Precalculus Examples

$\frac{5 x^{3} + 32 x - 4}{{(x^{2} + 6)}^{2}}$

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 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.

$\frac{A x + B}{{(x^{2} + 6)}^{2}}$

Step 1.2

$\frac{A x + B}{{(x^{2} + 6)}^{2}} + \frac{C x + D}{x^{2} + 6}$

Step 1.3

Multiply each fraction in the equation by the denominator of the original expression. In this case, the denominator is ${(x^{2} + 6)}^{2}$ .

$\frac{(5 x^{3} + 32 x - 4) {(x^{2} + 6)}^{2}}{{(x^{2} + 6)}^{2}} = \frac{(A x + B) {(x^{2} + 6)}^{2}}{{(x^{2} + 6)}^{2}} + \frac{(C x + D) {(x^{2} + 6)}^{2}}{x^{2} + 6}$

Step 1.4

Cancel the common factor of ${(x^{2} + 6)}^{2}$ .

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Step 1.4.1

Cancel the common factor.

$\frac{(5 x^{3} + 32 x - 4) {(x^{2} + 6)}^{2}}{{(x^{2} + 6)}^{2}} = \frac{(A x + B) {(x^{2} + 6)}^{2}}{{(x^{2} + 6)}^{2}} + \frac{(C x + D) {(x^{2} + 6)}^{2}}{x^{2} + 6}$

Step 1.4.2

Divide $5 x^{3} + 32 x - 4$ by $1$ .

$5 x^{3} + 32 x - 4 = \frac{(A x + B) {(x^{2} + 6)}^{2}}{{(x^{2} + 6)}^{2}} + \frac{(C x + D) {(x^{2} + 6)}^{2}}{x^{2} + 6}$

Step 1.5

Simplify each term.

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Step 1.5.1

Cancel the common factor of ${(x^{2} + 6)}^{2}$ .

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Step 1.5.1.1

Cancel the common factor.

$5 x^{3} + 32 x - 4 = \frac{(A x + B) {(x^{2} + 6)}^{2}}{{(x^{2} + 6)}^{2}} + \frac{(C x + D) {(x^{2} + 6)}^{2}}{x^{2} + 6}$

Step 1.5.1.2

Divide $A x + B$ by $1$ .

$5 x^{3} + 32 x - 4 = A x + B + \frac{(C x + D) {(x^{2} + 6)}^{2}}{x^{2} + 6}$

Step 1.5.2

Cancel the common factor of ${(x^{2} + 6)}^{2}$ and $x^{2} + 6$ .

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Step 1.5.2.1

Factor $x^{2} + 6$ out of $(C x + D) {(x^{2} + 6)}^{2}$ .

$5 x^{3} + 32 x - 4 = A x + B + \frac{(x^{2} + 6) ((C x + D) (x^{2} + 6))}{x^{2} + 6}$

Step 1.5.2.2

Cancel the common factors.

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Step 1.5.2.2.1

Multiply by $1$ .

$5 x^{3} + 32 x - 4 = A x + B + \frac{(x^{2} + 6) ((C x + D) (x^{2} + 6))}{(x^{2} + 6) \cdot 1}$

Step 1.5.2.2.2

Cancel the common factor.

$5 x^{3} + 32 x - 4 = A x + B + \frac{(x^{2} + 6) ((C x + D) (x^{2} + 6))}{(x^{2} + 6) \cdot 1}$

Step 1.5.2.2.3

Rewrite the expression.

$5 x^{3} + 32 x - 4 = A x + B + \frac{(C x + D) (x^{2} + 6)}{1}$

Step 1.5.2.2.4

Divide $(C x + D) (x^{2} + 6)$ by $1$ .

$5 x^{3} + 32 x - 4 = A x + B + (C x + D) (x^{2} + 6)$

Step 1.5.3

Expand $(C x + D) (x^{2} + 6)$ using the FOIL Method.

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Step 1.5.3.1

Apply the distributive property.

$5 x^{3} + 32 x - 4 = A x + B + C x (x^{2} + 6) + D (x^{2} + 6)$

Step 1.5.3.2

Apply the distributive property.

$5 x^{3} + 32 x - 4 = A x + B + C x \cdot x^{2} + C x \cdot 6 + D (x^{2} + 6)$

Step 1.5.3.3

Apply the distributive property.

$5 x^{3} + 32 x - 4 = A x + B + C x \cdot x^{2} + C x \cdot 6 + D x^{2} + D \cdot 6$

Step 1.5.4

Simplify each term.

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Step 1.5.4.1

Multiply $x$ by $x^{2}$ by adding the exponents.

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Step 1.5.4.1.1

Move $x^{2}$ .

$5 x^{3} + 32 x - 4 = A x + B + C (x^{2} x) + C x \cdot 6 + D x^{2} + D \cdot 6$

Step 1.5.4.1.2

Multiply $x^{2}$ by $x$ .

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Step 1.5.4.1.2.1

Raise $x$ to the power of $1$ .

$5 x^{3} + 32 x - 4 = A x + B + C (x^{2} x) + C x \cdot 6 + D x^{2} + D \cdot 6$

Step 1.5.4.1.2.2

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$5 x^{3} + 32 x - 4 = A x + B + C x^{2 + 1} + C x \cdot 6 + D x^{2} + D \cdot 6$

Step 1.5.4.1.3

Add $2$ and $1$ .

$5 x^{3} + 32 x - 4 = A x + B + C x^{3} + C x \cdot 6 + D x^{2} + D \cdot 6$

Step 1.5.4.2

Move $6$ to the left of $C x$ .

$5 x^{3} + 32 x - 4 = A x + B + C x^{3} + 6 \cdot (C x) + D x^{2} + D \cdot 6$

Step 1.5.4.3

Move $6$ to the left of $D$ .

$5 x^{3} + 32 x - 4 = A x + B + C x^{3} + 6 C x + D x^{2} + 6 D$

Step 1.6

Simplify the expression.

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Step 1.6.1

Move $B$ .

$5 x^{3} + 32 x - 4 = A x + C x^{3} + 6 C x + D x^{2} + B + 6 D$

Step 1.6.2

Move $6 C x$ .

$5 x^{3} + 32 x - 4 = A x + C x^{3} + D x^{2} + 6 C x + B + 6 D$

Step 1.6.3

Move $A x$ .

$5 x^{3} + 32 x - 4 = C x^{3} + D x^{2} + A x + 6 C x + B + 6 D$

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 $x^{3}$ from each side of the equation. For the equation to be equal, the equivalent coefficients on each side of the equation must be equal.

$5 = C$

Step 2.2

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.

$0 = D$

Step 2.3

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.

$32 = A + 6 C$

Step 2.4

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.

$- 4 = B + 6 D$

Step 2.5

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

$5 = C$

$0 = D$

$32 = A + 6 C$

$- 4 = B + 6 D$

$5 = C$

$0 = D$

$32 = A + 6 C$

$- 4 = B + 6 D$

Step 3

Solve the system of equations.

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Step 3.1

Rewrite the equation as $C = 5$ .

$C = 5$

$0 = D$

$32 = A + 6 C$

$- 4 = B + 6 D$

Step 3.2

Replace all occurrences of $C$ with $5$ in each equation.

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Step 3.2.1

Rewrite the equation as $D = 0$ .

$D = 0$

$C = 5$

$32 = A + 6 C$

$- 4 = B + 6 D$

Step 3.2.2

Replace all occurrences of $C$ in $32 = A + 6 C$ with $5$ .

$32 = A + 6 (5)$

$D = 0$

$C = 5$

$- 4 = B + 6 D$

Step 3.2.3

Simplify the right side.

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Step 3.2.3.1

Multiply $6$ by $5$ .

$32 = A + 30$

$D = 0$

$C = 5$

$- 4 = B + 6 D$

$32 = A + 30$

$D = 0$

$C = 5$

$- 4 = B + 6 D$

$32 = A + 30$

$D = 0$

$C = 5$

$- 4 = B + 6 D$

Step 3.3

Replace all occurrences of $D$ with $0$ in each equation.

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Step 3.3.1

Rewrite the equation as $A + 30 = 32$ .

$A + 30 = 32$

$D = 0$

$C = 5$

$- 4 = B + 6 D$

Step 3.3.2

Move all terms not containing $A$ to the right side of the equation.

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Step 3.3.2.1

Subtract $30$ from both sides of the equation.

$A = 32 - 30$

$D = 0$

$C = 5$

$- 4 = B + 6 D$

Step 3.3.2.2

Subtract $30$ from $32$ .

$A = 2$

$D = 0$

$C = 5$

$- 4 = B + 6 D$

$A = 2$

$D = 0$

$C = 5$

$- 4 = B + 6 D$

Step 3.3.3

Replace all occurrences of $D$ in $- 4 = B + 6 D$ with $0$ .

$- 4 = B + 6 (0)$

$A = 2$

$D = 0$

$C = 5$

Step 3.3.4

Simplify the right side.

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Step 3.3.4.1

Simplify $B + 6 (0)$ .

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Step 3.3.4.1.1

Multiply $6$ by $0$ .

$- 4 = B + 0$

$A = 2$

$D = 0$

$C = 5$

Step 3.3.4.1.2

Add $B$ and $0$ .

$- 4 = B$

$A = 2$

$D = 0$

$C = 5$

$- 4 = B$

$A = 2$

$D = 0$

$C = 5$

$- 4 = B$

$A = 2$

$D = 0$

$C = 5$

$- 4 = B$

$A = 2$

$D = 0$

$C = 5$

Step 3.4

Rewrite the equation as $B = - 4$ .

$B = - 4$

$A = 2$

$D = 0$

$C = 5$

Step 3.5

Solve the system of equations.

$B = - 4 A = 2 D = 0 C = 5$

Step 3.6

List all of the solutions.

$B = - 4, A = 2, D = 0, C = 5$

Step 4

Replace each of the partial fraction coefficients in $\frac{A x + B}{{(x^{2} + 6)}^{2}} + \frac{C x + D}{x^{2} + 6}$ with the values found for $A$ , $B$ , $C$ , and $D$ .

$\frac{2 x - 4}{{(x^{2} + 6)}^{2}} + \frac{5 x + 0}{x^{2} + 6}$

Step 5

Simplify.

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Step 5.1

Multiply $2$ by $x$ .

$\frac{2 x - 4}{{(x^{2} + 6)}^{2}} + \frac{(5) \cdot x + 0}{x^{2} + 6}$

Step 5.2

Add $(5) \cdot x$ and $0$ .

$\frac{2 x - 4}{{(x^{2} + 6)}^{2}} + \frac{5 x}{x^{2} + 6}$