Precalculus Examples

$\frac{10 x^{2} + 10 x + 3}{(x + 1) (x^{2} + 3 x + 3)}$

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 $A$ .

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

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 is 2nd order, $2$ terms are required in the numerator. The number of terms required in the numerator is always equal to the order of the factor in the denominator.

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

Step 1.3

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

$\frac{(10 x^{2} + 10 x + 3) (x + 1) (x^{2} + 3 x + 3)}{(x + 1) (x^{2} + 3 x + 3)} = \frac{(A) (x + 1) (x^{2} + 3 x + 3)}{x + 1} + \frac{(B x + C) (x + 1) (x^{2} + 3 x + 3)}{x^{2} + 3 x + 3}$

Step 1.4

Cancel the common factor of $x + 1$ .

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Step 1.4.1

Cancel the common factor.

$\frac{(10 x^{2} + 10 x + 3) (x + 1) (x^{2} + 3 x + 3)}{(x + 1) (x^{2} + 3 x + 3)} = \frac{(A) (x + 1) (x^{2} + 3 x + 3)}{x + 1} + \frac{(B x + C) (x + 1) (x^{2} + 3 x + 3)}{x^{2} + 3 x + 3}$

Step 1.4.2

Rewrite the expression.

$\frac{(10 x^{2} + 10 x + 3) (x^{2} + 3 x + 3)}{x^{2} + 3 x + 3} = \frac{(A) (x + 1) (x^{2} + 3 x + 3)}{x + 1} + \frac{(B x + C) (x + 1) (x^{2} + 3 x + 3)}{x^{2} + 3 x + 3}$

Step 1.5

Cancel the common factor of $x^{2} + 3 x + 3$ .

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Step 1.5.1

Cancel the common factor.

$\frac{(10 x^{2} + 10 x + 3) (x^{2} + 3 x + 3)}{x^{2} + 3 x + 3} = \frac{(A) (x + 1) (x^{2} + 3 x + 3)}{x + 1} + \frac{(B x + C) (x + 1) (x^{2} + 3 x + 3)}{x^{2} + 3 x + 3}$

Step 1.5.2

Divide $10 x^{2} + 10 x + 3$ by $1$ .

$10 x^{2} + 10 x + 3 = \frac{(A) (x + 1) (x^{2} + 3 x + 3)}{x + 1} + \frac{(B x + C) (x + 1) (x^{2} + 3 x + 3)}{x^{2} + 3 x + 3}$

Step 1.6

Simplify each term.

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Step 1.6.1

Cancel the common factor of $x + 1$ .

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Step 1.6.1.1

Cancel the common factor.

$10 x^{2} + 10 x + 3 = \frac{A (x + 1) (x^{2} + 3 x + 3)}{x + 1} + \frac{(B x + C) (x + 1) (x^{2} + 3 x + 3)}{x^{2} + 3 x + 3}$

Step 1.6.1.2

Divide $(A) (x^{2} + 3 x + 3)$ by $1$ .

$10 x^{2} + 10 x + 3 = (A) (x^{2} + 3 x + 3) + \frac{(B x + C) (x + 1) (x^{2} + 3 x + 3)}{x^{2} + 3 x + 3}$

Step 1.6.2

Apply the distributive property.

$10 x^{2} + 10 x + 3 = A x^{2} + A (3 x) + A \cdot 3 + \frac{(B x + C) (x + 1) (x^{2} + 3 x + 3)}{x^{2} + 3 x + 3}$

Step 1.6.3

Simplify.

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Step 1.6.3.1

Rewrite using the commutative property of multiplication.

$10 x^{2} + 10 x + 3 = A x^{2} + 3 A x + A \cdot 3 + \frac{(B x + C) (x + 1) (x^{2} + 3 x + 3)}{x^{2} + 3 x + 3}$

Step 1.6.3.2

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

$10 x^{2} + 10 x + 3 = A x^{2} + 3 A x + 3 \cdot A + \frac{(B x + C) (x + 1) (x^{2} + 3 x + 3)}{x^{2} + 3 x + 3}$

Step 1.6.4

Cancel the common factor of $x^{2} + 3 x + 3$ .

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Step 1.6.4.1

Cancel the common factor.

$10 x^{2} + 10 x + 3 = A x^{2} + 3 A x + 3 A + \frac{(B x + C) (x + 1) (x^{2} + 3 x + 3)}{x^{2} + 3 x + 3}$

Step 1.6.4.2

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

$10 x^{2} + 10 x + 3 = A x^{2} + 3 A x + 3 A + (B x + C) (x + 1)$

Step 1.6.5

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

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Step 1.6.5.1

Apply the distributive property.

$10 x^{2} + 10 x + 3 = A x^{2} + 3 A x + 3 A + B x (x + 1) + C (x + 1)$

Step 1.6.5.2

Apply the distributive property.

$10 x^{2} + 10 x + 3 = A x^{2} + 3 A x + 3 A + B x \cdot x + B x \cdot 1 + C (x + 1)$

Step 1.6.5.3

Apply the distributive property.

$10 x^{2} + 10 x + 3 = A x^{2} + 3 A x + 3 A + B x \cdot x + B x \cdot 1 + C x + C \cdot 1$

Step 1.6.6

Simplify each term.

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Step 1.6.6.1

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

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Step 1.6.6.1.1

Move $x$ .

$10 x^{2} + 10 x + 3 = A x^{2} + 3 A x + 3 A + B (x \cdot x) + B x \cdot 1 + C x + C \cdot 1$

Step 1.6.6.1.2

Multiply $x$ by $x$ .

$10 x^{2} + 10 x + 3 = A x^{2} + 3 A x + 3 A + B x^{2} + B x \cdot 1 + C x + C \cdot 1$

Step 1.6.6.2

Multiply $B$ by $1$ .

$10 x^{2} + 10 x + 3 = A x^{2} + 3 A x + 3 A + B x^{2} + B x + C x + C \cdot 1$

Step 1.6.6.3

Multiply $C$ by $1$ .

$10 x^{2} + 10 x + 3 = A x^{2} + 3 A x + 3 A + B x^{2} + B x + C x + C$

Step 1.7

Simplify the expression.

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Step 1.7.1

Move $A$ .

$10 x^{2} + 10 x + 3 = A x^{2} + 3 x A + 3 A + B x^{2} + B x + C x + C$

Step 1.7.2

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

$10 x^{2} + 10 x + 3 = A x^{2} + 3 x A + 3 A + B x^{2} + B x + C x + C$

Step 1.7.3

Reorder $B$ and $x$ .

$10 x^{2} + 10 x + 3 = A x^{2} + 3 x A + 3 A + B x^{2} + B x + C x + C$

Step 1.7.4

Move $3 A$ .

$10 x^{2} + 10 x + 3 = A x^{2} + 3 x A + B x^{2} + B x + C x + 3 A + C$

Step 1.7.5

Move $3 x A$ .

$10 x^{2} + 10 x + 3 = A x^{2} + B x^{2} + 3 x A + B x + C x + 3 A + C$

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.

$10 = A + B$

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.

$10 = 3 A + B + C$

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.

$3 = 3 A + C$

Step 2.4

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

$10 = A + B$

$10 = 3 A + B + C$

$3 = 3 A + C$

$10 = A + B$

$10 = 3 A + B + C$

$3 = 3 A + C$

Step 3

Solve the system of equations.

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Step 3.1

Solve for $A$ in $10 = A + B$ .

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Step 3.1.1

Rewrite the equation as $A + B = 10$ .

$A + B = 10$

$10 = 3 A + B + C$

$3 = 3 A + C$

Step 3.1.2

Subtract $B$ from both sides of the equation.

$A = 10 - B$

$10 = 3 A + B + C$

$3 = 3 A + C$

$A = 10 - B$

$10 = 3 A + B + C$

$3 = 3 A + C$

Step 3.2

Replace all occurrences of $A$ with $10 - B$ in each equation.

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Step 3.2.1

Replace all occurrences of $A$ in $10 = 3 A + B + C$ with $10 - B$ .

$10 = 3 (10 - B) + B + C$

$A = 10 - B$

$3 = 3 A + C$

Step 3.2.2

Simplify the right side.

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Step 3.2.2.1

Simplify $3 (10 - B) + B + C$ .

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Step 3.2.2.1.1

Simplify each term.

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Step 3.2.2.1.1.1

Apply the distributive property.

$10 = 3 \cdot 10 + 3 (- B) + B + C$

$A = 10 - B$

$3 = 3 A + C$

Step 3.2.2.1.1.2

Multiply $3$ by $10$ .

$10 = 30 + 3 (- B) + B + C$

$A = 10 - B$

$3 = 3 A + C$

Step 3.2.2.1.1.3

Multiply $- 1$ by $3$ .

$10 = 30 - 3 B + B + C$

$A = 10 - B$

$3 = 3 A + C$

$10 = 30 - 3 B + B + C$

$A = 10 - B$

$3 = 3 A + C$

Step 3.2.2.1.2

Add $- 3 B$ and $B$ .

$10 = 30 - 2 B + C$

$A = 10 - B$

$3 = 3 A + C$

$10 = 30 - 2 B + C$

$A = 10 - B$

$3 = 3 A + C$

$10 = 30 - 2 B + C$

$A = 10 - B$

$3 = 3 A + C$

Step 3.2.3

Replace all occurrences of $A$ in $3 = 3 A + C$ with $10 - B$ .

$3 = 3 (10 - B) + C$

$10 = 30 - 2 B + C$

$A = 10 - B$

Step 3.2.4

Simplify the right side.

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Step 3.2.4.1

Simplify each term.

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Step 3.2.4.1.1

Apply the distributive property.

$3 = 3 \cdot 10 + 3 (- B) + C$

$10 = 30 - 2 B + C$

$A = 10 - B$

Step 3.2.4.1.2

Multiply $3$ by $10$ .

$3 = 30 + 3 (- B) + C$

$10 = 30 - 2 B + C$

$A = 10 - B$

Step 3.2.4.1.3

Multiply $- 1$ by $3$ .

$3 = 30 - 3 B + C$

$10 = 30 - 2 B + C$

$A = 10 - B$

$3 = 30 - 3 B + C$

$10 = 30 - 2 B + C$

$A = 10 - B$

$3 = 30 - 3 B + C$

$10 = 30 - 2 B + C$

$A = 10 - B$

$3 = 30 - 3 B + C$

$10 = 30 - 2 B + C$

$A = 10 - B$

Step 3.3

Reorder $10$ and $- B$ .

$A = - B + 10$

$3 = 30 - 3 B + C$

$10 = 30 - 2 B + C$

Step 3.4

Solve for $C$ in $3 = 30 - 3 B + C$ .

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Step 3.4.1

Rewrite the equation as $30 - 3 B + C = 3$ .

$30 - 3 B + C = 3$

$A = - B + 10$

$10 = 30 - 2 B + C$

Step 3.4.2

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

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Step 3.4.2.1

Subtract $30$ from both sides of the equation.

$- 3 B + C = 3 - 30$

$A = - B + 10$

$10 = 30 - 2 B + C$

Step 3.4.2.2

Add $3 B$ to both sides of the equation.

$C = 3 - 30 + 3 B$

$A = - B + 10$

$10 = 30 - 2 B + C$

Step 3.4.2.3

Subtract $30$ from $3$ .

$C = - 27 + 3 B$

$A = - B + 10$

$10 = 30 - 2 B + C$

$C = - 27 + 3 B$

$A = - B + 10$

$10 = 30 - 2 B + C$

$C = - 27 + 3 B$

$A = - B + 10$

$10 = 30 - 2 B + C$

Step 3.5

Replace all occurrences of $C$ with $- 27 + 3 B$ in each equation.

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Step 3.5.1

Replace all occurrences of $C$ in $10 = 30 - 2 B + C$ with $- 27 + 3 B$ .

$10 = 30 - 2 B - 27 + 3 B$

$C = - 27 + 3 B$

$A = - B + 10$

Step 3.5.2

Simplify $10 = 30 - 2 B - 27 + 3 B$ .

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Step 3.5.2.1

Simplify the left side.

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Step 3.5.2.1.1

Remove parentheses.

$10 = 30 - 2 B - 27 + 3 B$

$C = - 27 + 3 B$

$A = - B + 10$

$10 = 30 - 2 B - 27 + 3 B$

$C = - 27 + 3 B$

$A = - B + 10$

Step 3.5.2.2

Simplify the right side.

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Step 3.5.2.2.1

Simplify $30 - 2 B - 27 + 3 B$ .

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Step 3.5.2.2.1.1

Subtract $27$ from $30$ .

$10 = - 2 B + 3 + 3 B$

$C = - 27 + 3 B$

$A = - B + 10$

Step 3.5.2.2.1.2

Add $- 2 B$ and $3 B$ .

$10 = B + 3$

$C = - 27 + 3 B$

$A = - B + 10$

$10 = B + 3$

$C = - 27 + 3 B$

$A = - B + 10$

$10 = B + 3$

$C = - 27 + 3 B$

$A = - B + 10$

$10 = B + 3$

$C = - 27 + 3 B$

$A = - B + 10$

$10 = B + 3$

$C = - 27 + 3 B$

$A = - B + 10$

Step 3.6

Solve for $B$ in $10 = B + 3$ .

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Step 3.6.1

Rewrite the equation as $B + 3 = 10$ .

$B + 3 = 10$

$C = - 27 + 3 B$

$A = - B + 10$

Step 3.6.2

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

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Step 3.6.2.1

Subtract $3$ from both sides of the equation.

$B = 10 - 3$

$C = - 27 + 3 B$

$A = - B + 10$

Step 3.6.2.2

Subtract $3$ from $10$ .

$B = 7$

$C = - 27 + 3 B$

$A = - B + 10$

$B = 7$

$C = - 27 + 3 B$

$A = - B + 10$

$B = 7$

$C = - 27 + 3 B$

$A = - B + 10$

Step 3.7

Replace all occurrences of $B$ with $7$ in each equation.

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Step 3.7.1

Replace all occurrences of $B$ in $C = - 27 + 3 B$ with $7$ .

$C = - 27 + 3 (7)$

$B = 7$

$A = - B + 10$

Step 3.7.2

Simplify the right side.

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Step 3.7.2.1

Simplify $- 27 + 3 (7)$ .

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Step 3.7.2.1.1

Multiply $3$ by $7$ .

$C = - 27 + 21$

$B = 7$

$A = - B + 10$

Step 3.7.2.1.2

Add $- 27$ and $21$ .

$C = - 6$

$B = 7$

$A = - B + 10$

$C = - 6$

$B = 7$

$A = - B + 10$

$C = - 6$

$B = 7$

$A = - B + 10$

Step 3.7.3

Replace all occurrences of $B$ in $A = - B + 10$ with $7$ .

$A = - (7) + 10$

$C = - 6$

$B = 7$

Step 3.7.4

Simplify the right side.

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Step 3.7.4.1

Simplify $- (7) + 10$ .

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Step 3.7.4.1.1

Multiply $- 1$ by $7$ .

$A = - 7 + 10$

$C = - 6$

$B = 7$

Step 3.7.4.1.2

Add $- 7$ and $10$ .

$A = 3$

$C = - 6$

$B = 7$

$A = 3$

$C = - 6$

$B = 7$

$A = 3$

$C = - 6$

$B = 7$

$A = 3$

$C = - 6$

$B = 7$

Step 3.8

List all of the solutions.

$A = 3, C = - 6, B = 7$

Step 4

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

$\frac{3}{x + 1} + \frac{7 x - 6}{x^{2} + 3 x + 3}$

Step 5

Multiply $7$ by $x$ .

$\frac{3}{x + 1} + \frac{7 x - 6}{x^{2} + 3 x + 3}$