Precalculus Examples

$\frac{4 x + 62}{{(x - 2)}^{2} (x + 3)}$

Step 1

Decompose the fraction and multiply through by the common denominator.

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Step 1.1

Factor $2$ out of $4 x + 62$ .

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Step 1.1.1

Factor $2$ out of $4 x$ .

$\frac{2 (2 x) + 62}{{(x - 2)}^{2} (x + 3)}$

Step 1.1.2

Factor $2$ out of $62$ .

$\frac{2 (2 x) + 2 (31)}{{(x - 2)}^{2} (x + 3)}$

Step 1.1.3

Factor $2$ out of $2 (2 x) + 2 (31)$ .

$\frac{2 (2 x + 31)}{{(x - 2)}^{2} (x + 3)}$

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 $A$ .

$\frac{A}{{(x - 2)}^{2}}$

Step 1.3

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

Step 1.4

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

Step 1.5

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

$\frac{2 (2 x + 31) {(x - 2)}^{2} (x + 3)}{{(x - 2)}^{2} (x + 3)} = \frac{(A) {(x - 2)}^{2} (x + 3)}{{(x - 2)}^{2}} + \frac{(B) {(x - 2)}^{2} (x + 3)}{x - 2} + \frac{(C) {(x - 2)}^{2} (x + 3)}{x + 3}$

Step 1.6

Cancel the common factor of ${(x - 2)}^{2}$ .

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Step 1.6.1

Cancel the common factor.

Step 1.6.2

Rewrite the expression.

$\frac{2 (2 x + 31) (x + 3)}{x + 3} = \frac{(A) {(x - 2)}^{2} (x + 3)}{{(x - 2)}^{2}} + \frac{(B) {(x - 2)}^{2} (x + 3)}{x - 2} + \frac{(C) {(x - 2)}^{2} (x + 3)}{x + 3}$

Step 1.7

Cancel the common factor of $x + 3$ .

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Step 1.7.1

Cancel the common factor.

$\frac{2 (2 x + 31) (x + 3)}{x + 3} = \frac{(A) {(x - 2)}^{2} (x + 3)}{{(x - 2)}^{2}} + \frac{(B) {(x - 2)}^{2} (x + 3)}{x - 2} + \frac{(C) {(x - 2)}^{2} (x + 3)}{x + 3}$

Step 1.7.2

Divide $2 (2 x + 31)$ by $1$ .

$2 (2 x + 31) = \frac{(A) {(x - 2)}^{2} (x + 3)}{{(x - 2)}^{2}} + \frac{(B) {(x - 2)}^{2} (x + 3)}{x - 2} + \frac{(C) {(x - 2)}^{2} (x + 3)}{x + 3}$

Step 1.8

Apply the distributive property.

$2 (2 x) + 2 \cdot 31 = \frac{(A) {(x - 2)}^{2} (x + 3)}{{(x - 2)}^{2}} + \frac{(B) {(x - 2)}^{2} (x + 3)}{x - 2} + \frac{(C) {(x - 2)}^{2} (x + 3)}{x + 3}$

Step 1.9

Multiply.

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Step 1.9.1

Multiply $2$ by $2$ .

$4 x + 2 \cdot 31 = \frac{(A) {(x - 2)}^{2} (x + 3)}{{(x - 2)}^{2}} + \frac{(B) {(x - 2)}^{2} (x + 3)}{x - 2} + \frac{(C) {(x - 2)}^{2} (x + 3)}{x + 3}$

Step 1.9.2

Multiply $2$ by $31$ .

$4 x + 62 = \frac{(A) {(x - 2)}^{2} (x + 3)}{{(x - 2)}^{2}} + \frac{(B) {(x - 2)}^{2} (x + 3)}{x - 2} + \frac{(C) {(x - 2)}^{2} (x + 3)}{x + 3}$

Step 1.10

Simplify each term.

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Step 1.10.1

Cancel the common factor of ${(x - 2)}^{2}$ .

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Step 1.10.1.1

Cancel the common factor.

$4 x + 62 = \frac{A {(x - 2)}^{2} (x + 3)}{{(x - 2)}^{2}} + \frac{(B) {(x - 2)}^{2} (x + 3)}{x - 2} + \frac{(C) {(x - 2)}^{2} (x + 3)}{x + 3}$

Step 1.10.1.2

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

$4 x + 62 = (A) (x + 3) + \frac{(B) {(x - 2)}^{2} (x + 3)}{x - 2} + \frac{(C) {(x - 2)}^{2} (x + 3)}{x + 3}$

Step 1.10.2

Apply the distributive property.

$4 x + 62 = A x + A \cdot 3 + \frac{(B) {(x - 2)}^{2} (x + 3)}{x - 2} + \frac{(C) {(x - 2)}^{2} (x + 3)}{x + 3}$

Step 1.10.3

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

$4 x + 62 = A x + 3 \cdot A + \frac{(B) {(x - 2)}^{2} (x + 3)}{x - 2} + \frac{(C) {(x - 2)}^{2} (x + 3)}{x + 3}$

Step 1.10.4

Cancel the common factor of ${(x - 2)}^{2}$ and $x - 2$ .

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Step 1.10.4.1

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

$4 x + 62 = A x + 3 A + \frac{(x - 2) (B (x - 2) (x + 3))}{x - 2} + \frac{(C) {(x - 2)}^{2} (x + 3)}{x + 3}$

Step 1.10.4.2

Cancel the common factors.

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Step 1.10.4.2.1

Multiply by $1$ .

$4 x + 62 = A x + 3 A + \frac{(x - 2) (B (x - 2) (x + 3))}{(x - 2) \cdot 1} + \frac{(C) {(x - 2)}^{2} (x + 3)}{x + 3}$

Step 1.10.4.2.2

Cancel the common factor.

$4 x + 62 = A x + 3 A + \frac{(x - 2) (B (x - 2) (x + 3))}{(x - 2) \cdot 1} + \frac{(C) {(x - 2)}^{2} (x + 3)}{x + 3}$

Step 1.10.4.2.3

Rewrite the expression.

$4 x + 62 = A x + 3 A + \frac{B (x - 2) (x + 3)}{1} + \frac{(C) {(x - 2)}^{2} (x + 3)}{x + 3}$

Step 1.10.4.2.4

Divide $B (x - 2) (x + 3)$ by $1$ .

$4 x + 62 = A x + 3 A + B (x - 2) (x + 3) + \frac{(C) {(x - 2)}^{2} (x + 3)}{x + 3}$

Step 1.10.5

Apply the distributive property.

$4 x + 62 = A x + 3 A + (B x + B \cdot - 2) (x + 3) + \frac{(C) {(x - 2)}^{2} (x + 3)}{x + 3}$

Step 1.10.6

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

$4 x + 62 = A x + 3 A + (B x - 2 \cdot B) (x + 3) + \frac{(C) {(x - 2)}^{2} (x + 3)}{x + 3}$

Step 1.10.7

Expand $(B x - 2 B) (x + 3)$ using the FOIL Method.

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Step 1.10.7.1

Apply the distributive property.

$4 x + 62 = A x + 3 A + B x (x + 3) - 2 B (x + 3) + \frac{(C) {(x - 2)}^{2} (x + 3)}{x + 3}$

Step 1.10.7.2

Apply the distributive property.

$4 x + 62 = A x + 3 A + B x \cdot x + B x \cdot 3 - 2 B (x + 3) + \frac{(C) {(x - 2)}^{2} (x + 3)}{x + 3}$

Step 1.10.7.3

Apply the distributive property.

$4 x + 62 = A x + 3 A + B x \cdot x + B x \cdot 3 - 2 B x - 2 B \cdot 3 + \frac{(C) {(x - 2)}^{2} (x + 3)}{x + 3}$

Step 1.10.8

Simplify and combine like terms.

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Step 1.10.8.1

Simplify each term.

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Step 1.10.8.1.1

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

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Step 1.10.8.1.1.1

Move $x$ .

$4 x + 62 = A x + 3 A + B (x \cdot x) + B x \cdot 3 - 2 B x - 2 B \cdot 3 + \frac{(C) {(x - 2)}^{2} (x + 3)}{x + 3}$

Step 1.10.8.1.1.2

Multiply $x$ by $x$ .

$4 x + 62 = A x + 3 A + B x^{2} + B x \cdot 3 - 2 B x - 2 B \cdot 3 + \frac{(C) {(x - 2)}^{2} (x + 3)}{x + 3}$

Step 1.10.8.1.2

Move $3$ to the left of $B x$ .

$4 x + 62 = A x + 3 A + B x^{2} + 3 \cdot (B x) - 2 B x - 2 B \cdot 3 + \frac{(C) {(x - 2)}^{2} (x + 3)}{x + 3}$

Step 1.10.8.1.3

Multiply $3$ by $- 2$ .

$4 x + 62 = A x + 3 A + B x^{2} + 3 B x - 2 B x - 6 B + \frac{(C) {(x - 2)}^{2} (x + 3)}{x + 3}$

Step 1.10.8.2

Subtract $2 B x$ from $3 B x$ .

$4 x + 62 = A x + 3 A + B x^{2} + 1 B x - 6 B + \frac{(C) {(x - 2)}^{2} (x + 3)}{x + 3}$

Step 1.10.9

Multiply $B$ by $1$ .

$4 x + 62 = A x + 3 A + B x^{2} + B x - 6 B + \frac{(C) {(x - 2)}^{2} (x + 3)}{x + 3}$

Step 1.10.10

Cancel the common factor of $x + 3$ .

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Step 1.10.10.1

Cancel the common factor.

$4 x + 62 = A x + 3 A + B x^{2} + B x - 6 B + \frac{(C) {(x - 2)}^{2} (x + 3)}{x + 3}$

Step 1.10.10.2

Divide $(C) {(x - 2)}^{2}$ by $1$ .

$4 x + 62 = A x + 3 A + B x^{2} + B x - 6 B + (C) {(x - 2)}^{2}$

Step 1.10.11

Rewrite ${(x - 2)}^{2}$ as $(x - 2) (x - 2)$ .

$4 x + 62 = A x + 3 A + B x^{2} + B x - 6 B + C ((x - 2) (x - 2))$

Step 1.10.12

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

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Step 1.10.12.1

Apply the distributive property.

$4 x + 62 = A x + 3 A + B x^{2} + B x - 6 B + C (x (x - 2) - 2 (x - 2))$

Step 1.10.12.2

Apply the distributive property.

$4 x + 62 = A x + 3 A + B x^{2} + B x - 6 B + C (x \cdot x + x \cdot - 2 - 2 (x - 2))$

Step 1.10.12.3

Apply the distributive property.

$4 x + 62 = A x + 3 A + B x^{2} + B x - 6 B + C (x \cdot x + x \cdot - 2 - 2 x - 2 \cdot - 2)$

Step 1.10.13

Simplify and combine like terms.

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Step 1.10.13.1

Simplify each term.

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Step 1.10.13.1.1

Multiply $x$ by $x$ .

$4 x + 62 = A x + 3 A + B x^{2} + B x - 6 B + C (x^{2} + x \cdot - 2 - 2 x - 2 \cdot - 2)$

Step 1.10.13.1.2

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

$4 x + 62 = A x + 3 A + B x^{2} + B x - 6 B + C (x^{2} - 2 \cdot x - 2 x - 2 \cdot - 2)$

Step 1.10.13.1.3

Multiply $- 2$ by $- 2$ .

$4 x + 62 = A x + 3 A + B x^{2} + B x - 6 B + C (x^{2} - 2 x - 2 x + 4)$

Step 1.10.13.2

Subtract $2 x$ from $- 2 x$ .

$4 x + 62 = A x + 3 A + B x^{2} + B x - 6 B + C (x^{2} - 4 x + 4)$

Step 1.10.14

Apply the distributive property.

$4 x + 62 = A x + 3 A + B x^{2} + B x - 6 B + C x^{2} + C (- 4 x) + C \cdot 4$

Step 1.10.15

Simplify.

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Step 1.10.15.1

Rewrite using the commutative property of multiplication.

$4 x + 62 = A x + 3 A + B x^{2} + B x - 6 B + C x^{2} - 4 C x + C \cdot 4$

Step 1.10.15.2

Move $4$ to the left of $C$ .

$4 x + 62 = A x + 3 A + B x^{2} + B x - 6 B + C x^{2} - 4 C x + 4 \cdot C$

$4 x + 62 = A x + 3 A + B x^{2} + B x - 6 B + C x^{2} - 4 C x + 4 C$

Step 1.11

Simplify the expression.

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Step 1.11.1

Reorder $B$ and $x$ .

$4 x + 62 = A x + 3 A + B x^{2} + B x - 6 B + C x^{2} - 4 C x + 4 C$

Step 1.11.2

Move $- 6 B$ .

$4 x + 62 = A x + 3 A + B x^{2} + B x + C x^{2} - 4 C x - 6 B + 4 C$

Step 1.11.3

Move $3 A$ .

$4 x + 62 = A x + B x^{2} + B x + C x^{2} - 4 C x + 3 A - 6 B + 4 C$

Step 1.11.4

Move $B x$ .

$4 x + 62 = A x + B x^{2} + C x^{2} + B x - 4 C x + 3 A - 6 B + 4 C$

Step 1.11.5

Move $A x$ .

$4 x + 62 = B x^{2} + C x^{2} + A x + B x - 4 C x + 3 A - 6 B + 4 C$

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.

$0 = B + C$

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.

$4 = A + B - 4 C$

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.

$62 = 3 A - 6 B + 4 C$

Step 2.4

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

$0 = B + C$

$4 = A + B - 4 C$

$62 = 3 A - 6 B + 4 C$

$0 = B + C$

$4 = A + B - 4 C$

$62 = 3 A - 6 B + 4 C$

Step 3

Solve the system of equations.

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Step 3.1

Solve for $B$ in $0 = B + C$ .

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Step 3.1.1

Rewrite the equation as $B + C = 0$ .

$B + C = 0$

$4 = A + B - 4 C$

$62 = 3 A - 6 B + 4 C$

Step 3.1.2

Subtract $C$ from both sides of the equation.

$B = - C$

$4 = A + B - 4 C$

$62 = 3 A - 6 B + 4 C$

$B = - C$

$4 = A + B - 4 C$

$62 = 3 A - 6 B + 4 C$

Step 3.2

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

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Step 3.2.1

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

$4 = A - C - 4 C$

$B = - C$

$62 = 3 A - 6 B + 4 C$

Step 3.2.2

Simplify $4 = A - C - 4 C$ .

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Step 3.2.2.1

Simplify the left side.

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Step 3.2.2.1.1

Remove parentheses.

$4 = A - C - 4 C$

$B = - C$

$62 = 3 A - 6 B + 4 C$

$4 = A - C - 4 C$

$B = - C$

$62 = 3 A - 6 B + 4 C$

Step 3.2.2.2

Simplify the right side.

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Step 3.2.2.2.1

Subtract $4 C$ from $- C$ .

$4 = A - 5 C$

$B = - C$

$62 = 3 A - 6 B + 4 C$

$4 = A - 5 C$

$B = - C$

$62 = 3 A - 6 B + 4 C$

$4 = A - 5 C$

$B = - C$

$62 = 3 A - 6 B + 4 C$

Step 3.2.3

Replace all occurrences of $B$ in $62 = 3 A - 6 B + 4 C$ with $- C$ .

$62 = 3 A - 6 (- C) + 4 C$

$4 = A - 5 C$

$B = - C$

Step 3.2.4

Simplify the right side.

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Step 3.2.4.1

Simplify $3 A - 6 (- C) + 4 C$ .

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Step 3.2.4.1.1

Multiply $- 1$ by $- 6$ .

$62 = 3 A + 6 C + 4 C$

$4 = A - 5 C$

$B = - C$

Step 3.2.4.1.2

Add $6 C$ and $4 C$ .

$62 = 3 A + 10 C$

$4 = A - 5 C$

$B = - C$

$62 = 3 A + 10 C$

$4 = A - 5 C$

$B = - C$

$62 = 3 A + 10 C$

$4 = A - 5 C$

$B = - C$

$62 = 3 A + 10 C$

$4 = A - 5 C$

$B = - C$

Step 3.3

Solve for $A$ in $4 = A - 5 C$ .

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Step 3.3.1

Rewrite the equation as $A - 5 C = 4$ .

$A - 5 C = 4$

$62 = 3 A + 10 C$

$B = - C$

Step 3.3.2

Add $5 C$ to both sides of the equation.

$A = 4 + 5 C$

$62 = 3 A + 10 C$

$B = - C$

$A = 4 + 5 C$

$62 = 3 A + 10 C$

$B = - C$

Step 3.4

Replace all occurrences of $A$ with $4 + 5 C$ in each equation.

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Step 3.4.1

Replace all occurrences of $A$ in $62 = 3 A + 10 C$ with $4 + 5 C$ .

$62 = 3 (4 + 5 C) + 10 C$

$A = 4 + 5 C$

$B = - C$

Step 3.4.2

Simplify the right side.

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Step 3.4.2.1

Simplify $3 (4 + 5 C) + 10 C$ .

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Step 3.4.2.1.1

Simplify each term.

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Step 3.4.2.1.1.1

Apply the distributive property.

$62 = 3 \cdot 4 + 3 (5 C) + 10 C$

$A = 4 + 5 C$

$B = - C$

Step 3.4.2.1.1.2

Multiply $3$ by $4$ .

$62 = 12 + 3 (5 C) + 10 C$

$A = 4 + 5 C$

$B = - C$

Step 3.4.2.1.1.3

Multiply $5$ by $3$ .

$62 = 12 + 15 C + 10 C$

$A = 4 + 5 C$

$B = - C$

$62 = 12 + 15 C + 10 C$

$A = 4 + 5 C$

$B = - C$

Step 3.4.2.1.2

Add $15 C$ and $10 C$ .

$62 = 12 + 25 C$

$A = 4 + 5 C$

$B = - C$

$62 = 12 + 25 C$

$A = 4 + 5 C$

$B = - C$

$62 = 12 + 25 C$

$A = 4 + 5 C$

$B = - C$

$62 = 12 + 25 C$

$A = 4 + 5 C$

$B = - C$

Step 3.5

Solve for $C$ in $62 = 12 + 25 C$ .

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Step 3.5.1

Rewrite the equation as $12 + 25 C = 62$ .

$12 + 25 C = 62$

$A = 4 + 5 C$

$B = - C$

Step 3.5.2

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

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Step 3.5.2.1

Subtract $12$ from both sides of the equation.

$25 C = 62 - 12$

$A = 4 + 5 C$

$B = - C$

Step 3.5.2.2

Subtract $12$ from $62$ .

$25 C = 50$

$A = 4 + 5 C$

$B = - C$

$25 C = 50$

$A = 4 + 5 C$

$B = - C$

Step 3.5.3

Divide each term in $25 C = 50$ by $25$ and simplify.

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Step 3.5.3.1

Divide each term in $25 C = 50$ by $25$ .

$\frac{25 C}{25} = \frac{50}{25}$

$A = 4 + 5 C$

$B = - C$

Step 3.5.3.2

Simplify the left side.

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Step 3.5.3.2.1

Cancel the common factor of $25$ .

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Step 3.5.3.2.1.1

Cancel the common factor.

$\frac{25 C}{25} = \frac{50}{25}$

$A = 4 + 5 C$

$B = - C$

Step 3.5.3.2.1.2

Divide $C$ by $1$ .

$C = \frac{50}{25}$

$A = 4 + 5 C$

$B = - C$

$C = \frac{50}{25}$

$A = 4 + 5 C$

$B = - C$

$C = \frac{50}{25}$

$A = 4 + 5 C$

$B = - C$

Step 3.5.3.3

Simplify the right side.

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Step 3.5.3.3.1

Divide $50$ by $25$ .

$C = 2$

$A = 4 + 5 C$

$B = - C$

$C = 2$

$A = 4 + 5 C$

$B = - C$

$C = 2$

$A = 4 + 5 C$

$B = - C$

$C = 2$

$A = 4 + 5 C$

$B = - C$

Step 3.6

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

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Step 3.6.1

Replace all occurrences of $C$ in $A = 4 + 5 C$ with $2$ .

$A = 4 + 5 (2)$

$C = 2$

$B = - C$

Step 3.6.2

Simplify the right side.

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Step 3.6.2.1

Simplify $4 + 5 (2)$ .

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Step 3.6.2.1.1

Multiply $5$ by $2$ .

$A = 4 + 10$

$C = 2$

$B = - C$

Step 3.6.2.1.2

Add $4$ and $10$ .

$A = 14$

$C = 2$

$B = - C$

$A = 14$

$C = 2$

$B = - C$

$A = 14$

$C = 2$

$B = - C$

Step 3.6.3

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

$B = - (2)$

$A = 14$

$C = 2$

Step 3.6.4

Simplify the right side.

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Step 3.6.4.1

Multiply $- 1$ by $2$ .

$B = - 2$

$A = 14$

$C = 2$

$B = - 2$

$A = 14$

$C = 2$

$B = - 2$

$A = 14$

$C = 2$

Step 3.7

List all of the solutions.

$B = - 2, A = 14, C = 2$

Step 4

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

$\frac{14}{{(x - 2)}^{2}} + \frac{- 2}{x - 2} + \frac{2}{x + 3}$