Precalculus Examples

$\frac{6 x^{2} - 8 x + 3}{{(1 - x)}^{3}}$

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 $A$ .

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

Step 1.2

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

Step 1.3

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

Step 1.4

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

$\frac{(6 x^{2} - 8 x + 3) {(1 - x)}^{3}}{{(1 - x)}^{3}} = \frac{(A) {(1 - x)}^{3}}{{(1 - x)}^{3}} + \frac{(B) {(1 - x)}^{3}}{{(1 - x)}^{2}} + \frac{(C) {(1 - x)}^{3}}{1 - x}$

Step 1.5

Cancel the common factor of ${(1 - x)}^{3}$ .

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Step 1.5.1

Cancel the common factor.

$\frac{(6 x^{2} - 8 x + 3) {(1 - x)}^{3}}{{(1 - x)}^{3}} = \frac{(A) {(1 - x)}^{3}}{{(1 - x)}^{3}} + \frac{(B) {(1 - x)}^{3}}{{(1 - x)}^{2}} + \frac{(C) {(1 - x)}^{3}}{1 - x}$

Step 1.5.2

Divide $6 x^{2} - 8 x + 3$ by $1$ .

$6 x^{2} - 8 x + 3 = \frac{(A) {(1 - x)}^{3}}{{(1 - x)}^{3}} + \frac{(B) {(1 - x)}^{3}}{{(1 - x)}^{2}} + \frac{(C) {(1 - x)}^{3}}{1 - x}$

Step 1.6

Simplify each term.

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Step 1.6.1

Cancel the common factor of ${(1 - x)}^{3}$ .

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Step 1.6.1.1

Cancel the common factor.

$6 x^{2} - 8 x + 3 = \frac{A {(1 - x)}^{3}}{{(1 - x)}^{3}} + \frac{(B) {(1 - x)}^{3}}{{(1 - x)}^{2}} + \frac{(C) {(1 - x)}^{3}}{1 - x}$

Step 1.6.1.2

Divide $A$ by $1$ .

$6 x^{2} - 8 x + 3 = A + \frac{(B) {(1 - x)}^{3}}{{(1 - x)}^{2}} + \frac{(C) {(1 - x)}^{3}}{1 - x}$

Step 1.6.2

Cancel the common factor of ${(1 - x)}^{3}$ and ${(1 - x)}^{2}$ .

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Step 1.6.2.1

Factor ${(1 - x)}^{2}$ out of $(B) {(1 - x)}^{3}$ .

$6 x^{2} - 8 x + 3 = A + \frac{{(1 - x)}^{2} (B (1 - x))}{{(1 - x)}^{2}} + \frac{(C) {(1 - x)}^{3}}{1 - x}$

Step 1.6.2.2

Cancel the common factors.

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Step 1.6.2.2.1

Multiply by $1$ .

$6 x^{2} - 8 x + 3 = A + \frac{{(1 - x)}^{2} (B (1 - x))}{{(1 - x)}^{2} \cdot 1} + \frac{(C) {(1 - x)}^{3}}{1 - x}$

Step 1.6.2.2.2

Cancel the common factor.

$6 x^{2} - 8 x + 3 = A + \frac{{(1 - x)}^{2} (B (1 - x))}{{(1 - x)}^{2} \cdot 1} + \frac{(C) {(1 - x)}^{3}}{1 - x}$

Step 1.6.2.2.3

Rewrite the expression.

$6 x^{2} - 8 x + 3 = A + \frac{B (1 - x)}{1} + \frac{(C) {(1 - x)}^{3}}{1 - x}$

Step 1.6.2.2.4

Divide $B (1 - x)$ by $1$ .

$6 x^{2} - 8 x + 3 = A + B (1 - x) + \frac{(C) {(1 - x)}^{3}}{1 - x}$

Step 1.6.3

Apply the distributive property.

$6 x^{2} - 8 x + 3 = A + B \cdot 1 + B (- x) + \frac{(C) {(1 - x)}^{3}}{1 - x}$

Step 1.6.4

Multiply $B$ by $1$ .

$6 x^{2} - 8 x + 3 = A + B + B (- x) + \frac{(C) {(1 - x)}^{3}}{1 - x}$

Step 1.6.5

Rewrite using the commutative property of multiplication.

$6 x^{2} - 8 x + 3 = A + B - B x + \frac{(C) {(1 - x)}^{3}}{1 - x}$

Step 1.6.6

Cancel the common factor of ${(1 - x)}^{3}$ and $1 - x$ .

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Step 1.6.6.1

Factor $1 - x$ out of $(C) {(1 - x)}^{3}$ .

$6 x^{2} - 8 x + 3 = A + B - B x + \frac{(1 - x) (C {(1 - x)}^{2})}{1 - x}$

Step 1.6.6.2

Cancel the common factors.

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Step 1.6.6.2.1

Multiply by $1$ .

$6 x^{2} - 8 x + 3 = A + B - B x + \frac{(1 - x) (C {(1 - x)}^{2})}{(1 - x) \cdot 1}$

Step 1.6.6.2.2

Cancel the common factor.

$6 x^{2} - 8 x + 3 = A + B - B x + \frac{(1 - x) (C {(1 - x)}^{2})}{(1 - x) \cdot 1}$

Step 1.6.6.2.3

Rewrite the expression.

$6 x^{2} - 8 x + 3 = A + B - B x + \frac{C {(1 - x)}^{2}}{1}$

Step 1.6.6.2.4

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

$6 x^{2} - 8 x + 3 = A + B - B x + C {(1 - x)}^{2}$

Step 1.6.7

Rewrite ${(1 - x)}^{2}$ as $(1 - x) (1 - x)$ .

$6 x^{2} - 8 x + 3 = A + B - B x + C ((1 - x) (1 - x))$

Step 1.6.8

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

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Step 1.6.8.1

Apply the distributive property.

$6 x^{2} - 8 x + 3 = A + B - B x + C (1 (1 - x) - x (1 - x))$

Step 1.6.8.2

Apply the distributive property.

$6 x^{2} - 8 x + 3 = A + B - B x + C (1 \cdot 1 + 1 (- x) - x (1 - x))$

Step 1.6.8.3

Apply the distributive property.

$6 x^{2} - 8 x + 3 = A + B - B x + C (1 \cdot 1 + 1 (- x) - x \cdot 1 - x (- x))$

Step 1.6.9

Simplify and combine like terms.

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Step 1.6.9.1

Simplify each term.

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Step 1.6.9.1.1

Multiply $1$ by $1$ .

$6 x^{2} - 8 x + 3 = A + B - B x + C (1 + 1 (- x) - x \cdot 1 - x (- x))$

Step 1.6.9.1.2

Multiply $- x$ by $1$ .

$6 x^{2} - 8 x + 3 = A + B - B x + C (1 - x - x \cdot 1 - x (- x))$

Step 1.6.9.1.3

Multiply $- 1$ by $1$ .

$6 x^{2} - 8 x + 3 = A + B - B x + C (1 - x - x - x (- x))$

Step 1.6.9.1.4

Rewrite using the commutative property of multiplication.

$6 x^{2} - 8 x + 3 = A + B - B x + C (1 - x - x - 1 \cdot (- 1 x \cdot x))$

Step 1.6.9.1.5

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

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Step 1.6.9.1.5.1

Move $x$ .

$6 x^{2} - 8 x + 3 = A + B - B x + C (1 - x - x - 1 \cdot (- 1 (x \cdot x)))$

Step 1.6.9.1.5.2

Multiply $x$ by $x$ .

$6 x^{2} - 8 x + 3 = A + B - B x + C (1 - x - x - 1 \cdot (- 1 x^{2}))$

Step 1.6.9.1.6

Multiply $- 1$ by $- 1$ .

$6 x^{2} - 8 x + 3 = A + B - B x + C (1 - x - x + 1 x^{2})$

Step 1.6.9.1.7

Multiply $x^{2}$ by $1$ .

$6 x^{2} - 8 x + 3 = A + B - B x + C (1 - x - x + x^{2})$

Step 1.6.9.2

Subtract $x$ from $- x$ .

$6 x^{2} - 8 x + 3 = A + B - B x + C (1 - 2 x + x^{2})$

Step 1.6.10

Apply the distributive property.

$6 x^{2} - 8 x + 3 = A + B - B x + C \cdot 1 + C (- 2 x) + C x^{2}$

Step 1.6.11

Simplify.

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Step 1.6.11.1

Multiply $C$ by $1$ .

$6 x^{2} - 8 x + 3 = A + B - B x + C + C (- 2 x) + C x^{2}$

Step 1.6.11.2

Rewrite using the commutative property of multiplication.

$6 x^{2} - 8 x + 3 = A + B - B x + C - 2 C x + C x^{2}$

Step 1.7

Simplify the expression.

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Step 1.7.1

Move $B$ .

$6 x^{2} - 8 x + 3 = A + B - x B + C - 2 C x + C x^{2}$

Step 1.7.2

Move $C$ .

$6 x^{2} - 8 x + 3 = A + B - x B - 2 C x + C x^{2} + C$

Step 1.7.3

Move $B$ .

$6 x^{2} - 8 x + 3 = A - x B - 2 C x + C x^{2} + B + C$

Step 1.7.4

Move $A$ .

$6 x^{2} - 8 x + 3 = - x B - 2 C x + C x^{2} + A + B + C$

Step 1.7.5

Move $- 2 C x$ .

$6 x^{2} - 8 x + 3 = - x B + C x^{2} - 2 C x + A + B + C$

Step 1.7.6

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

$6 x^{2} - 8 x + 3 = C x^{2} - x B - 2 C x + A + B + C$

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.

$6 = C$

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.

$- 8 = - 1 B - 2 C$

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.

$3 = A + B + C$

Step 2.4

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

$6 = C$

$- 8 = - 1 B - 2 C$

$3 = A + B + C$

$6 = C$

$- 8 = - 1 B - 2 C$

$3 = A + B + C$

Step 3

Solve the system of equations.

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Step 3.1

Rewrite the equation as $C = 6$ .

$C = 6$

$- 8 = - 1 B - 2 C$

$3 = A + B + C$

Step 3.2

Replace all occurrences of $C$ with $6$ in each equation.

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Step 3.2.1

Replace all occurrences of $C$ in $- 8 = - 1 B - 2 C$ with $6$ .

$- 8 = - 1 B - 2 \cdot 6$

$C = 6$

$3 = A + B + C$

Step 3.2.2

Simplify the right side.

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Step 3.2.2.1

Simplify each term.

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Step 3.2.2.1.1

Rewrite $- 1 B$ as $- B$ .

$- 8 = - B - 2 \cdot 6$

$C = 6$

$3 = A + B + C$

Step 3.2.2.1.2

Multiply $- 2$ by $6$ .

$- 8 = - B - 12$

$C = 6$

$3 = A + B + C$

$- 8 = - B - 12$

$C = 6$

$3 = A + B + C$

$- 8 = - B - 12$

$C = 6$

$3 = A + B + C$

Step 3.2.3

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

$3 = A + B + 6$

$- 8 = - B - 12$

$C = 6$

Step 3.2.4

Simplify the left side.

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Step 3.2.4.1

Remove parentheses.

$3 = A + B + 6$

$- 8 = - B - 12$

$C = 6$

$3 = A + B + 6$

$- 8 = - B - 12$

$C = 6$

$3 = A + B + 6$

$- 8 = - B - 12$

$C = 6$

Step 3.3

Solve for $B$ in $- 8 = - B - 12$ .

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Step 3.3.1

Rewrite the equation as $- B - 12 = - 8$ .

$- B - 12 = - 8$

$3 = A + B + 6$

$C = 6$

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

Add $12$ to both sides of the equation.

$- B = - 8 + 12$

$3 = A + B + 6$

$C = 6$

Step 3.3.2.2

Add $- 8$ and $12$ .

$- B = 4$

$3 = A + B + 6$

$C = 6$

$- B = 4$

$3 = A + B + 6$

$C = 6$

Step 3.3.3

Divide each term in $- B = 4$ by $- 1$ and simplify.

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Step 3.3.3.1

Divide each term in $- B = 4$ by $- 1$ .

$\frac{- B}{- 1} = \frac{4}{- 1}$

$3 = A + B + 6$

$C = 6$

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.

$\frac{B}{1} = \frac{4}{- 1}$

$3 = A + B + 6$

$C = 6$

Step 3.3.3.2.2

Divide $B$ by $1$ .

$B = \frac{4}{- 1}$

$3 = A + B + 6$

$C = 6$

$B = \frac{4}{- 1}$

$3 = A + B + 6$

$C = 6$

Step 3.3.3.3

Simplify the right side.

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Step 3.3.3.3.1

Divide $4$ by $- 1$ .

$B = - 4$

$3 = A + B + 6$

$C = 6$

$B = - 4$

$3 = A + B + 6$

$C = 6$

$B = - 4$

$3 = A + B + 6$

$C = 6$

$B = - 4$

$3 = A + B + 6$

$C = 6$

Step 3.4

Replace all occurrences of $B$ with $- 4$ in each equation.

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Step 3.4.1

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

$3 = A - 4 + 6$

$B = - 4$

$C = 6$

Step 3.4.2

Simplify $3 = A - 4 + 6$ .

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Step 3.4.2.1

Simplify the left side.

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Step 3.4.2.1.1

Remove parentheses.

$3 = A - 4 + 6$

$B = - 4$

$C = 6$

$3 = A - 4 + 6$

$B = - 4$

$C = 6$

Step 3.4.2.2

Simplify the right side.

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Step 3.4.2.2.1

Add $- 4$ and $6$ .

$3 = A + 2$

$B = - 4$

$C = 6$

$3 = A + 2$

$B = - 4$

$C = 6$

$3 = A + 2$

$B = - 4$

$C = 6$

$3 = A + 2$

$B = - 4$

$C = 6$

Step 3.5

Solve for $A$ in $3 = A + 2$ .

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Step 3.5.1

Rewrite the equation as $A + 2 = 3$ .

$A + 2 = 3$

$B = - 4$

$C = 6$

Step 3.5.2

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

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Step 3.5.2.1

Subtract $2$ from both sides of the equation.

$A = 3 - 2$

$B = - 4$

$C = 6$

Step 3.5.2.2

Subtract $2$ from $3$ .

$A = 1$

$B = - 4$

$C = 6$

$A = 1$

$B = - 4$

$C = 6$

$A = 1$

$B = - 4$

$C = 6$

Step 3.6

Solve the system of equations.

$A = 1 B = - 4 C = 6$

Step 3.7

List all of the solutions.

$A = 1, B = - 4, C = 6$

Step 4

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

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