Precalculus Examples

$\frac{7 x^{2} - x - 16}{x (x + 1) (x - 1)}$

Step 1

Decompose the fraction and multiply through by the common denominator.

Tap for more steps...

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 $B$ .

$\frac{A}{x} + \frac{B}{x + 1}$

Step 1.2

$\frac{A}{x} + \frac{B}{x + 1} + \frac{C}{x - 1}$

Step 1.3

Multiply each fraction in the equation by the denominator of the original expression. In this case, the denominator is $x (x + 1) (x - 1)$ .

$\frac{(7 x^{2} - x - 16) (x (x + 1) (x - 1))}{x (x + 1) (x - 1)} = \frac{A (x (x + 1) (x - 1))}{x} + \frac{(B) (x (x + 1) (x - 1))}{x + 1} + \frac{(C) (x (x + 1) (x - 1))}{x - 1}$

Step 1.4

Cancel the common factor of $x$ .

Tap for more steps...

Step 1.4.1

Cancel the common factor.

$\frac{(7 x^{2} - x - 16) (x (x + 1) (x - 1))}{x (x + 1) (x - 1)} = \frac{A (x (x + 1) (x - 1))}{x} + \frac{(B) (x (x + 1) (x - 1))}{x + 1} + \frac{(C) (x (x + 1) (x - 1))}{x - 1}$

Step 1.4.2

Rewrite the expression.

$\frac{(7 x^{2} - x - 16) ((x + 1) (x - 1))}{(x + 1) (x - 1)} = \frac{A (x (x + 1) (x - 1))}{x} + \frac{(B) (x (x + 1) (x - 1))}{x + 1} + \frac{(C) (x (x + 1) (x - 1))}{x - 1}$

Step 1.5

Cancel the common factor of $x + 1$ .

Tap for more steps...

Step 1.5.1

Cancel the common factor.

$\frac{(7 x^{2} - x - 16) ((x + 1) (x - 1))}{(x + 1) (x - 1)} = \frac{A (x (x + 1) (x - 1))}{x} + \frac{(B) (x (x + 1) (x - 1))}{x + 1} + \frac{(C) (x (x + 1) (x - 1))}{x - 1}$

Step 1.5.2

Rewrite the expression.

$\frac{(7 x^{2} - x - 16) (x - 1)}{x - 1} = \frac{A (x (x + 1) (x - 1))}{x} + \frac{(B) (x (x + 1) (x - 1))}{x + 1} + \frac{(C) (x (x + 1) (x - 1))}{x - 1}$

Step 1.6

Cancel the common factor of $x - 1$ .

Tap for more steps...

Step 1.6.1

Cancel the common factor.

$\frac{(7 x^{2} - x - 16) (x - 1)}{x - 1} = \frac{A (x (x + 1) (x - 1))}{x} + \frac{(B) (x (x + 1) (x - 1))}{x + 1} + \frac{(C) (x (x + 1) (x - 1))}{x - 1}$

Step 1.6.2

Divide $7 x^{2} - x - 16$ by $1$ .

$7 x^{2} - x - 16 = \frac{A (x (x + 1) (x - 1))}{x} + \frac{(B) (x (x + 1) (x - 1))}{x + 1} + \frac{(C) (x (x + 1) (x - 1))}{x - 1}$

Step 1.7

Simplify each term.

Tap for more steps...

Step 1.7.1

Cancel the common factor of $x$ .

Tap for more steps...

Step 1.7.1.1

Cancel the common factor.

$7 x^{2} - x - 16 = \frac{A (x (x + 1) (x - 1))}{x} + \frac{(B) (x (x + 1) (x - 1))}{x + 1} + \frac{(C) (x (x + 1) (x - 1))}{x - 1}$

Step 1.7.1.2

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

$7 x^{2} - x - 16 = A ((x + 1) (x - 1)) + \frac{(B) (x (x + 1) (x - 1))}{x + 1} + \frac{(C) (x (x + 1) (x - 1))}{x - 1}$

Step 1.7.2

Expand $(x + 1) (x - 1)$ using the FOIL Method.

Tap for more steps...

Step 1.7.2.1

Apply the distributive property.

$7 x^{2} - x - 16 = A (x (x - 1) + 1 (x - 1)) + \frac{(B) (x (x + 1) (x - 1))}{x + 1} + \frac{(C) (x (x + 1) (x - 1))}{x - 1}$

Step 1.7.2.2

Apply the distributive property.

$7 x^{2} - x - 16 = A (x \cdot x + x \cdot - 1 + 1 (x - 1)) + \frac{(B) (x (x + 1) (x - 1))}{x + 1} + \frac{(C) (x (x + 1) (x - 1))}{x - 1}$

Step 1.7.2.3

Apply the distributive property.

$7 x^{2} - x - 16 = A (x \cdot x + x \cdot - 1 + 1 x + 1 \cdot - 1) + \frac{(B) (x (x + 1) (x - 1))}{x + 1} + \frac{(C) (x (x + 1) (x - 1))}{x - 1}$

Step 1.7.3

Simplify and combine like terms.

Tap for more steps...

Step 1.7.3.1

Simplify each term.

Tap for more steps...

Step 1.7.3.1.1

Multiply $x$ by $x$ .

$7 x^{2} - x - 16 = A (x^{2} + x \cdot - 1 + 1 x + 1 \cdot - 1) + \frac{(B) (x (x + 1) (x - 1))}{x + 1} + \frac{(C) (x (x + 1) (x - 1))}{x - 1}$

Step 1.7.3.1.2

Move $- 1$ to the left of $x$ .

$7 x^{2} - x - 16 = A (x^{2} - 1 \cdot x + 1 x + 1 \cdot - 1) + \frac{(B) (x (x + 1) (x - 1))}{x + 1} + \frac{(C) (x (x + 1) (x - 1))}{x - 1}$

Step 1.7.3.1.3

Rewrite $- 1 x$ as $- x$ .

$7 x^{2} - x - 16 = A (x^{2} - x + 1 x + 1 \cdot - 1) + \frac{(B) (x (x + 1) (x - 1))}{x + 1} + \frac{(C) (x (x + 1) (x - 1))}{x - 1}$

Step 1.7.3.1.4

Multiply $x$ by $1$ .

$7 x^{2} - x - 16 = A (x^{2} - x + x + 1 \cdot - 1) + \frac{(B) (x (x + 1) (x - 1))}{x + 1} + \frac{(C) (x (x + 1) (x - 1))}{x - 1}$

Step 1.7.3.1.5

Multiply $- 1$ by $1$ .

$7 x^{2} - x - 16 = A (x^{2} - x + x - 1) + \frac{(B) (x (x + 1) (x - 1))}{x + 1} + \frac{(C) (x (x + 1) (x - 1))}{x - 1}$

Step 1.7.3.2

Add $- x$ and $x$ .

$7 x^{2} - x - 16 = A (x^{2} + 0 - 1) + \frac{(B) (x (x + 1) (x - 1))}{x + 1} + \frac{(C) (x (x + 1) (x - 1))}{x - 1}$

Step 1.7.3.3

Add $x^{2}$ and $0$ .

$7 x^{2} - x - 16 = A (x^{2} - 1) + \frac{(B) (x (x + 1) (x - 1))}{x + 1} + \frac{(C) (x (x + 1) (x - 1))}{x - 1}$

Step 1.7.4

Apply the distributive property.

$7 x^{2} - x - 16 = A x^{2} + A \cdot - 1 + \frac{(B) (x (x + 1) (x - 1))}{x + 1} + \frac{(C) (x (x + 1) (x - 1))}{x - 1}$

Step 1.7.5

Move $- 1$ to the left of $A$ .

$7 x^{2} - x - 16 = A x^{2} - 1 \cdot A + \frac{(B) (x (x + 1) (x - 1))}{x + 1} + \frac{(C) (x (x + 1) (x - 1))}{x - 1}$

Step 1.7.6

Rewrite $- 1 A$ as $- A$ .

$7 x^{2} - x - 16 = A x^{2} - A + \frac{(B) (x (x + 1) (x - 1))}{x + 1} + \frac{(C) (x (x + 1) (x - 1))}{x - 1}$

Step 1.7.7

Cancel the common factor of $x + 1$ .

Tap for more steps...

Step 1.7.7.1

Cancel the common factor.

$7 x^{2} - x - 16 = A x^{2} - A + \frac{B (x (x + 1) (x - 1))}{x + 1} + \frac{(C) (x (x + 1) (x - 1))}{x - 1}$

Step 1.7.7.2

Divide $(B) (x (x - 1))$ by $1$ .

$7 x^{2} - x - 16 = A x^{2} - A + (B) (x (x - 1)) + \frac{(C) (x (x + 1) (x - 1))}{x - 1}$

Step 1.7.8

Apply the distributive property.

$7 x^{2} - x - 16 = A x^{2} - A + B (x \cdot x + x \cdot - 1) + \frac{(C) (x (x + 1) (x - 1))}{x - 1}$

Step 1.7.9

Multiply $x$ by $x$ .

$7 x^{2} - x - 16 = A x^{2} - A + B (x^{2} + x \cdot - 1) + \frac{(C) (x (x + 1) (x - 1))}{x - 1}$

Step 1.7.10

Move $- 1$ to the left of $x$ .

$7 x^{2} - x - 16 = A x^{2} - A + B (x^{2} - 1 \cdot x) + \frac{(C) (x (x + 1) (x - 1))}{x - 1}$

Step 1.7.11

Rewrite $- 1 x$ as $- x$ .

$7 x^{2} - x - 16 = A x^{2} - A + B (x^{2} - x) + \frac{(C) (x (x + 1) (x - 1))}{x - 1}$

Step 1.7.12

Apply the distributive property.

$7 x^{2} - x - 16 = A x^{2} - A + B x^{2} + B (- x) + \frac{(C) (x (x + 1) (x - 1))}{x - 1}$

Step 1.7.13

Rewrite using the commutative property of multiplication.

$7 x^{2} - x - 16 = A x^{2} - A + B x^{2} - B x + \frac{(C) (x (x + 1) (x - 1))}{x - 1}$

Step 1.7.14

Cancel the common factor of $x - 1$ .

Tap for more steps...

Step 1.7.14.1

Cancel the common factor.

$7 x^{2} - x - 16 = A x^{2} - A + B x^{2} - B x + \frac{C (x (x + 1) (x - 1))}{x - 1}$

Step 1.7.14.2

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

$7 x^{2} - x - 16 = A x^{2} - A + B x^{2} - B x + (C) (x (x + 1))$

Step 1.7.15

Apply the distributive property.

$7 x^{2} - x - 16 = A x^{2} - A + B x^{2} - B x + C (x \cdot x + x \cdot 1)$

Step 1.7.16

Multiply $x$ by $x$ .

$7 x^{2} - x - 16 = A x^{2} - A + B x^{2} - B x + C (x^{2} + x \cdot 1)$

Step 1.7.17

Multiply $x$ by $1$ .

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

Step 1.7.18

Apply the distributive property.

$7 x^{2} - x - 16 = A x^{2} - A + B x^{2} - B x + C x^{2} + C x$

Step 1.8

Simplify the expression.

Tap for more steps...

Step 1.8.1

Move $B$ .

$7 x^{2} - x - 16 = A x^{2} - A + B x^{2} - x B + C x^{2} + C x$

Step 1.8.2

Move $- A$ .

$7 x^{2} - x - 16 = A x^{2} + B x^{2} - x B + C x^{2} + C x - A$

Step 1.8.3

Move $- x B$ .

$7 x^{2} - x - 16 = A x^{2} + B x^{2} + C x^{2} - x B + C x - A$

Step 2

Create equations for the partial fraction variables and use them to set up a system of equations.

Tap for more steps...

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.

$7 = A + B + C$

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.

$- 1 = - 1 B + C$

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.

$- 16 = - 1 A$

Step 2.4

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

$7 = A + B + C$

$- 1 = - 1 B + C$

$- 16 = - 1 A$

$7 = A + B + C$

$- 1 = - 1 B + C$

$- 16 = - 1 A$

Step 3

Solve the system of equations.

Tap for more steps...

Step 3.1

Solve for $A$ in $- 16 = - 1 A$ .

Tap for more steps...

Step 3.1.1

Rewrite the equation as $- 1 A = - 16$ .

$- 1 A = - 16$

$7 = A + B + C$

$- 1 = - 1 B + C$

Step 3.1.2

Divide each term in $- 1 A = - 16$ by $- 1$ and simplify.

Tap for more steps...

Step 3.1.2.1

Divide each term in $- 1 A = - 16$ by $- 1$ .

$\frac{- 1 A}{- 1} = \frac{- 16}{- 1}$

$7 = A + B + C$

$- 1 = - 1 B + C$

Step 3.1.2.2

Simplify the left side.

Tap for more steps...

Step 3.1.2.2.1

Dividing two negative values results in a positive value.

$\frac{A}{1} = \frac{- 16}{- 1}$

$7 = A + B + C$

$- 1 = - 1 B + C$

Step 3.1.2.2.2

Divide $A$ by $1$ .

$A = \frac{- 16}{- 1}$

$7 = A + B + C$

$- 1 = - 1 B + C$

$A = \frac{- 16}{- 1}$

$7 = A + B + C$

$- 1 = - 1 B + C$

Step 3.1.2.3

Simplify the right side.

Tap for more steps...

Step 3.1.2.3.1

Divide $- 16$ by $- 1$ .

$A = 16$

$7 = A + B + C$

$- 1 = - 1 B + C$

$A = 16$

$7 = A + B + C$

$- 1 = - 1 B + C$

$A = 16$

$7 = A + B + C$

$- 1 = - 1 B + C$

$A = 16$

$7 = A + B + C$

$- 1 = - 1 B + C$

Step 3.2

Replace all occurrences of $A$ with $16$ in each equation.

Tap for more steps...

Step 3.2.1

Replace all occurrences of $A$ in $7 = A + B + C$ with $16$ .

$7 = (16) + B + C$

$A = 16$

$- 1 = - 1 B + C$

Step 3.2.2

Simplify the right side.

Tap for more steps...

Step 3.2.2.1

Remove parentheses.

$7 = 16 + B + C$

$A = 16$

$- 1 = - 1 B + C$

$7 = 16 + B + C$

$A = 16$

$- 1 = - 1 B + C$

$7 = 16 + B + C$

$A = 16$

$- 1 = - 1 B + C$

Step 3.3

Solve for $B$ in $7 = 16 + B + C$ .

Tap for more steps...

Step 3.3.1

Rewrite the equation as $16 + B + C = 7$ .

$16 + B + C = 7$

$A = 16$

$- 1 = - 1 B + C$

Step 3.3.2

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

Tap for more steps...

Step 3.3.2.1

Subtract $16$ from both sides of the equation.

$B + C = 7 - 16$

$A = 16$

$- 1 = - 1 B + C$

Step 3.3.2.2

Subtract $C$ from both sides of the equation.

$B = 7 - 16 - C$

$A = 16$

$- 1 = - 1 B + C$

Step 3.3.2.3

Subtract $16$ from $7$ .

$B = - 9 - C$

$A = 16$

$- 1 = - 1 B + C$

$B = - 9 - C$

$A = 16$

$- 1 = - 1 B + C$

$B = - 9 - C$

$A = 16$

$- 1 = - 1 B + C$

Step 3.4

Replace all occurrences of $B$ with $- 9 - C$ in each equation.

Tap for more steps...

Step 3.4.1

Replace all occurrences of $B$ in $- 1 = - 1 B + C$ with $- 9 - C$ .

$- 1 = - 1 (- 9 - C) + C$

$B = - 9 - C$

$A = 16$

Step 3.4.2

Simplify the right side.

Tap for more steps...

Step 3.4.2.1

Simplify $- 1 (- 9 - C) + C$ .

Tap for more steps...

Step 3.4.2.1.1

Simplify each term.

Tap for more steps...

Step 3.4.2.1.1.1

Apply the distributive property.

$- 1 = - 1 \cdot - 9 - 1 (- C) + C$

$B = - 9 - C$

$A = 16$

Step 3.4.2.1.1.2

Multiply $- 1$ by $- 9$ .

$- 1 = 9 - 1 (- C) + C$

$B = - 9 - C$

$A = 16$

Step 3.4.2.1.1.3

Multiply $- 1 (- C)$ .

Tap for more steps...

Step 3.4.2.1.1.3.1

Multiply $- 1$ by $- 1$ .

$- 1 = 9 + 1 C + C$

$B = - 9 - C$

$A = 16$

Step 3.4.2.1.1.3.2

Multiply $C$ by $1$ .

$- 1 = 9 + C + C$

$B = - 9 - C$

$A = 16$

$- 1 = 9 + C + C$

$B = - 9 - C$

$A = 16$

$- 1 = 9 + C + C$

$B = - 9 - C$

$A = 16$

Step 3.4.2.1.2

Add $C$ and $C$ .

$- 1 = 9 + 2 C$

$B = - 9 - C$

$A = 16$

$- 1 = 9 + 2 C$

$B = - 9 - C$

$A = 16$

$- 1 = 9 + 2 C$

$B = - 9 - C$

$A = 16$

$- 1 = 9 + 2 C$

$B = - 9 - C$

$A = 16$

Step 3.5

Solve for $C$ in $- 1 = 9 + 2 C$ .

Tap for more steps...

Step 3.5.1

Rewrite the equation as $9 + 2 C = - 1$ .

$9 + 2 C = - 1$

$B = - 9 - C$

$A = 16$

Step 3.5.2

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

Tap for more steps...

Step 3.5.2.1

Subtract $9$ from both sides of the equation.

$2 C = - 1 - 9$

$B = - 9 - C$

$A = 16$

Step 3.5.2.2

Subtract $9$ from $- 1$ .

$2 C = - 10$

$B = - 9 - C$

$A = 16$

$2 C = - 10$

$B = - 9 - C$

$A = 16$

Step 3.5.3

Divide each term in $2 C = - 10$ by $2$ and simplify.

Tap for more steps...

Step 3.5.3.1

Divide each term in $2 C = - 10$ by $2$ .

$\frac{2 C}{2} = \frac{- 10}{2}$

$B = - 9 - C$

$A = 16$

Step 3.5.3.2

Simplify the left side.

Tap for more steps...

Step 3.5.3.2.1

Cancel the common factor of $2$ .

Tap for more steps...

Step 3.5.3.2.1.1

Cancel the common factor.

$\frac{2 C}{2} = \frac{- 10}{2}$

$B = - 9 - C$

$A = 16$

Step 3.5.3.2.1.2

Divide $C$ by $1$ .

$C = \frac{- 10}{2}$

$B = - 9 - C$

$A = 16$

$C = \frac{- 10}{2}$

$B = - 9 - C$

$A = 16$

$C = \frac{- 10}{2}$

$B = - 9 - C$

$A = 16$

Step 3.5.3.3

Simplify the right side.

Tap for more steps...

Step 3.5.3.3.1

Divide $- 10$ by $2$ .

$C = - 5$

$B = - 9 - C$

$A = 16$

$C = - 5$

$B = - 9 - C$

$A = 16$

$C = - 5$

$B = - 9 - C$

$A = 16$

$C = - 5$

$B = - 9 - C$

$A = 16$

Step 3.6

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

Tap for more steps...

Step 3.6.1

Replace all occurrences of $C$ in $B = - 9 - C$ with $- 5$ .

$B = - 9 - (- 5)$

$C = - 5$

$A = 16$

Step 3.6.2

Simplify the right side.

Tap for more steps...

Step 3.6.2.1

Simplify $- 9 - (- 5)$ .

Tap for more steps...

Step 3.6.2.1.1

Multiply $- 1$ by $- 5$ .

$B = - 9 + 5$

$C = - 5$

$A = 16$

Step 3.6.2.1.2

Add $- 9$ and $5$ .

$B = - 4$

$C = - 5$

$A = 16$

$B = - 4$

$C = - 5$

$A = 16$

$B = - 4$

$C = - 5$

$A = 16$

$B = - 4$

$C = - 5$

$A = 16$

Step 3.7

List all of the solutions.

$B = - 4, C = - 5, A = 16$

Step 4

Replace each of the partial fraction coefficients in $\frac{A}{x} + \frac{B}{x + 1} + \frac{C}{x - 1}$ with the values found for $A$ , $B$ , and $C$ .

$\frac{16}{x} + \frac{- 4}{x + 1} + \frac{- 5}{x - 1}$