Precalculus Examples

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

Step 1

Decompose the fraction and multiply through by the common denominator.

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Step 1.1

Factor by grouping.

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Step 1.1.1

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 $a \cdot c = 2 \cdot 1 = 2$ and whose sum is $b = 3$ .

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Step 1.1.1.1

Factor $3$ out of $3 x$ .

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

Step 1.1.1.2

Rewrite $3$ as $1$ plus $2$

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

Step 1.1.1.3

Apply the distributive property.

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

Step 1.1.1.4

Multiply $x$ by $1$ .

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

Step 1.1.2

Factor out the greatest common factor from each group.

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Step 1.1.2.1

Group the first two terms and the last two terms.

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

Step 1.1.2.2

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

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

Step 1.1.3

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

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

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 $A$ .

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

Step 1.3

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

Step 1.4

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

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

Step 1.5

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

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Step 1.5.1

Cancel the common factor.

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

Step 1.5.2

Rewrite the expression.

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

Step 1.6

Cancel the common factor of $x + 1$ .

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Step 1.6.1

Cancel the common factor.

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

Step 1.6.2

Divide $x + 2$ by $1$ .

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

Step 1.7

Simplify each term.

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Step 1.7.1

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

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Step 1.7.1.1

Cancel the common factor.

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

Step 1.7.1.2

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

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

Step 1.7.2

Apply the distributive property.

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

Step 1.7.3

Multiply $A$ by $1$ .

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

Step 1.7.4

Cancel the common factor of $x + 1$ .

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Step 1.7.4.1

Cancel the common factor.

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

Step 1.7.4.2

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

$x + 2 = A x + A + (B) (2 x + 1)$

Step 1.7.5

Apply the distributive property.

$x + 2 = A x + A + B (2 x) + B \cdot 1$

Step 1.7.6

Rewrite using the commutative property of multiplication.

$x + 2 = A x + A + 2 B x + B \cdot 1$

Step 1.7.7

Multiply $B$ by $1$ .

$x + 2 = A x + A + 2 B x + B$

Step 1.8

Simplify the expression.

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Step 1.8.1

Move $B$ .

$x + 2 = A x + A + 2 x B + B$

Step 1.8.2

Move $A$ .

$x + 2 = A x + 2 x B + A + B$

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

$1 = A + 2 B$

Step 2.2

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.

$2 = A + B$

Step 2.3

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

$1 = A + 2 B$

$2 = A + B$

$1 = A + 2 B$

$2 = A + B$

Step 3

Solve the system of equations.

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Step 3.1

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

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Step 3.1.1

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

$A + 2 B = 1$

$2 = A + B$

Step 3.1.2

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

$A = 1 - 2 B$

$2 = A + B$

$A = 1 - 2 B$

$2 = A + B$

Step 3.2

Replace all occurrences of $A$ with $1 - 2 B$ in each equation.

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Step 3.2.1

Replace all occurrences of $A$ in $2 = A + B$ with $1 - 2 B$ .

$2 = (1 - 2 B) + B$

$A = 1 - 2 B$

Step 3.2.2

Simplify the right side.

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Step 3.2.2.1

Add $- 2 B$ and $B$ .

$2 = 1 - B$

$A = 1 - 2 B$

$2 = 1 - B$

$A = 1 - 2 B$

$2 = 1 - B$

$A = 1 - 2 B$

Step 3.3

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

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Step 3.3.1

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

$1 - B = 2$

$A = 1 - 2 B$

Step 3.3.2

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

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Step 3.3.2.1

Subtract $1$ from both sides of the equation.

$- B = 2 - 1$

$A = 1 - 2 B$

Step 3.3.2.2

Subtract $1$ from $2$ .

$- B = 1$

$A = 1 - 2 B$

$- B = 1$

$A = 1 - 2 B$

Step 3.3.3

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

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Step 3.3.3.1

Divide each term in $- B = 1$ by $- 1$ .

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

$A = 1 - 2 B$

Step 3.3.3.2

Simplify the left side.

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Step 3.3.3.2.1

Dividing two negative values results in a positive value.

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

$A = 1 - 2 B$

Step 3.3.3.2.2

Divide $B$ by $1$ .

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

$A = 1 - 2 B$

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

$A = 1 - 2 B$

Step 3.3.3.3

Simplify the right side.

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Step 3.3.3.3.1

Divide $1$ by $- 1$ .

$B = - 1$

$A = 1 - 2 B$

$B = - 1$

$A = 1 - 2 B$

$B = - 1$

$A = 1 - 2 B$

$B = - 1$

$A = 1 - 2 B$

Step 3.4

Replace all occurrences of $B$ with $- 1$ in each equation.

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Step 3.4.1

Replace all occurrences of $B$ in $A = 1 - 2 B$ with $- 1$ .

$A = 1 - 2 \cdot - 1$

$B = - 1$

Step 3.4.2

Simplify the right side.

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Step 3.4.2.1

Simplify $1 - 2 \cdot - 1$ .

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Step 3.4.2.1.1

Multiply $- 2$ by $- 1$ .

$A = 1 + 2$

$B = - 1$

Step 3.4.2.1.2

Add $1$ and $2$ .

$A = 3$

$B = - 1$

$A = 3$

$B = - 1$

$A = 3$

$B = - 1$

$A = 3$

$B = - 1$

Step 3.5

List all of the solutions.

$A = 3, B = - 1$

Step 4

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

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