Precalculus Examples

$\frac{6 x^{2} + 19 x}{(x + 4) (x^{2} + 4)}$

Step 1

Decompose the fraction and multiply through by the common denominator.

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Step 1.1

Factor $x$ out of $6 x^{2} + 19 x$ .

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Step 1.1.1

Factor $x$ out of $6 x^{2}$ .

$\frac{x (6 x) + 19 x}{(x + 4) (x^{2} + 4)}$

Step 1.1.2

Factor $x$ out of $19 x$ .

$\frac{x (6 x) + x \cdot 19}{(x + 4) (x^{2} + 4)}$

Step 1.1.3

Factor $x$ out of $x (6 x) + x \cdot 19$ .

$\frac{x (6 x + 19)}{(x + 4) (x^{2} + 4)}$

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 $A$ .

$\frac{A}{x + 4}$

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.

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

Step 1.4

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

$\frac{x (6 x + 19) (x + 4) (x^{2} + 4)}{(x + 4) (x^{2} + 4)} = \frac{(A) (x + 4) (x^{2} + 4)}{x + 4} + \frac{(B x + C) (x + 4) (x^{2} + 4)}{x^{2} + 4}$

Step 1.5

Simplify terms.

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Step 1.5.1

Cancel the common factor of $x + 4$ .

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Step 1.5.1.1

Cancel the common factor.

$\frac{x (6 x + 19) (x + 4) (x^{2} + 4)}{(x + 4) (x^{2} + 4)} = \frac{(A) (x + 4) (x^{2} + 4)}{x + 4} + \frac{(B x + C) (x + 4) (x^{2} + 4)}{x^{2} + 4}$

Step 1.5.1.2

Rewrite the expression.

$\frac{x (6 x + 19) (x^{2} + 4)}{x^{2} + 4} = \frac{(A) (x + 4) (x^{2} + 4)}{x + 4} + \frac{(B x + C) (x + 4) (x^{2} + 4)}{x^{2} + 4}$

Step 1.5.2

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

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Step 1.5.2.1

Cancel the common factor.

$\frac{x (6 x + 19) (x^{2} + 4)}{x^{2} + 4} = \frac{(A) (x + 4) (x^{2} + 4)}{x + 4} + \frac{(B x + C) (x + 4) (x^{2} + 4)}{x^{2} + 4}$

Step 1.5.2.2

Divide $x (6 x + 19)$ by $1$ .

$x (6 x + 19) = \frac{(A) (x + 4) (x^{2} + 4)}{x + 4} + \frac{(B x + C) (x + 4) (x^{2} + 4)}{x^{2} + 4}$

Step 1.5.3

Apply the distributive property.

$x (6 x) + x \cdot 19 = \frac{(A) (x + 4) (x^{2} + 4)}{x + 4} + \frac{(B x + C) (x + 4) (x^{2} + 4)}{x^{2} + 4}$

Step 1.5.4

Reorder.

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Step 1.5.4.1

Rewrite using the commutative property of multiplication.

$6 x \cdot x + x \cdot 19 = \frac{(A) (x + 4) (x^{2} + 4)}{x + 4} + \frac{(B x + C) (x + 4) (x^{2} + 4)}{x^{2} + 4}$

Step 1.5.4.2

Move $19$ to the left of $x$ .

$6 x \cdot x + 19 \cdot x = \frac{(A) (x + 4) (x^{2} + 4)}{x + 4} + \frac{(B x + C) (x + 4) (x^{2} + 4)}{x^{2} + 4}$

Step 1.6

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

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Step 1.6.1

Move $x$ .

$6 (x \cdot x) + 19 \cdot x = \frac{(A) (x + 4) (x^{2} + 4)}{x + 4} + \frac{(B x + C) (x + 4) (x^{2} + 4)}{x^{2} + 4}$

Step 1.6.2

Multiply $x$ by $x$ .

$6 x^{2} + 19 \cdot x = \frac{(A) (x + 4) (x^{2} + 4)}{x + 4} + \frac{(B x + C) (x + 4) (x^{2} + 4)}{x^{2} + 4}$

$6 x^{2} + 19 x = \frac{(A) (x + 4) (x^{2} + 4)}{x + 4} + \frac{(B x + C) (x + 4) (x^{2} + 4)}{x^{2} + 4}$

Step 1.7

Simplify each term.

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Step 1.7.1

Cancel the common factor of $x + 4$ .

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Step 1.7.1.1

Cancel the common factor.

$6 x^{2} + 19 x = \frac{A (x + 4) (x^{2} + 4)}{x + 4} + \frac{(B x + C) (x + 4) (x^{2} + 4)}{x^{2} + 4}$

Step 1.7.1.2

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

$6 x^{2} + 19 x = (A) (x^{2} + 4) + \frac{(B x + C) (x + 4) (x^{2} + 4)}{x^{2} + 4}$

Step 1.7.2

Apply the distributive property.

$6 x^{2} + 19 x = A x^{2} + A \cdot 4 + \frac{(B x + C) (x + 4) (x^{2} + 4)}{x^{2} + 4}$

Step 1.7.3

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

$6 x^{2} + 19 x = A x^{2} + 4 \cdot A + \frac{(B x + C) (x + 4) (x^{2} + 4)}{x^{2} + 4}$

Step 1.7.4

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

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Step 1.7.4.1

Cancel the common factor.

$6 x^{2} + 19 x = A x^{2} + 4 A + \frac{(B x + C) (x + 4) (x^{2} + 4)}{x^{2} + 4}$

Step 1.7.4.2

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

$6 x^{2} + 19 x = A x^{2} + 4 A + (B x + C) (x + 4)$

Step 1.7.5

Expand $(B x + C) (x + 4)$ using the FOIL Method.

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Step 1.7.5.1

Apply the distributive property.

$6 x^{2} + 19 x = A x^{2} + 4 A + B x (x + 4) + C (x + 4)$

Step 1.7.5.2

Apply the distributive property.

$6 x^{2} + 19 x = A x^{2} + 4 A + B x \cdot x + B x \cdot 4 + C (x + 4)$

Step 1.7.5.3

Apply the distributive property.

$6 x^{2} + 19 x = A x^{2} + 4 A + B x \cdot x + B x \cdot 4 + C x + C \cdot 4$

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 $x$ .

$6 x^{2} + 19 x = A x^{2} + 4 A + B (x \cdot x) + B x \cdot 4 + C x + C \cdot 4$

Step 1.7.6.1.2

Multiply $x$ by $x$ .

$6 x^{2} + 19 x = A x^{2} + 4 A + B x^{2} + B x \cdot 4 + C x + C \cdot 4$

Step 1.7.6.2

Move $4$ to the left of $B x$ .

$6 x^{2} + 19 x = A x^{2} + 4 A + B x^{2} + 4 \cdot (B x) + C x + C \cdot 4$

Step 1.7.6.3

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

$6 x^{2} + 19 x = A x^{2} + 4 A + B x^{2} + 4 B x + C x + 4 C$

Step 1.8

Simplify the expression.

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Step 1.8.1

Move $B$ .

$6 x^{2} + 19 x = A x^{2} + 4 A + B x^{2} + 4 x B + C x + 4 C$

Step 1.8.2

Move $4 A$ .

$6 x^{2} + 19 x = A x^{2} + B x^{2} + 4 x B + C x + 4 A + 4 C$

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.

$6 = A + B$

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.

$19 = 4 B + C$

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.

$0 = 4 A + 4 C$

Step 2.4

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

$6 = A + B$

$19 = 4 B + C$

$0 = 4 A + 4 C$

$6 = A + B$

$19 = 4 B + C$

$0 = 4 A + 4 C$

Step 3

Solve the system of equations.

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Step 3.1

Solve for $A$ in $6 = A + B$ .

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Step 3.1.1

Rewrite the equation as $A + B = 6$ .

$A + B = 6$

$19 = 4 B + C$

$0 = 4 A + 4 C$

Step 3.1.2

Subtract $B$ from both sides of the equation.

$A = 6 - B$

$19 = 4 B + C$

$0 = 4 A + 4 C$

$A = 6 - B$

$19 = 4 B + C$

$0 = 4 A + 4 C$

Step 3.2

Replace all occurrences of $A$ with $6 - B$ in each equation.

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Step 3.2.1

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

$0 = 4 (6 - B) + 4 C$

$A = 6 - B$

$19 = 4 B + C$

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

Apply the distributive property.

$0 = 4 \cdot 6 + 4 (- B) + 4 C$

$A = 6 - B$

$19 = 4 B + C$

Step 3.2.2.1.2

Multiply $4$ by $6$ .

$0 = 24 + 4 (- B) + 4 C$

$A = 6 - B$

$19 = 4 B + C$

Step 3.2.2.1.3

Multiply $- 1$ by $4$ .

$0 = 24 - 4 B + 4 C$

$A = 6 - B$

$19 = 4 B + C$

$0 = 24 - 4 B + 4 C$

$A = 6 - B$

$19 = 4 B + C$

$0 = 24 - 4 B + 4 C$

$A = 6 - B$

$19 = 4 B + C$

$0 = 24 - 4 B + 4 C$

$A = 6 - B$

$19 = 4 B + C$

Step 3.3

Reorder $6$ and $- B$ .

$A = - B + 6$

$0 = 24 - 4 B + 4 C$

$19 = 4 B + C$

Step 3.4

Solve for $C$ in $19 = 4 B + C$ .

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Step 3.4.1

Rewrite the equation as $4 B + C = 19$ .

$4 B + C = 19$

$A = - B + 6$

$0 = 24 - 4 B + 4 C$

Step 3.4.2

Subtract $4 B$ from both sides of the equation.

$C = 19 - 4 B$

$A = - B + 6$

$0 = 24 - 4 B + 4 C$

$C = 19 - 4 B$

$A = - B + 6$

$0 = 24 - 4 B + 4 C$

Step 3.5

Replace all occurrences of $C$ with $19 - 4 B$ in each equation.

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Step 3.5.1

Replace all occurrences of $C$ in $0 = 24 - 4 B + 4 C$ with $19 - 4 B$ .

$0 = 24 - 4 B + 4 (19 - 4 B)$

$C = 19 - 4 B$

$A = - B + 6$

Step 3.5.2

Simplify the right side.

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Step 3.5.2.1

Simplify $24 - 4 B + 4 (19 - 4 B)$ .

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Step 3.5.2.1.1

Simplify each term.

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Step 3.5.2.1.1.1

Apply the distributive property.

$0 = 24 - 4 B + 4 \cdot 19 + 4 (- 4 B)$

$C = 19 - 4 B$

$A = - B + 6$

Step 3.5.2.1.1.2

Multiply $4$ by $19$ .

$0 = 24 - 4 B + 76 + 4 (- 4 B)$

$C = 19 - 4 B$

$A = - B + 6$

Step 3.5.2.1.1.3

Multiply $- 4$ by $4$ .

$0 = 24 - 4 B + 76 - 16 B$

$C = 19 - 4 B$

$A = - B + 6$

$0 = 24 - 4 B + 76 - 16 B$

$C = 19 - 4 B$

$A = - B + 6$

Step 3.5.2.1.2

Simplify by adding terms.

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Step 3.5.2.1.2.1

Add $24$ and $76$ .

$0 = - 4 B + 100 - 16 B$

$C = 19 - 4 B$

$A = - B + 6$

Step 3.5.2.1.2.2

Subtract $16 B$ from $- 4 B$ .

$0 = - 20 B + 100$

$C = 19 - 4 B$

$A = - B + 6$

$0 = - 20 B + 100$

$C = 19 - 4 B$

$A = - B + 6$

$0 = - 20 B + 100$

$C = 19 - 4 B$

$A = - B + 6$

$0 = - 20 B + 100$

$C = 19 - 4 B$

$A = - B + 6$

$0 = - 20 B + 100$

$C = 19 - 4 B$

$A = - B + 6$

Step 3.6

Solve for $B$ in $0 = - 20 B + 100$ .

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Step 3.6.1

Rewrite the equation as $- 20 B + 100 = 0$ .

$- 20 B + 100 = 0$

$C = 19 - 4 B$

$A = - B + 6$

Step 3.6.2

Subtract $100$ from both sides of the equation.

$- 20 B = - 100$

$C = 19 - 4 B$

$A = - B + 6$

Step 3.6.3

Divide each term in $- 20 B = - 100$ by $- 20$ and simplify.

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Step 3.6.3.1

Divide each term in $- 20 B = - 100$ by $- 20$ .

$\frac{- 20 B}{- 20} = \frac{- 100}{- 20}$

$C = 19 - 4 B$

$A = - B + 6$

Step 3.6.3.2

Simplify the left side.

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Step 3.6.3.2.1

Cancel the common factor of $- 20$ .

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Step 3.6.3.2.1.1

Cancel the common factor.

$\frac{- 20 B}{- 20} = \frac{- 100}{- 20}$

$C = 19 - 4 B$

$A = - B + 6$

Step 3.6.3.2.1.2

Divide $B$ by $1$ .

$B = \frac{- 100}{- 20}$

$C = 19 - 4 B$

$A = - B + 6$

$B = \frac{- 100}{- 20}$

$C = 19 - 4 B$

$A = - B + 6$

$B = \frac{- 100}{- 20}$

$C = 19 - 4 B$

$A = - B + 6$

Step 3.6.3.3

Simplify the right side.

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Step 3.6.3.3.1

Divide $- 100$ by $- 20$ .

$B = 5$

$C = 19 - 4 B$

$A = - B + 6$

$B = 5$

$C = 19 - 4 B$

$A = - B + 6$

$B = 5$

$C = 19 - 4 B$

$A = - B + 6$

$B = 5$

$C = 19 - 4 B$

$A = - B + 6$

Step 3.7

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

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Step 3.7.1

Replace all occurrences of $B$ in $C = 19 - 4 B$ with $5$ .

$C = 19 - 4 \cdot 5$

$B = 5$

$A = - B + 6$

Step 3.7.2

Simplify the right side.

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Step 3.7.2.1

Simplify $19 - 4 \cdot 5$ .

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Step 3.7.2.1.1

Multiply $- 4$ by $5$ .

$C = 19 - 20$

$B = 5$

$A = - B + 6$

Step 3.7.2.1.2

Subtract $20$ from $19$ .

$C = - 1$

$B = 5$

$A = - B + 6$

$C = - 1$

$B = 5$

$A = - B + 6$

$C = - 1$

$B = 5$

$A = - B + 6$

Step 3.7.3

Replace all occurrences of $B$ in $A = - B + 6$ with $5$ .

$A = - (5) + 6$

$C = - 1$

$B = 5$

Step 3.7.4

Simplify the right side.

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Step 3.7.4.1

Simplify $- (5) + 6$ .

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Step 3.7.4.1.1

Multiply $- 1$ by $5$ .

$A = - 5 + 6$

$C = - 1$

$B = 5$

Step 3.7.4.1.2

Add $- 5$ and $6$ .

$A = 1$

$C = - 1$

$B = 5$

$A = 1$

$C = - 1$

$B = 5$

$A = 1$

$C = - 1$

$B = 5$

$A = 1$

$C = - 1$

$B = 5$

Step 3.8

List all of the solutions.

$A = 1, C = - 1, B = 5$

Step 4

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

$\frac{1}{x + 4} + \frac{5 x - 1}{x^{2} + 4}$

Step 5

Multiply $5$ by $x$ .

$\frac{1}{x + 4} + \frac{5 x - 1}{x^{2} + 4}$