Precalculus Examples

$\frac{x^{2} + 2 x - 1}{2 x^{3} + 3 x^{2} - 2 x}$

Step 1

Decompose the fraction and multiply through by the common denominator.

Tap for more steps...

Step 1.1

Factor the fraction.

Tap for more steps...

Step 1.1.1

Factor $x$ out of $2 x^{3} + 3 x^{2} - 2 x$ .

Tap for more steps...

Step 1.1.1.1

Factor $x$ out of $2 x^{3}$ .

$\frac{x^{2} + 2 x - 1}{x (2 x^{2}) + 3 x^{2} - 2 x}$

Step 1.1.1.2

Factor $x$ out of $3 x^{2}$ .

$\frac{x^{2} + 2 x - 1}{x (2 x^{2}) + x (3 x) - 2 x}$

Step 1.1.1.3

Factor $x$ out of $- 2 x$ .

$\frac{x^{2} + 2 x - 1}{x (2 x^{2}) + x (3 x) + x \cdot - 2}$

Step 1.1.1.4

Factor $x$ out of $x (2 x^{2}) + x (3 x)$ .

$\frac{x^{2} + 2 x - 1}{x (2 x^{2} + 3 x) + x \cdot - 2}$

Step 1.1.1.5

Factor $x$ out of $x (2 x^{2} + 3 x) + x \cdot - 2$ .

$\frac{x^{2} + 2 x - 1}{x (2 x^{2} + 3 x - 2)}$

Step 1.1.2

Factor.

Tap for more steps...

Step 1.1.2.1

Factor by grouping.

Tap for more steps...

Step 1.1.2.1.1

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = 2 \cdot - 2 = - 4$ and whose sum is $b = 3$ .

Tap for more steps...

Step 1.1.2.1.1.1

Factor $3$ out of $3 x$ .

$\frac{x^{2} + 2 x - 1}{x (2 x^{2} + 3 (x) - 2)}$

Step 1.1.2.1.1.2

Rewrite $3$ as $- 1$ plus $4$

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

Step 1.1.2.1.1.3

Apply the distributive property.

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

Step 1.1.2.1.2

Factor out the greatest common factor from each group.

Tap for more steps...

Step 1.1.2.1.2.1

Group the first two terms and the last two terms.

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

Step 1.1.2.1.2.2

Factor out the greatest common factor (GCF) from each group.

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

Step 1.1.2.1.3

Factor the polynomial by factoring out the greatest common factor, $2 x - 1$ .

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

Step 1.1.2.2

Remove unnecessary parentheses.

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

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 $B$ .

$\frac{A}{x} + \frac{B}{2 x - 1}$

Step 1.3

$\frac{A}{x} + \frac{B}{2 x - 1} + \frac{C}{x + 2}$

Step 1.4

Multiply each fraction in the equation by the denominator of the original expression. In this case, the denominator is $x (2 x - 1) (x + 2)$ .

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

Step 1.5

Cancel the common factor of $x$ .

Tap for more steps...

Step 1.5.1

Cancel the common factor.

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

Step 1.5.2

Rewrite the expression.

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

Step 1.6

Cancel the common factor of $2 x - 1$ .

Tap for more steps...

Step 1.6.1

Cancel the common factor.

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

Step 1.6.2

Rewrite the expression.

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

Step 1.7

Cancel the common factor of $x + 2$ .

Tap for more steps...

Step 1.7.1

Cancel the common factor.

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

Step 1.7.2

Divide $x^{2} + 2 x - 1$ by $1$ .

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

Step 1.8

Simplify each term.

Tap for more steps...

Step 1.8.1

Cancel the common factor of $x$ .

Tap for more steps...

Step 1.8.1.1

Cancel the common factor.

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

Step 1.8.1.2

Divide $A ((2 x - 1) (x + 2))$ by $1$ .

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

Step 1.8.2

Expand $(2 x - 1) (x + 2)$ using the FOIL Method.

Tap for more steps...

Step 1.8.2.1

Apply the distributive property.

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

Step 1.8.2.2

Apply the distributive property.

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

Step 1.8.2.3

Apply the distributive property.

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

Step 1.8.3

Simplify and combine like terms.

Tap for more steps...

Step 1.8.3.1

Simplify each term.

Tap for more steps...

Step 1.8.3.1.1

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

Tap for more steps...

Step 1.8.3.1.1.1

Move $x$ .

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

Step 1.8.3.1.1.2

Multiply $x$ by $x$ .

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

Step 1.8.3.1.2

Multiply $2$ by $2$ .

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

Step 1.8.3.1.3

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

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

Step 1.8.3.1.4

Multiply $- 1$ by $2$ .

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

Step 1.8.3.2

Subtract $x$ from $4 x$ .

$x^{2} + 2 x - 1 = A (2 x^{2} + 3 x - 2) + \frac{(B) (x (2 x - 1) (x + 2))}{2 x - 1} + \frac{(C) (x (2 x - 1) (x + 2))}{x + 2}$

Step 1.8.4

Apply the distributive property.

$x^{2} + 2 x - 1 = A (2 x^{2}) + A (3 x) + A \cdot - 2 + \frac{(B) (x (2 x - 1) (x + 2))}{2 x - 1} + \frac{(C) (x (2 x - 1) (x + 2))}{x + 2}$

Step 1.8.5

Simplify.

Tap for more steps...

Step 1.8.5.1

Rewrite using the commutative property of multiplication.

$x^{2} + 2 x - 1 = 2 A x^{2} + A (3 x) + A \cdot - 2 + \frac{(B) (x (2 x - 1) (x + 2))}{2 x - 1} + \frac{(C) (x (2 x - 1) (x + 2))}{x + 2}$

Step 1.8.5.2

Rewrite using the commutative property of multiplication.

$x^{2} + 2 x - 1 = 2 A x^{2} + 3 A x + A \cdot - 2 + \frac{(B) (x (2 x - 1) (x + 2))}{2 x - 1} + \frac{(C) (x (2 x - 1) (x + 2))}{x + 2}$

Step 1.8.5.3

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

$x^{2} + 2 x - 1 = 2 A x^{2} + 3 A x - 2 \cdot A + \frac{(B) (x (2 x - 1) (x + 2))}{2 x - 1} + \frac{(C) (x (2 x - 1) (x + 2))}{x + 2}$

Step 1.8.6

Cancel the common factor of $2 x - 1$ .

Tap for more steps...

Step 1.8.6.1

Cancel the common factor.

$x^{2} + 2 x - 1 = 2 A x^{2} + 3 A x - 2 A + \frac{B (x (2 x - 1) (x + 2))}{2 x - 1} + \frac{(C) (x (2 x - 1) (x + 2))}{x + 2}$

Step 1.8.6.2

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

$x^{2} + 2 x - 1 = 2 A x^{2} + 3 A x - 2 A + (B) (x (x + 2)) + \frac{(C) (x (2 x - 1) (x + 2))}{x + 2}$

Step 1.8.7

Apply the distributive property.

$x^{2} + 2 x - 1 = 2 A x^{2} + 3 A x - 2 A + B (x \cdot x + x \cdot 2) + \frac{(C) (x (2 x - 1) (x + 2))}{x + 2}$

Step 1.8.8

Multiply $x$ by $x$ .

$x^{2} + 2 x - 1 = 2 A x^{2} + 3 A x - 2 A + B (x^{2} + x \cdot 2) + \frac{(C) (x (2 x - 1) (x + 2))}{x + 2}$

Step 1.8.9

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

$x^{2} + 2 x - 1 = 2 A x^{2} + 3 A x - 2 A + B (x^{2} + 2 \cdot x) + \frac{(C) (x (2 x - 1) (x + 2))}{x + 2}$

Step 1.8.10

Apply the distributive property.

$x^{2} + 2 x - 1 = 2 A x^{2} + 3 A x - 2 A + B x^{2} + B (2 x) + \frac{(C) (x (2 x - 1) (x + 2))}{x + 2}$

Step 1.8.11

Rewrite using the commutative property of multiplication.

$x^{2} + 2 x - 1 = 2 A x^{2} + 3 A x - 2 A + B x^{2} + 2 B x + \frac{(C) (x (2 x - 1) (x + 2))}{x + 2}$

Step 1.8.12

Cancel the common factor of $x + 2$ .

Tap for more steps...

Step 1.8.12.1

Cancel the common factor.

$x^{2} + 2 x - 1 = 2 A x^{2} + 3 A x - 2 A + B x^{2} + 2 B x + \frac{C (x (2 x - 1) (x + 2))}{x + 2}$

Step 1.8.12.2

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

$x^{2} + 2 x - 1 = 2 A x^{2} + 3 A x - 2 A + B x^{2} + 2 B x + (C) (x (2 x - 1))$

Step 1.8.13

Apply the distributive property.

$x^{2} + 2 x - 1 = 2 A x^{2} + 3 A x - 2 A + B x^{2} + 2 B x + C (x (2 x) + x \cdot - 1)$

Step 1.8.14

Rewrite using the commutative property of multiplication.

$x^{2} + 2 x - 1 = 2 A x^{2} + 3 A x - 2 A + B x^{2} + 2 B x + C (2 x \cdot x + x \cdot - 1)$

Step 1.8.15

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

$x^{2} + 2 x - 1 = 2 A x^{2} + 3 A x - 2 A + B x^{2} + 2 B x + C (2 x \cdot x - 1 \cdot x)$

Step 1.8.16

Simplify each term.

Tap for more steps...

Step 1.8.16.1

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

Tap for more steps...

Step 1.8.16.1.1

Move $x$ .

$x^{2} + 2 x - 1 = 2 A x^{2} + 3 A x - 2 A + B x^{2} + 2 B x + C (2 (x \cdot x) - 1 \cdot x)$

Step 1.8.16.1.2

Multiply $x$ by $x$ .

$x^{2} + 2 x - 1 = 2 A x^{2} + 3 A x - 2 A + B x^{2} + 2 B x + C (2 x^{2} - 1 \cdot x)$

Step 1.8.16.2

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

$x^{2} + 2 x - 1 = 2 A x^{2} + 3 A x - 2 A + B x^{2} + 2 B x + C (2 x^{2} - x)$

Step 1.8.17

Apply the distributive property.

$x^{2} + 2 x - 1 = 2 A x^{2} + 3 A x - 2 A + B x^{2} + 2 B x + C (2 x^{2}) + C (- x)$

Step 1.8.18

Rewrite using the commutative property of multiplication.

$x^{2} + 2 x - 1 = 2 A x^{2} + 3 A x - 2 A + B x^{2} + 2 B x + 2 C x^{2} + C (- x)$

Step 1.8.19

Rewrite using the commutative property of multiplication.

$x^{2} + 2 x - 1 = 2 A x^{2} + 3 A x - 2 A + B x^{2} + 2 B x + 2 C x^{2} - C x$

Step 1.9

Simplify the expression.

Tap for more steps...

Step 1.9.1

Move $A$ .

$x^{2} + 2 x - 1 = 2 x^{2} A + 3 A x - 2 A + B x^{2} + 2 B x + 2 C x^{2} - C x$

Step 1.9.2

Move $A$ .

$x^{2} + 2 x - 1 = 2 x^{2} A + 3 x A - 2 A + B x^{2} + 2 B x + 2 C x^{2} - C x$

Step 1.9.3

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

$x^{2} + 2 x - 1 = 2 x^{2} A + 3 x A - 2 A + B x^{2} + 2 B x + 2 C x^{2} - C x$

Step 1.9.4

Move $B$ .

$x^{2} + 2 x - 1 = 2 x^{2} A + 3 x A - 2 A + B x^{2} + 2 x B + 2 C x^{2} - C x$

Step 1.9.5

Move $- 2 A$ .

$x^{2} + 2 x - 1 = 2 x^{2} A + 3 x A + B x^{2} + 2 x B + 2 C x^{2} - C x - 2 A$

Step 1.9.6

Move $2 x B$ .

$x^{2} + 2 x - 1 = 2 x^{2} A + 3 x A + B x^{2} + 2 C x^{2} + 2 x B - C x - 2 A$

Step 1.9.7

Move $3 x A$ .

$x^{2} + 2 x - 1 = 2 x^{2} A + B x^{2} + 2 C x^{2} + 3 x A + 2 x B - C x - 2 A$

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.

$1 = 2 A + B + 2 C$

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.

$2 = 3 A + 2 B - 1 C$

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.

$- 1 = - 2 A$

Step 2.4

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

$1 = 2 A + B + 2 C$

$2 = 3 A + 2 B - 1 C$

$- 1 = - 2 A$

$1 = 2 A + B + 2 C$

$2 = 3 A + 2 B - 1 C$

$- 1 = - 2 A$

Step 3

Solve the system of equations.

Tap for more steps...

Step 3.1

Solve for $A$ in $- 1 = - 2 A$ .

Tap for more steps...

Step 3.1.1

Rewrite the equation as $- 2 A = - 1$ .

$- 2 A = - 1$

$1 = 2 A + B + 2 C$

$2 = 3 A + 2 B - 1 C$

Step 3.1.2

Divide each term in $- 2 A = - 1$ by $- 2$ and simplify.

Tap for more steps...

Step 3.1.2.1

Divide each term in $- 2 A = - 1$ by $- 2$ .

$\frac{- 2 A}{- 2} = \frac{- 1}{- 2}$

$1 = 2 A + B + 2 C$

$2 = 3 A + 2 B - 1 C$

Step 3.1.2.2

Simplify the left side.

Tap for more steps...

Step 3.1.2.2.1

Cancel the common factor of $- 2$ .

Tap for more steps...

Step 3.1.2.2.1.1

Cancel the common factor.

$\frac{- 2 A}{- 2} = \frac{- 1}{- 2}$

$1 = 2 A + B + 2 C$

$2 = 3 A + 2 B - 1 C$

Step 3.1.2.2.1.2

Divide $A$ by $1$ .

$A = \frac{- 1}{- 2}$

$1 = 2 A + B + 2 C$

$2 = 3 A + 2 B - 1 C$

$A = \frac{- 1}{- 2}$

$1 = 2 A + B + 2 C$

$2 = 3 A + 2 B - 1 C$

$A = \frac{- 1}{- 2}$

$1 = 2 A + B + 2 C$

$2 = 3 A + 2 B - 1 C$

Step 3.1.2.3

Simplify the right side.

Tap for more steps...

Step 3.1.2.3.1

Dividing two negative values results in a positive value.

$A = \frac{1}{2}$

$1 = 2 A + B + 2 C$

$2 = 3 A + 2 B - 1 C$

$A = \frac{1}{2}$

$1 = 2 A + B + 2 C$

$2 = 3 A + 2 B - 1 C$

$A = \frac{1}{2}$

$1 = 2 A + B + 2 C$

$2 = 3 A + 2 B - 1 C$

$A = \frac{1}{2}$

$1 = 2 A + B + 2 C$

$2 = 3 A + 2 B - 1 C$

Step 3.2

Replace all occurrences of $A$ with $\frac{1}{2}$ in each equation.

Tap for more steps...

Step 3.2.1

Replace all occurrences of $A$ in $1 = 2 A + B + 2 C$ with $\frac{1}{2}$ .

$1 = 2 (\frac{1}{2}) + B + 2 C$

$A = \frac{1}{2}$

$2 = 3 A + 2 B - 1 C$

Step 3.2.2

Simplify the right side.

Tap for more steps...

Step 3.2.2.1

Cancel the common factor of $2$ .

Tap for more steps...

Step 3.2.2.1.1

Cancel the common factor.

$1 = 2 (\frac{1}{2}) + B + 2 C$

$A = \frac{1}{2}$

$2 = 3 A + 2 B - 1 C$

Step 3.2.2.1.2

Rewrite the expression.

$1 = 1 + B + 2 C$

$A = \frac{1}{2}$

$2 = 3 A + 2 B - 1 C$

$1 = 1 + B + 2 C$

$A = \frac{1}{2}$

$2 = 3 A + 2 B - 1 C$

$1 = 1 + B + 2 C$

$A = \frac{1}{2}$

$2 = 3 A + 2 B - 1 C$

Step 3.2.3

Replace all occurrences of $A$ in $2 = 3 A + 2 B - 1 C$ with $\frac{1}{2}$ .

$2 = 3 (\frac{1}{2}) + 2 B - 1 C$

$1 = 1 + B + 2 C$

$A = \frac{1}{2}$

Step 3.2.4

Simplify the right side.

Tap for more steps...

Step 3.2.4.1

Simplify each term.

Tap for more steps...

Step 3.2.4.1.1

Combine $3$ and $\frac{1}{2}$ .

$2 = \frac{3}{2} + 2 B - 1 C$

$1 = 1 + B + 2 C$

$A = \frac{1}{2}$

Step 3.2.4.1.2

Rewrite $- 1 C$ as $- C$ .

$2 = \frac{3}{2} + 2 B - C$

$1 = 1 + B + 2 C$

$A = \frac{1}{2}$

$2 = \frac{3}{2} + 2 B - C$

$1 = 1 + B + 2 C$

$A = \frac{1}{2}$

$2 = \frac{3}{2} + 2 B - C$

$1 = 1 + B + 2 C$

$A = \frac{1}{2}$

$2 = \frac{3}{2} + 2 B - C$

$1 = 1 + B + 2 C$

$A = \frac{1}{2}$

Step 3.3

Solve for $B$ in $1 = 1 + B + 2 C$ .

Tap for more steps...

Step 3.3.1

Rewrite the equation as $1 + B + 2 C = 1$ .

$1 + B + 2 C = 1$

$2 = \frac{3}{2} + 2 B - C$

$A = \frac{1}{2}$

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 $1$ from both sides of the equation.

$B + 2 C = 1 - 1$

$2 = \frac{3}{2} + 2 B - C$

$A = \frac{1}{2}$

Step 3.3.2.2

Subtract $2 C$ from both sides of the equation.

$B = 1 - 1 - 2 C$

$2 = \frac{3}{2} + 2 B - C$

$A = \frac{1}{2}$

Step 3.3.2.3

Subtract $1$ from $1$ .

$B = 0 - 2 C$

$2 = \frac{3}{2} + 2 B - C$

$A = \frac{1}{2}$

Step 3.3.2.4

Subtract $2 C$ from $0$ .

$B = - 2 C$

$2 = \frac{3}{2} + 2 B - C$

$A = \frac{1}{2}$

$B = - 2 C$

$2 = \frac{3}{2} + 2 B - C$

$A = \frac{1}{2}$

$B = - 2 C$

$2 = \frac{3}{2} + 2 B - C$

$A = \frac{1}{2}$

Step 3.4

Replace all occurrences of $B$ with $- 2 C$ in each equation.

Tap for more steps...

Step 3.4.1

Replace all occurrences of $B$ in $2 = \frac{3}{2} + 2 B - C$ with $- 2 C$ .

$2 = \frac{3}{2} + 2 (- 2 C) - C$

$B = - 2 C$

$A = \frac{1}{2}$

Step 3.4.2

Simplify the right side.

Tap for more steps...

Step 3.4.2.1

Simplify $\frac{3}{2} + 2 (- 2 C) - C$ .

Tap for more steps...

Step 3.4.2.1.1

Multiply $- 2$ by $2$ .

$2 = \frac{3}{2} - 4 C - C$

$B = - 2 C$

$A = \frac{1}{2}$

Step 3.4.2.1.2

Subtract $C$ from $- 4 C$ .

$2 = \frac{3}{2} - 5 C$

$B = - 2 C$

$A = \frac{1}{2}$

$2 = \frac{3}{2} - 5 C$

$B = - 2 C$

$A = \frac{1}{2}$

$2 = \frac{3}{2} - 5 C$

$B = - 2 C$

$A = \frac{1}{2}$

$2 = \frac{3}{2} - 5 C$

$B = - 2 C$

$A = \frac{1}{2}$

Step 3.5

Solve for $C$ in $2 = \frac{3}{2} - 5 C$ .

Tap for more steps...

Step 3.5.1

Rewrite the equation as $\frac{3}{2} - 5 C = 2$ .

$\frac{3}{2} - 5 C = 2$

$B = - 2 C$

$A = \frac{1}{2}$

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 $\frac{3}{2}$ from both sides of the equation.

$- 5 C = 2 - \frac{3}{2}$

$B = - 2 C$

$A = \frac{1}{2}$

Step 3.5.2.2

To write $2$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$- 5 C = 2 \cdot \frac{2}{2} - \frac{3}{2}$

$B = - 2 C$

$A = \frac{1}{2}$

Step 3.5.2.3

Combine $2$ and $\frac{2}{2}$ .

$- 5 C = \frac{2 \cdot 2}{2} - \frac{3}{2}$

$B = - 2 C$

$A = \frac{1}{2}$

Step 3.5.2.4

Combine the numerators over the common denominator.

$- 5 C = \frac{2 \cdot 2 - 3}{2}$

$B = - 2 C$

$A = \frac{1}{2}$

Step 3.5.2.5

Simplify the numerator.

Tap for more steps...

Step 3.5.2.5.1

Multiply $2$ by $2$ .

$- 5 C = \frac{4 - 3}{2}$

$B = - 2 C$

$A = \frac{1}{2}$

Step 3.5.2.5.2

Subtract $3$ from $4$ .

$- 5 C = \frac{1}{2}$

$B = - 2 C$

$A = \frac{1}{2}$

$- 5 C = \frac{1}{2}$

$B = - 2 C$

$A = \frac{1}{2}$

$- 5 C = \frac{1}{2}$

$B = - 2 C$

$A = \frac{1}{2}$

Step 3.5.3

Divide each term in $- 5 C = \frac{1}{2}$ by $- 5$ and simplify.

Tap for more steps...

Step 3.5.3.1

Divide each term in $- 5 C = \frac{1}{2}$ by $- 5$ .

$\frac{- 5 C}{- 5} = \frac{\frac{1}{2}}{- 5}$

$B = - 2 C$

$A = \frac{1}{2}$

Step 3.5.3.2

Simplify the left side.

Tap for more steps...

Step 3.5.3.2.1

Cancel the common factor of $- 5$ .

Tap for more steps...

Step 3.5.3.2.1.1

Cancel the common factor.

$\frac{- 5 C}{- 5} = \frac{\frac{1}{2}}{- 5}$

$B = - 2 C$

$A = \frac{1}{2}$

Step 3.5.3.2.1.2

Divide $C$ by $1$ .

$C = \frac{\frac{1}{2}}{- 5}$

$B = - 2 C$

$A = \frac{1}{2}$

$C = \frac{\frac{1}{2}}{- 5}$

$B = - 2 C$

$A = \frac{1}{2}$

$C = \frac{\frac{1}{2}}{- 5}$

$B = - 2 C$

$A = \frac{1}{2}$

Step 3.5.3.3

Simplify the right side.

Tap for more steps...

Step 3.5.3.3.1

Multiply the numerator by the reciprocal of the denominator.

$C = \frac{1}{2} \cdot \frac{1}{- 5}$

$B = - 2 C$

$A = \frac{1}{2}$

Step 3.5.3.3.2

Move the negative in front of the fraction.

$C = \frac{1}{2} \cdot (- \frac{1}{5})$

$B = - 2 C$

$A = \frac{1}{2}$

Step 3.5.3.3.3

Multiply $\frac{1}{2} (- \frac{1}{5})$ .

Tap for more steps...

Step 3.5.3.3.3.1

Multiply $\frac{1}{2}$ by $\frac{1}{5}$ .

$C = - \frac{1}{2 \cdot 5}$

$B = - 2 C$

$A = \frac{1}{2}$

Step 3.5.3.3.3.2

Multiply $2$ by $5$ .

$C = - \frac{1}{10}$

$B = - 2 C$

$A = \frac{1}{2}$

$C = - \frac{1}{10}$

$B = - 2 C$

$A = \frac{1}{2}$

$C = - \frac{1}{10}$

$B = - 2 C$

$A = \frac{1}{2}$

$C = - \frac{1}{10}$

$B = - 2 C$

$A = \frac{1}{2}$

$C = - \frac{1}{10}$

$B = - 2 C$

$A = \frac{1}{2}$

Step 3.6

Replace all occurrences of $C$ with $- \frac{1}{10}$ in each equation.

Tap for more steps...

Step 3.6.1

Replace all occurrences of $C$ in $B = - 2 C$ with $- \frac{1}{10}$ .

$B = - 2 (- \frac{1}{10})$

$C = - \frac{1}{10}$

$A = \frac{1}{2}$

Step 3.6.2

Simplify the right side.

Tap for more steps...

Step 3.6.2.1

Simplify $- 2 (- \frac{1}{10})$ .

Tap for more steps...

Step 3.6.2.1.1

Cancel the common factor of $2$ .

Tap for more steps...

Step 3.6.2.1.1.1

Move the leading negative in $- \frac{1}{10}$ into the numerator.

$B = - 2 (\frac{- 1}{10})$

$C = - \frac{1}{10}$

$A = \frac{1}{2}$

Step 3.6.2.1.1.2

Factor $2$ out of $- 2$ .

$B = 2 (- 1) (\frac{- 1}{10})$

$C = - \frac{1}{10}$

$A = \frac{1}{2}$

Step 3.6.2.1.1.3

Factor $2$ out of $10$ .

$B = 2 \cdot (- 1 \frac{- 1}{2 \cdot 5})$

$C = - \frac{1}{10}$

$A = \frac{1}{2}$

Step 3.6.2.1.1.4

Cancel the common factor.

$B = 2 \cdot (- 1 \frac{- 1}{2 \cdot 5})$

$C = - \frac{1}{10}$

$A = \frac{1}{2}$

Step 3.6.2.1.1.5

Rewrite the expression.

$B = - 1 (\frac{- 1}{5})$

$C = - \frac{1}{10}$

$A = \frac{1}{2}$

$B = - 1 (\frac{- 1}{5})$

$C = - \frac{1}{10}$

$A = \frac{1}{2}$

Step 3.6.2.1.2

Move the negative in front of the fraction.

$B = - 1 (- \frac{1}{5})$

$C = - \frac{1}{10}$

$A = \frac{1}{2}$

Step 3.6.2.1.3

Multiply $- 1 (- \frac{1}{5})$ .

Tap for more steps...

Step 3.6.2.1.3.1

Multiply $- 1$ by $- 1$ .

$B = 1 (\frac{1}{5})$

$C = - \frac{1}{10}$

$A = \frac{1}{2}$

Step 3.6.2.1.3.2

Multiply $\frac{1}{5}$ by $1$ .

$B = \frac{1}{5}$

$C = - \frac{1}{10}$

$A = \frac{1}{2}$

$B = \frac{1}{5}$

$C = - \frac{1}{10}$

$A = \frac{1}{2}$

$B = \frac{1}{5}$

$C = - \frac{1}{10}$

$A = \frac{1}{2}$

$B = \frac{1}{5}$

$C = - \frac{1}{10}$

$A = \frac{1}{2}$

$B = \frac{1}{5}$

$C = - \frac{1}{10}$

$A = \frac{1}{2}$

Step 3.7

List all of the solutions.

$B = \frac{1}{5}, C = - \frac{1}{10}, A = \frac{1}{2}$

Step 4

Replace each of the partial fraction coefficients in $\frac{A}{x} + \frac{B}{2 x - 1} + \frac{C}{x + 2}$ with the values found for $A$ , $B$ , and $C$ .

$\frac{\frac{1}{2}}{x} + \frac{\frac{1}{5}}{2 x - 1} + \frac{- \frac{1}{10}}{x + 2}$