Precalculus Examples

$\frac{4}{2 x^{2} - 5 x - 3}$

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 $a \cdot c = 2 \cdot - 3 = - 6$ and whose sum is $b = - 5$ .

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Step 1.1.1.1

Factor $- 5$ out of $- 5 x$ .

$\frac{4}{2 x^{2} - 5 (x) - 3}$

Step 1.1.1.2

Rewrite $- 5$ as $1$ plus $- 6$

$\frac{4}{2 x^{2} + (1 - 6) x - 3}$

Step 1.1.1.3

Apply the distributive property.

$\frac{4}{2 x^{2} + 1 x - 6 x - 3}$

Step 1.1.1.4

Multiply $x$ by $1$ .

$\frac{4}{2 x^{2} + x - 6 x - 3}$

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.

$\frac{4}{(2 x^{2} + x) - 6 x - 3}$

Step 1.1.2.2

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

$\frac{4}{x (2 x + 1) - 3 (2 x + 1)}$

Step 1.1.3

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

$\frac{4}{(2 x + 1) (x - 3)}$

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

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

Step 1.4

Multiply each fraction in the equation by the denominator of the original expression. In this case, the denominator is $(2 x + 1) (x - 3)$ .

$\frac{4 (2 x + 1) (x - 3)}{(2 x + 1) (x - 3)} = \frac{(A) (2 x + 1) (x - 3)}{2 x + 1} + \frac{(B) (2 x + 1) (x - 3)}{x - 3}$

Step 1.5

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

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Step 1.5.1

Cancel the common factor.

$\frac{4 (2 x + 1) (x - 3)}{(2 x + 1) (x - 3)} = \frac{(A) (2 x + 1) (x - 3)}{2 x + 1} + \frac{(B) (2 x + 1) (x - 3)}{x - 3}$

Step 1.5.2

Rewrite the expression.

$\frac{4 (x - 3)}{x - 3} = \frac{(A) (2 x + 1) (x - 3)}{2 x + 1} + \frac{(B) (2 x + 1) (x - 3)}{x - 3}$

Step 1.6

Cancel the common factor of $x - 3$ .

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Step 1.6.1

Cancel the common factor.

$\frac{4 (x - 3)}{x - 3} = \frac{(A) (2 x + 1) (x - 3)}{2 x + 1} + \frac{(B) (2 x + 1) (x - 3)}{x - 3}$

Step 1.6.2

Divide $4$ by $1$ .

$4 = \frac{(A) (2 x + 1) (x - 3)}{2 x + 1} + \frac{(B) (2 x + 1) (x - 3)}{x - 3}$

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.

$4 = \frac{A (2 x + 1) (x - 3)}{2 x + 1} + \frac{(B) (2 x + 1) (x - 3)}{x - 3}$

Step 1.7.1.2

Divide $(A) (x - 3)$ by $1$ .

$4 = (A) (x - 3) + \frac{(B) (2 x + 1) (x - 3)}{x - 3}$

Step 1.7.2

Apply the distributive property.

$4 = A x + A \cdot - 3 + \frac{(B) (2 x + 1) (x - 3)}{x - 3}$

Step 1.7.3

Move $- 3$ to the left of $A$ .

$4 = A x - 3 \cdot A + \frac{(B) (2 x + 1) (x - 3)}{x - 3}$

Step 1.7.4

Cancel the common factor of $x - 3$ .

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Step 1.7.4.1

Cancel the common factor.

$4 = A x - 3 A + \frac{(B) (2 x + 1) (x - 3)}{x - 3}$

Step 1.7.4.2

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

$4 = A x - 3 A + (B) (2 x + 1)$

Step 1.7.5

Apply the distributive property.

$4 = A x - 3 A + B (2 x) + B \cdot 1$

Step 1.7.6

Rewrite using the commutative property of multiplication.

$4 = A x - 3 A + 2 B x + B \cdot 1$

Step 1.7.7

Multiply $B$ by $1$ .

$4 = A x - 3 A + 2 B x + B$

Step 1.8

Simplify the expression.

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Step 1.8.1

Move $B$ .

$4 = A x - 3 A + 2 x B + B$

Step 1.8.2

Move $- 3 A$ .

$4 = A x + 2 x B - 3 A + B$

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.

$0 = A + 2 B$

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.

$4 = - 3 A + B$

Step 2.3

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

$0 = A + 2 B$

$4 = - 3 A + B$

$0 = A + 2 B$

$4 = - 3 A + B$

Step 3

Solve the system of equations.

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Step 3.1

Solve for $A$ in $0 = A + 2 B$ .

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Step 3.1.1

Rewrite the equation as $A + 2 B = 0$ .

$A + 2 B = 0$

$4 = - 3 A + B$

Step 3.1.2

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

$A = - 2 B$

$4 = - 3 A + B$

$A = - 2 B$

$4 = - 3 A + B$

Step 3.2

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

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Step 3.2.1

Replace all occurrences of $A$ in $4 = - 3 A + B$ with $- 2 B$ .

$4 = - 3 (- 2 B) + B$

$A = - 2 B$

Step 3.2.2

Simplify the right side.

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Step 3.2.2.1

Simplify $- 3 (- 2 B) + B$ .

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Step 3.2.2.1.1

Multiply $- 2$ by $- 3$ .

$4 = 6 B + B$

$A = - 2 B$

Step 3.2.2.1.2

Add $6 B$ and $B$ .

$4 = 7 B$

$A = - 2 B$

$4 = 7 B$

$A = - 2 B$

$4 = 7 B$

$A = - 2 B$

$4 = 7 B$

$A = - 2 B$

Step 3.3

Solve for $B$ in $4 = 7 B$ .

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Step 3.3.1

Rewrite the equation as $7 B = 4$ .

$7 B = 4$

$A = - 2 B$

Step 3.3.2

Divide each term in $7 B = 4$ by $7$ and simplify.

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Step 3.3.2.1

Divide each term in $7 B = 4$ by $7$ .

$\frac{7 B}{7} = \frac{4}{7}$

$A = - 2 B$

Step 3.3.2.2

Simplify the left side.

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Step 3.3.2.2.1

Cancel the common factor of $7$ .

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Step 3.3.2.2.1.1

Cancel the common factor.

$\frac{7 B}{7} = \frac{4}{7}$

$A = - 2 B$

Step 3.3.2.2.1.2

Divide $B$ by $1$ .

$B = \frac{4}{7}$

$A = - 2 B$

$B = \frac{4}{7}$

$A = - 2 B$

$B = \frac{4}{7}$

$A = - 2 B$

$B = \frac{4}{7}$

$A = - 2 B$

$B = \frac{4}{7}$

$A = - 2 B$

Step 3.4

Replace all occurrences of $B$ with $\frac{4}{7}$ in each equation.

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Step 3.4.1

Replace all occurrences of $B$ in $A = - 2 B$ with $\frac{4}{7}$ .

$A = - 2 (\frac{4}{7})$

$B = \frac{4}{7}$

Step 3.4.2

Simplify the right side.

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Step 3.4.2.1

Simplify $- 2 (\frac{4}{7})$ .

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Step 3.4.2.1.1

Multiply $- 2 (\frac{4}{7})$ .

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Step 3.4.2.1.1.1

Combine $- 2$ and $\frac{4}{7}$ .

$A = \frac{- 2 \cdot 4}{7}$

$B = \frac{4}{7}$

Step 3.4.2.1.1.2

Multiply $- 2$ by $4$ .

$A = \frac{- 8}{7}$

$B = \frac{4}{7}$

$A = \frac{- 8}{7}$

$B = \frac{4}{7}$

Step 3.4.2.1.2

Move the negative in front of the fraction.

$A = - \frac{8}{7}$

$B = \frac{4}{7}$

$A = - \frac{8}{7}$

$B = \frac{4}{7}$

$A = - \frac{8}{7}$

$B = \frac{4}{7}$

$A = - \frac{8}{7}$

$B = \frac{4}{7}$

Step 3.5

List all of the solutions.

$A = - \frac{8}{7}, B = \frac{4}{7}$

Step 4

Replace each of the partial fraction coefficients in $\frac{A}{2 x + 1} + \frac{B}{x - 3}$ with the values found for $A$ and $B$ .

$\frac{- \frac{8}{7}}{2 x + 1} + \frac{\frac{4}{7}}{x - 3}$