Precalculus Examples

$\frac{7 x + 116}{{(x - 4)}^{2} (x + 5)}$

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 $A$ .

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

Step 1.2

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

Step 1.3

$\frac{A}{{(x - 4)}^{2}} + \frac{B}{x - 4} + \frac{C}{x + 5}$

Step 1.4

Multiply each fraction in the equation by the denominator of the original expression. In this case, the denominator is ${(x - 4)}^{2} (x + 5)$ .

$\frac{(7 x + 116) {(x - 4)}^{2} (x + 5)}{{(x - 4)}^{2} (x + 5)} = \frac{(A) {(x - 4)}^{2} (x + 5)}{{(x - 4)}^{2}} + \frac{(B) {(x - 4)}^{2} (x + 5)}{x - 4} + \frac{(C) {(x - 4)}^{2} (x + 5)}{x + 5}$

Step 1.5

Cancel the common factor of ${(x - 4)}^{2}$ .

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Step 1.5.1

Cancel the common factor.

$\frac{(7 x + 116) {(x - 4)}^{2} (x + 5)}{{(x - 4)}^{2} (x + 5)} = \frac{(A) {(x - 4)}^{2} (x + 5)}{{(x - 4)}^{2}} + \frac{(B) {(x - 4)}^{2} (x + 5)}{x - 4} + \frac{(C) {(x - 4)}^{2} (x + 5)}{x + 5}$

Step 1.5.2

Rewrite the expression.

$\frac{(7 x + 116) (x + 5)}{x + 5} = \frac{(A) {(x - 4)}^{2} (x + 5)}{{(x - 4)}^{2}} + \frac{(B) {(x - 4)}^{2} (x + 5)}{x - 4} + \frac{(C) {(x - 4)}^{2} (x + 5)}{x + 5}$

Step 1.6

Cancel the common factor of $x + 5$ .

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Step 1.6.1

Cancel the common factor.

$\frac{(7 x + 116) (x + 5)}{x + 5} = \frac{(A) {(x - 4)}^{2} (x + 5)}{{(x - 4)}^{2}} + \frac{(B) {(x - 4)}^{2} (x + 5)}{x - 4} + \frac{(C) {(x - 4)}^{2} (x + 5)}{x + 5}$

Step 1.6.2

Divide $7 x + 116$ by $1$ .

$7 x + 116 = \frac{(A) {(x - 4)}^{2} (x + 5)}{{(x - 4)}^{2}} + \frac{(B) {(x - 4)}^{2} (x + 5)}{x - 4} + \frac{(C) {(x - 4)}^{2} (x + 5)}{x + 5}$

Step 1.7

Simplify each term.

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Step 1.7.1

Cancel the common factor of ${(x - 4)}^{2}$ .

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Step 1.7.1.1

Cancel the common factor.

$7 x + 116 = \frac{A {(x - 4)}^{2} (x + 5)}{{(x - 4)}^{2}} + \frac{(B) {(x - 4)}^{2} (x + 5)}{x - 4} + \frac{(C) {(x - 4)}^{2} (x + 5)}{x + 5}$

Step 1.7.1.2

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

$7 x + 116 = (A) (x + 5) + \frac{(B) {(x - 4)}^{2} (x + 5)}{x - 4} + \frac{(C) {(x - 4)}^{2} (x + 5)}{x + 5}$

Step 1.7.2

Apply the distributive property.

$7 x + 116 = A x + A \cdot 5 + \frac{(B) {(x - 4)}^{2} (x + 5)}{x - 4} + \frac{(C) {(x - 4)}^{2} (x + 5)}{x + 5}$

Step 1.7.3

Move $5$ to the left of $A$ .

$7 x + 116 = A x + 5 \cdot A + \frac{(B) {(x - 4)}^{2} (x + 5)}{x - 4} + \frac{(C) {(x - 4)}^{2} (x + 5)}{x + 5}$

Step 1.7.4

Cancel the common factor of ${(x - 4)}^{2}$ and $x - 4$ .

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Step 1.7.4.1

Factor $x - 4$ out of $(B) {(x - 4)}^{2} (x + 5)$ .

$7 x + 116 = A x + 5 A + \frac{(x - 4) (B (x - 4) (x + 5))}{x - 4} + \frac{(C) {(x - 4)}^{2} (x + 5)}{x + 5}$

Step 1.7.4.2

Cancel the common factors.

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Step 1.7.4.2.1

Multiply by $1$ .

$7 x + 116 = A x + 5 A + \frac{(x - 4) (B (x - 4) (x + 5))}{(x - 4) \cdot 1} + \frac{(C) {(x - 4)}^{2} (x + 5)}{x + 5}$

Step 1.7.4.2.2

Cancel the common factor.

$7 x + 116 = A x + 5 A + \frac{(x - 4) (B (x - 4) (x + 5))}{(x - 4) \cdot 1} + \frac{(C) {(x - 4)}^{2} (x + 5)}{x + 5}$

Step 1.7.4.2.3

Rewrite the expression.

$7 x + 116 = A x + 5 A + \frac{B (x - 4) (x + 5)}{1} + \frac{(C) {(x - 4)}^{2} (x + 5)}{x + 5}$

Step 1.7.4.2.4

Divide $B (x - 4) (x + 5)$ by $1$ .

$7 x + 116 = A x + 5 A + B (x - 4) (x + 5) + \frac{(C) {(x - 4)}^{2} (x + 5)}{x + 5}$

Step 1.7.5

Apply the distributive property.

$7 x + 116 = A x + 5 A + (B x + B \cdot - 4) (x + 5) + \frac{(C) {(x - 4)}^{2} (x + 5)}{x + 5}$

Step 1.7.6

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

$7 x + 116 = A x + 5 A + (B x - 4 \cdot B) (x + 5) + \frac{(C) {(x - 4)}^{2} (x + 5)}{x + 5}$

Step 1.7.7

Expand $(B x - 4 B) (x + 5)$ using the FOIL Method.

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Step 1.7.7.1

Apply the distributive property.

$7 x + 116 = A x + 5 A + B x (x + 5) - 4 B (x + 5) + \frac{(C) {(x - 4)}^{2} (x + 5)}{x + 5}$

Step 1.7.7.2

Apply the distributive property.

$7 x + 116 = A x + 5 A + B x \cdot x + B x \cdot 5 - 4 B (x + 5) + \frac{(C) {(x - 4)}^{2} (x + 5)}{x + 5}$

Step 1.7.7.3

Apply the distributive property.

$7 x + 116 = A x + 5 A + B x \cdot x + B x \cdot 5 - 4 B x - 4 B \cdot 5 + \frac{(C) {(x - 4)}^{2} (x + 5)}{x + 5}$

Step 1.7.8

Simplify and combine like terms.

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Step 1.7.8.1

Simplify each term.

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Step 1.7.8.1.1

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

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Step 1.7.8.1.1.1

Move $x$ .

$7 x + 116 = A x + 5 A + B (x \cdot x) + B x \cdot 5 - 4 B x - 4 B \cdot 5 + \frac{(C) {(x - 4)}^{2} (x + 5)}{x + 5}$

Step 1.7.8.1.1.2

Multiply $x$ by $x$ .

$7 x + 116 = A x + 5 A + B x^{2} + B x \cdot 5 - 4 B x - 4 B \cdot 5 + \frac{(C) {(x - 4)}^{2} (x + 5)}{x + 5}$

Step 1.7.8.1.2

Move $5$ to the left of $B x$ .

$7 x + 116 = A x + 5 A + B x^{2} + 5 \cdot (B x) - 4 B x - 4 B \cdot 5 + \frac{(C) {(x - 4)}^{2} (x + 5)}{x + 5}$

Step 1.7.8.1.3

Multiply $5$ by $- 4$ .

$7 x + 116 = A x + 5 A + B x^{2} + 5 B x - 4 B x - 20 B + \frac{(C) {(x - 4)}^{2} (x + 5)}{x + 5}$

Step 1.7.8.2

Subtract $4 B x$ from $5 B x$ .

$7 x + 116 = A x + 5 A + B x^{2} + 1 B x - 20 B + \frac{(C) {(x - 4)}^{2} (x + 5)}{x + 5}$

Step 1.7.9

Multiply $B$ by $1$ .

$7 x + 116 = A x + 5 A + B x^{2} + B x - 20 B + \frac{(C) {(x - 4)}^{2} (x + 5)}{x + 5}$

Step 1.7.10

Cancel the common factor of $x + 5$ .

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Step 1.7.10.1

Cancel the common factor.

$7 x + 116 = A x + 5 A + B x^{2} + B x - 20 B + \frac{(C) {(x - 4)}^{2} (x + 5)}{x + 5}$

Step 1.7.10.2

Divide $(C) {(x - 4)}^{2}$ by $1$ .

$7 x + 116 = A x + 5 A + B x^{2} + B x - 20 B + (C) {(x - 4)}^{2}$

Step 1.7.11

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

$7 x + 116 = A x + 5 A + B x^{2} + B x - 20 B + C ((x - 4) (x - 4))$

Step 1.7.12

Expand $(x - 4) (x - 4)$ using the FOIL Method.

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Step 1.7.12.1

Apply the distributive property.

$7 x + 116 = A x + 5 A + B x^{2} + B x - 20 B + C (x (x - 4) - 4 (x - 4))$

Step 1.7.12.2

Apply the distributive property.

$7 x + 116 = A x + 5 A + B x^{2} + B x - 20 B + C (x \cdot x + x \cdot - 4 - 4 (x - 4))$

Step 1.7.12.3

Apply the distributive property.

$7 x + 116 = A x + 5 A + B x^{2} + B x - 20 B + C (x \cdot x + x \cdot - 4 - 4 x - 4 \cdot - 4)$

Step 1.7.13

Simplify and combine like terms.

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Step 1.7.13.1

Simplify each term.

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Step 1.7.13.1.1

Multiply $x$ by $x$ .

$7 x + 116 = A x + 5 A + B x^{2} + B x - 20 B + C (x^{2} + x \cdot - 4 - 4 x - 4 \cdot - 4)$

Step 1.7.13.1.2

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

$7 x + 116 = A x + 5 A + B x^{2} + B x - 20 B + C (x^{2} - 4 \cdot x - 4 x - 4 \cdot - 4)$

Step 1.7.13.1.3

Multiply $- 4$ by $- 4$ .

$7 x + 116 = A x + 5 A + B x^{2} + B x - 20 B + C (x^{2} - 4 x - 4 x + 16)$

Step 1.7.13.2

Subtract $4 x$ from $- 4 x$ .

$7 x + 116 = A x + 5 A + B x^{2} + B x - 20 B + C (x^{2} - 8 x + 16)$

Step 1.7.14

Apply the distributive property.

$7 x + 116 = A x + 5 A + B x^{2} + B x - 20 B + C x^{2} + C (- 8 x) + C \cdot 16$

Step 1.7.15

Simplify.

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Step 1.7.15.1

Rewrite using the commutative property of multiplication.

$7 x + 116 = A x + 5 A + B x^{2} + B x - 20 B + C x^{2} - 8 C x + C \cdot 16$

Step 1.7.15.2

Move $16$ to the left of $C$ .

$7 x + 116 = A x + 5 A + B x^{2} + B x - 20 B + C x^{2} - 8 C x + 16 \cdot C$

$7 x + 116 = A x + 5 A + B x^{2} + B x - 20 B + C x^{2} - 8 C x + 16 C$

Step 1.8

Simplify the expression.

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Step 1.8.1

Reorder $B$ and $x$ .

$7 x + 116 = A x + 5 A + B x^{2} + B x - 20 B + C x^{2} - 8 C x + 16 C$

Step 1.8.2

Move $- 20 B$ .

$7 x + 116 = A x + 5 A + B x^{2} + B x + C x^{2} - 8 C x - 20 B + 16 C$

Step 1.8.3

Move $5 A$ .

$7 x + 116 = A x + B x^{2} + B x + C x^{2} - 8 C x + 5 A - 20 B + 16 C$

Step 1.8.4

Move $B x$ .

$7 x + 116 = A x + B x^{2} + C x^{2} + B x - 8 C x + 5 A - 20 B + 16 C$

Step 1.8.5

Move $A x$ .

$7 x + 116 = B x^{2} + C x^{2} + A x + B x - 8 C x + 5 A - 20 B + 16 C$

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.

$0 = B + C$

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.

$7 = A + B - 8 C$

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.

$116 = 5 A - 20 B + 16 C$

Step 2.4

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

$0 = B + C$

$7 = A + B - 8 C$

$116 = 5 A - 20 B + 16 C$

$0 = B + C$

$7 = A + B - 8 C$

$116 = 5 A - 20 B + 16 C$

Step 3

Solve the system of equations.

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Step 3.1

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

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Step 3.1.1

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

$B + C = 0$

$7 = A + B - 8 C$

$116 = 5 A - 20 B + 16 C$

Step 3.1.2

Subtract $C$ from both sides of the equation.

$B = - C$

$7 = A + B - 8 C$

$116 = 5 A - 20 B + 16 C$

$B = - C$

$7 = A + B - 8 C$

$116 = 5 A - 20 B + 16 C$

Step 3.2

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

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Step 3.2.1

Replace all occurrences of $B$ in $7 = A + B - 8 C$ with $- C$ .

$7 = A - C - 8 C$

$B = - C$

$116 = 5 A - 20 B + 16 C$

Step 3.2.2

Simplify $7 = A - C - 8 C$ .

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Step 3.2.2.1

Simplify the left side.

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Step 3.2.2.1.1

Remove parentheses.

$7 = A - C - 8 C$

$B = - C$

$116 = 5 A - 20 B + 16 C$

$7 = A - C - 8 C$

$B = - C$

$116 = 5 A - 20 B + 16 C$

Step 3.2.2.2

Simplify the right side.

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Step 3.2.2.2.1

Subtract $8 C$ from $- C$ .

$7 = A - 9 C$

$B = - C$

$116 = 5 A - 20 B + 16 C$

$7 = A - 9 C$

$B = - C$

$116 = 5 A - 20 B + 16 C$

$7 = A - 9 C$

$B = - C$

$116 = 5 A - 20 B + 16 C$

Step 3.2.3

Replace all occurrences of $B$ in $116 = 5 A - 20 B + 16 C$ with $- C$ .

$116 = 5 A - 20 (- C) + 16 C$

$7 = A - 9 C$

$B = - C$

Step 3.2.4

Simplify the right side.

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Step 3.2.4.1

Simplify $5 A - 20 (- C) + 16 C$ .

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Step 3.2.4.1.1

Multiply $- 1$ by $- 20$ .

$116 = 5 A + 20 C + 16 C$

$7 = A - 9 C$

$B = - C$

Step 3.2.4.1.2

Add $20 C$ and $16 C$ .

$116 = 5 A + 36 C$

$7 = A - 9 C$

$B = - C$

$116 = 5 A + 36 C$

$7 = A - 9 C$

$B = - C$

$116 = 5 A + 36 C$

$7 = A - 9 C$

$B = - C$

$116 = 5 A + 36 C$

$7 = A - 9 C$

$B = - C$

Step 3.3

Solve for $A$ in $7 = A - 9 C$ .

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Step 3.3.1

Rewrite the equation as $A - 9 C = 7$ .

$A - 9 C = 7$

$116 = 5 A + 36 C$

$B = - C$

Step 3.3.2

Add $9 C$ to both sides of the equation.

$A = 7 + 9 C$

$116 = 5 A + 36 C$

$B = - C$

$A = 7 + 9 C$

$116 = 5 A + 36 C$

$B = - C$

Step 3.4

Replace all occurrences of $A$ with $7 + 9 C$ in each equation.

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Step 3.4.1

Replace all occurrences of $A$ in $116 = 5 A + 36 C$ with $7 + 9 C$ .

$116 = 5 (7 + 9 C) + 36 C$

$A = 7 + 9 C$

$B = - C$

Step 3.4.2

Simplify the right side.

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Step 3.4.2.1

Simplify $5 (7 + 9 C) + 36 C$ .

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Step 3.4.2.1.1

Simplify each term.

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Step 3.4.2.1.1.1

Apply the distributive property.

$116 = 5 \cdot 7 + 5 (9 C) + 36 C$

$A = 7 + 9 C$

$B = - C$

Step 3.4.2.1.1.2

Multiply $5$ by $7$ .

$116 = 35 + 5 (9 C) + 36 C$

$A = 7 + 9 C$

$B = - C$

Step 3.4.2.1.1.3

Multiply $9$ by $5$ .

$116 = 35 + 45 C + 36 C$

$A = 7 + 9 C$

$B = - C$

$116 = 35 + 45 C + 36 C$

$A = 7 + 9 C$

$B = - C$

Step 3.4.2.1.2

Add $45 C$ and $36 C$ .

$116 = 35 + 81 C$

$A = 7 + 9 C$

$B = - C$

$116 = 35 + 81 C$

$A = 7 + 9 C$

$B = - C$

$116 = 35 + 81 C$

$A = 7 + 9 C$

$B = - C$

$116 = 35 + 81 C$

$A = 7 + 9 C$

$B = - C$

Step 3.5

Solve for $C$ in $116 = 35 + 81 C$ .

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Step 3.5.1

Rewrite the equation as $35 + 81 C = 116$ .

$35 + 81 C = 116$

$A = 7 + 9 C$

$B = - C$

Step 3.5.2

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

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Step 3.5.2.1

Subtract $35$ from both sides of the equation.

$81 C = 116 - 35$

$A = 7 + 9 C$

$B = - C$

Step 3.5.2.2

Subtract $35$ from $116$ .

$81 C = 81$

$A = 7 + 9 C$

$B = - C$

$81 C = 81$

$A = 7 + 9 C$

$B = - C$

Step 3.5.3

Divide each term in $81 C = 81$ by $81$ and simplify.

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Step 3.5.3.1

Divide each term in $81 C = 81$ by $81$ .

$\frac{81 C}{81} = \frac{81}{81}$

$A = 7 + 9 C$

$B = - C$

Step 3.5.3.2

Simplify the left side.

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Step 3.5.3.2.1

Cancel the common factor of $81$ .

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Step 3.5.3.2.1.1

Cancel the common factor.

$\frac{81 C}{81} = \frac{81}{81}$

$A = 7 + 9 C$

$B = - C$

Step 3.5.3.2.1.2

Divide $C$ by $1$ .

$C = \frac{81}{81}$

$A = 7 + 9 C$

$B = - C$

$C = \frac{81}{81}$

$A = 7 + 9 C$

$B = - C$

$C = \frac{81}{81}$

$A = 7 + 9 C$

$B = - C$

Step 3.5.3.3

Simplify the right side.

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Step 3.5.3.3.1

Divide $81$ by $81$ .

$C = 1$

$A = 7 + 9 C$

$B = - C$

$C = 1$

$A = 7 + 9 C$

$B = - C$

$C = 1$

$A = 7 + 9 C$

$B = - C$

$C = 1$

$A = 7 + 9 C$

$B = - C$

Step 3.6

Replace all occurrences of $C$ with $1$ in each equation.

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Step 3.6.1

Replace all occurrences of $C$ in $A = 7 + 9 C$ with $1$ .

$A = 7 + 9 (1)$

$C = 1$

$B = - C$

Step 3.6.2

Simplify the right side.

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Step 3.6.2.1

Simplify $7 + 9 (1)$ .

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Step 3.6.2.1.1

Multiply $9$ by $1$ .

$A = 7 + 9$

$C = 1$

$B = - C$

Step 3.6.2.1.2

Add $7$ and $9$ .

$A = 16$

$C = 1$

$B = - C$

$A = 16$

$C = 1$

$B = - C$

$A = 16$

$C = 1$

$B = - C$

Step 3.6.3

Replace all occurrences of $C$ in $B = - C$ with $1$ .

$B = - (1)$

$A = 16$

$C = 1$

Step 3.6.4

Simplify the right side.

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Step 3.6.4.1

Multiply $- 1$ by $1$ .

$B = - 1$

$A = 16$

$C = 1$

$B = - 1$

$A = 16$

$C = 1$

$B = - 1$

$A = 16$

$C = 1$

Step 3.7

List all of the solutions.

$B = - 1, A = 16, C = 1$

Step 4

Replace each of the partial fraction coefficients in $\frac{A}{{(x - 4)}^{2}} + \frac{B}{x - 4} + \frac{C}{x + 5}$ with the values found for $A$ , $B$ , and $C$ .

$\frac{16}{{(x - 4)}^{2}} + \frac{- 1}{x - 4} + \frac{1}{x + 5}$