x + 4}$

Step 1.5

Cancel the common factor of $x - 2$ .

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Step 1.5.1

Cancel the common factor.

$\frac{9 (x - 2) (x^{2} + 2 x + 4)}{(x - 2) (x^{2} + 2 x + 4)} = \frac{(A) (x - 2) (x^{2} + 2 x + 4)}{x - 2} + \frac{(B x + C) (x - 2) (x^{2} + 2 x + 4)}{x^{2} + 2 x + 4}$

Step 1.5.2

Rewrite the expression.

$\frac{9 (x^{2} + 2 x + 4)}{x^{2} + 2 x + 4} = \frac{(A) (x - 2) (x^{2} + 2 x + 4)}{x - 2} + \frac{(B x + C) (x - 2) (x^{2} + 2 x + 4)}{x^{2} + 2 x + 4}$

Step 1.6

Cancel the common factor of $x^{2} + 2 x + 4$ .

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Step 1.6.1

Cancel the common factor.

$\frac{9 (x^{2} + 2 x + 4)}{x^{2} + 2 x + 4} = \frac{(A) (x - 2) (x^{2} + 2 x + 4)}{x - 2} + \frac{(B x + C) (x - 2) (x^{2} + 2 x + 4)}{x^{2} + 2 x + 4}$

Step 1.6.2

Divide $9$ by $1$ .

$9 = \frac{(A) (x - 2) (x^{2} + 2 x + 4)}{x - 2} + \frac{(B x + C) (x - 2) (x^{2} + 2 x + 4)}{x^{2} + 2 x + 4}$

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.

$9 = \frac{A (x - 2) (x^{2} + 2 x + 4)}{x - 2} + \frac{(B x + C) (x - 2) (x^{2} + 2 x + 4)}{x^{2} + 2 x + 4}$

Step 1.7.1.2

Divide $(A) (x^{2} + 2 x + 4)$ by $1$ .

$9 = (A) (x^{2} + 2 x + 4) + \frac{(B x + C) (x - 2) (x^{2} + 2 x + 4)}{x^{2} + 2 x + 4}$

Step 1.7.2

Apply the distributive property.

$9 = A x^{2} + A (2 x) + A \cdot 4 + \frac{(B x + C) (x - 2) (x^{2} + 2 x + 4)}{x^{2} + 2 x + 4}$

Step 1.7.3

Simplify.

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Step 1.7.3.1

Rewrite using the commutative property of multiplication.

$9 = A x^{2} + 2 A x + A \cdot 4 + \frac{(B x + C) (x - 2) (x^{2} + 2 x + 4)}{x^{2} + 2 x + 4}$

Step 1.7.3.2

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

$9 = A x^{2} + 2 A x + 4 \cdot A + \frac{(B x + C) (x - 2) (x^{2} + 2 x + 4)}{x^{2} + 2 x + 4}$

Step 1.7.4

Cancel the common factor of $x^{2} + 2 x + 4$ .

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Step 1.7.4.1

Cancel the common factor.

$9 = A x^{2} + 2 A x + 4 A + \frac{(B x + C) (x - 2) (x^{2} + 2 x + 4)}{x^{2} + 2 x + 4}$

Step 1.7.4.2

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

$9 = A x^{2} + 2 A x + 4 A + (B x + C) (x - 2)$

Step 1.7.5

Expand $(B x + C) (x - 2)$ using the FOIL Method.

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Step 1.7.5.1

Apply the distributive property.

$9 = A x^{2} + 2 A x + 4 A + B x (x - 2) + C (x - 2)$

Step 1.7.5.2

Apply the distributive property.

$9 = A x^{2} + 2 A x + 4 A + B x \cdot x + B x \cdot - 2 + C (x - 2)$

Step 1.7.5.3

Apply the distributive property.

$9 = A x^{2} + 2 A x + 4 A + B x \cdot x + B x \cdot - 2 + C x + C \cdot - 2$

Step 1.7.6

Simplify each term.

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Step 1.7.6.1

Multiply $x$ by $x$ by adding the exponents.

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Step 1.7.6.1.1

Move $x$ .

$9 = A x^{2} + 2 A x + 4 A + B (x \cdot x) + B x \cdot - 2 + C x + C \cdot - 2$

Step 1.7.6.1.2

Multiply $x$ by $x$ .

$9 = A x^{2} + 2 A x + 4 A + B x^{2} + B x \cdot - 2 + C x + C \cdot - 2$

Step 1.7.6.2

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

$9 = A x^{2} + 2 A x + 4 A + B x^{2} - 2 \cdot (B x) + C x + C \cdot - 2$

Step 1.7.6.3

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

$9 = A x^{2} + 2 A x + 4 A + B x^{2} - 2 B x + C x - 2 C$

Step 1.8

Simplify the expression.

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Step 1.8.1

Move $A$ .

$9 = A x^{2} + 2 x A + 4 A + B x^{2} - 2 B x + C x - 2 C$

Step 1.8.2

Reorder $B$ and $x^{2}$ .

$9 = A x^{2} + 2 x A + 4 A + B x^{2} - 2 B x + C x - 2 C$

Step 1.8.3

Move $B$ .

$9 = A x^{2} + 2 x A + 4 A + B x^{2} - 2 x B + C x - 2 C$

Step 1.8.4

Move $4 A$ .

$9 = A x^{2} + 2 x A + B x^{2} - 2 x B + C x + 4 A - 2 C$

Step 1.8.5

Move $2 x A$ .

$9 = A x^{2} + B x^{2} + 2 x A - 2 x B + C x + 4 A - 2 C$

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 $x^{2}$ from each side of the equation. For the equation to be equal, the equivalent coefficients on each side of the equation must be equal.

$0 = A + B$

Step 2.2

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.

$0 = 2 A - 2 B + C$

Step 2.3

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.

$9 = 4 A - 2 C$

Step 2.4

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

$0 = A + B$

$0 = 2 A - 2 B + C$

$9 = 4 A - 2 C$

$0 = A + B$

$0 = 2 A - 2 B + C$

$9 = 4 A - 2 C$

Step 3

Solve the system of equations.

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Step 3.1

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

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Step 3.1.1

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

$A + B = 0$

$0 = 2 A - 2 B + C$

$9 = 4 A - 2 C$

Step 3.1.2

Subtract $B$ from both sides of the equation.

$A = - B$

$0 = 2 A - 2 B + C$

$9 = 4 A - 2 C$

$A = - B$

$0 = 2 A - 2 B + C$

$9 = 4 A - 2 C$

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 $0 = 2 A - 2 B + C$ with $- B$ .

$0 = 2 (- B) - 2 B + C$

$A = - B$

$9 = 4 A - 2 C$

Step 3.2.2

Simplify the right side.

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Step 3.2.2.1

Simplify $2 (- B) - 2 B + C$ .

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Step 3.2.2.1.1

Multiply $- 1$ by $2$ .

$0 = - 2 B - 2 B + C$

$A = - B$

$9 = 4 A - 2 C$

Step 3.2.2.1.2

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

$0 = - 4 B + C$

$A = - B$

$9 = 4 A - 2 C$

$0 = - 4 B + C$

$A = - B$

$9 = 4 A - 2 C$

$0 = - 4 B + C$

$A = - B$

$9 = 4 A - 2 C$

Step 3.2.3

Replace all occurrences of $A$ in $9 = 4 A - 2 C$ with $- B$ .

$9 = 4 (- B) - 2 C$

$0 = - 4 B + C$

$A = - B$

Step 3.2.4

Simplify the right side.

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Step 3.2.4.1

Multiply $- 1$ by $4$ .

$9 = - 4 B - 2 C$

$0 = - 4 B + C$

$A = - B$

$9 = - 4 B - 2 C$

$0 = - 4 B + C$

$A = - B$

$9 = - 4 B - 2 C$

$0 = - 4 B + C$

$A = - B$

Step 3.3

Solve for $C$ in $0 = - 4 B + C$ .

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Step 3.3.1

Rewrite the equation as $- 4 B + C = 0$ .

$- 4 B + C = 0$

$9 = - 4 B - 2 C$

$A = - B$

Step 3.3.2

Add $4 B$ to both sides of the equation.

$C = 4 B$

$9 = - 4 B - 2 C$

$A = - B$

$C = 4 B$

$9 = - 4 B - 2 C$

$A = - B$

Step 3.4

Replace all occurrences of $C$ with $4 B$ in each equation.

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Step 3.4.1

Replace all occurrences of $C$ in $9 = - 4 B - 2 C$ with $4 B$ .

$9 = - 4 B - 2 (4 B)$

$C = 4 B$

$A = - B$

Step 3.4.2

Simplify the right side.

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Step 3.4.2.1

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

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Step 3.4.2.1.1

Multiply $4$ by $- 2$ .

$9 = - 4 B - 8 B$

$C = 4 B$

$A = - B$

Step 3.4.2.1.2

Subtract $8 B$ from $- 4 B$ .

$9 = - 12 B$

$C = 4 B$

$A = - B$

$9 = - 12 B$

$C = 4 B$

$A = - B$

$9 = - 12 B$

$C = 4 B$

$A = - B$

$9 = - 12 B$

$C = 4 B$

$A = - B$

Step 3.5

Solve for $B$ in $9 = - 12 B$ .

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Step 3.5.1

Rewrite the equation as $- 12 B = 9$ .

$- 12 B = 9$

$C = 4 B$

$A = - B$

Step 3.5.2

Divide each term in $- 12 B = 9$ by $- 12$ and simplify.

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Step 3.5.2.1

Divide each term in $- 12 B = 9$ by $- 12$ .

$\frac{- 12 B}{- 12} = \frac{9}{- 12}$

$C = 4 B$

$A = - B$

Step 3.5.2.2

Simplify the left side.

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Step 3.5.2.2.1

Cancel the common factor of $- 12$ .

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Step 3.5.2.2.1.1

Cancel the common factor.

$\frac{- 12 B}{- 12} = \frac{9}{- 12}$

$C = 4 B$

$A = - B$

Step 3.5.2.2.1.2

Divide $B$ by $1$ .

$B = \frac{9}{- 12}$

$C = 4 B$

$A = - B$

$B = \frac{9}{- 12}$

$C = 4 B$

$A = - B$

$B = \frac{9}{- 12}$

$C = 4 B$

$A = - B$

Step 3.5.2.3

Simplify the right side.

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Step 3.5.2.3.1

Cancel the common factor of $9$ and $- 12$ .

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Step 3.5.2.3.1.1

Factor $3$ out of $9$ .

$B = \frac{3 (3)}{- 12}$

$C = 4 B$

$A = - B$

Step 3.5.2.3.1.2

Cancel the common factors.

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Step 3.5.2.3.1.2.1

Factor $3$ out of $- 12$ .

$B = \frac{3 \cdot 3}{3 \cdot - 4}$

$C = 4 B$

$A = - B$

Step 3.5.2.3.1.2.2

Cancel the common factor.

$B = \frac{3 \cdot 3}{3 \cdot - 4}$

$C = 4 B$

$A = - B$

Step 3.5.2.3.1.2.3

Rewrite the expression.

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

$C = 4 B$

$A = - B$

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

$C = 4 B$

$A = - B$

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

$C = 4 B$

$A = - B$

Step 3.5.2.3.2

Move the negative in front of the fraction.

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

$C = 4 B$

$A = - B$

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

$C = 4 B$

$A = - B$

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

$C = 4 B$

$A = - B$

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

$C = 4 B$

$A = - B$

Step 3.6

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

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Step 3.6.1

Replace all occurrences of $B$ in $C = 4 B$ with $- \frac{3}{4}$ .

$C = 4 (- \frac{3}{4})$

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

$A = - B$

Step 3.6.2

Simplify the right side.

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Step 3.6.2.1

Cancel the common factor of $4$ .

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Step 3.6.2.1.1

Move the leading negative in $- \frac{3}{4}$ into the numerator.

$C = 4 (\frac{- 3}{4})$

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

$A = - B$

Step 3.6.2.1.2

Cancel the common factor.

$C = 4 (\frac{- 3}{4})$

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

$A = - B$

Step 3.6.2.1.3

Rewrite the expression.

$C = - 3$

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

$A = - B$

$C = - 3$

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

$A = - B$

$C = - 3$

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

$A = - B$

Step 3.6.3

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

$A = - (- \frac{3}{4})$

$C = - 3$

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

Step 3.6.4

Simplify the right side.

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Step 3.6.4.1

Multiply $- (- \frac{3}{4})$ .

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Step 3.6.4.1.1

Multiply $- 1$ by $- 1$ .

$A = 1 (\frac{3}{4})$

$C = - 3$

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

Step 3.6.4.1.2

Multiply $\frac{3}{4}$ by $1$ .

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

$C = - 3$

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

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

$C = - 3$

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

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

$C = - 3$

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

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

$C = - 3$

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

Step 3.7

List all of the solutions.

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

Step 4

Replace each of the partial fraction coefficients in $\frac{A}{x - 2} + \frac{B x + C}{x^{2} + 2 x + 4}$ with the values found for $A$ , $B$ , and $C$ .

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

Step 5

Simplify each term.

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Step 5.1

Combine $x$ and $\frac{3}{4}$ .

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

Step 5.2

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

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