Precalculus Examples

$\frac{27 - 4 x}{x^{3} - 6 x^{2} + 9 x}$

Step 1

Decompose the fraction and multiply through by the common denominator.

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Step 1.1

Factor the fraction.

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Step 1.1.1

Factor $x$ out of $x^{3} - 6 x^{2} + 9 x$ .

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Step 1.1.1.1

Factor $x$ out of $x^{3}$ .

$\frac{27 - 4 x}{x \cdot x^{2} - 6 x^{2} + 9 x}$

Step 1.1.1.2

Factor $x$ out of $- 6 x^{2}$ .

$\frac{27 - 4 x}{x \cdot x^{2} + x (- 6 x) + 9 x}$

Step 1.1.1.3

Factor $x$ out of $9 x$ .

$\frac{27 - 4 x}{x \cdot x^{2} + x (- 6 x) + x \cdot 9}$

Step 1.1.1.4

Factor $x$ out of $x \cdot x^{2} + x (- 6 x)$ .

$\frac{27 - 4 x}{x (x^{2} - 6 x) + x \cdot 9}$

Step 1.1.1.5

Factor $x$ out of $x (x^{2} - 6 x) + x \cdot 9$ .

$\frac{27 - 4 x}{x (x^{2} - 6 x + 9)}$

Step 1.1.2

Factor using the perfect square rule.

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Step 1.1.2.1

Rewrite $9$ as $3^{2}$ .

$\frac{27 - 4 x}{x (x^{2} - 6 x + 3^{2})}$

Step 1.1.2.2

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$6 x = 2 \cdot x \cdot 3$

Step 1.1.2.3

Rewrite the polynomial.

$\frac{27 - 4 x}{x (x^{2} - 2 \cdot x \cdot 3 + 3^{2})}$

Step 1.1.2.4

Factor using the perfect square trinomial rule $a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ , where $a = x$ and $b = 3$ .

$\frac{27 - 4 x}{x {(x - 3)}^{2}}$

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 $B$ .

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

Step 1.3

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

Step 1.4

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

$\frac{(27 - 4 x) (x {(x - 3)}^{2})}{x {(x - 3)}^{2}} = \frac{A (x {(x - 3)}^{2})}{x} + \frac{(B) (x {(x - 3)}^{2})}{{(x - 3)}^{2}} + \frac{(C) (x {(x - 3)}^{2})}{x - 3}$

Step 1.5

Cancel the common factor of $x$ .

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Step 1.5.1

Cancel the common factor.

$\frac{(27 - 4 x) (x {(x - 3)}^{2})}{x {(x - 3)}^{2}} = \frac{A (x {(x - 3)}^{2})}{x} + \frac{(B) (x {(x - 3)}^{2})}{{(x - 3)}^{2}} + \frac{(C) (x {(x - 3)}^{2})}{x - 3}$

Step 1.5.2

Rewrite the expression.

$\frac{(27 - 4 x) ({(x - 3)}^{2})}{{(x - 3)}^{2}} = \frac{A (x {(x - 3)}^{2})}{x} + \frac{(B) (x {(x - 3)}^{2})}{{(x - 3)}^{2}} + \frac{(C) (x {(x - 3)}^{2})}{x - 3}$

Step 1.6

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

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Step 1.6.1

Cancel the common factor.

$\frac{(27 - 4 x) {(x - 3)}^{2}}{{(x - 3)}^{2}} = \frac{A (x {(x - 3)}^{2})}{x} + \frac{(B) (x {(x - 3)}^{2})}{{(x - 3)}^{2}} + \frac{(C) (x {(x - 3)}^{2})}{x - 3}$

Step 1.6.2

Divide $27 - 4 x$ by $1$ .

$27 - 4 x = \frac{A (x {(x - 3)}^{2})}{x} + \frac{(B) (x {(x - 3)}^{2})}{{(x - 3)}^{2}} + \frac{(C) (x {(x - 3)}^{2})}{x - 3}$

Step 1.7

Reorder $27$ and $- 4 x$ .

$- 4 x + 27 = \frac{A (x {(x - 3)}^{2})}{x} + \frac{(B) (x {(x - 3)}^{2})}{{(x - 3)}^{2}} + \frac{(C) (x {(x - 3)}^{2})}{x - 3}$

Step 1.8

Simplify each term.

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Step 1.8.1

Cancel the common factor of $x$ .

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Step 1.8.1.1

Cancel the common factor.

$- 4 x + 27 = \frac{A (x {(x - 3)}^{2})}{x} + \frac{(B) (x {(x - 3)}^{2})}{{(x - 3)}^{2}} + \frac{(C) (x {(x - 3)}^{2})}{x - 3}$

Step 1.8.1.2

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

$- 4 x + 27 = A ({(x - 3)}^{2}) + \frac{(B) (x {(x - 3)}^{2})}{{(x - 3)}^{2}} + \frac{(C) (x {(x - 3)}^{2})}{x - 3}$

Step 1.8.2

Rewrite ${(x - 3)}^{2}$ as $(x - 3) (x - 3)$ .

$- 4 x + 27 = A ((x - 3) (x - 3)) + \frac{(B) (x {(x - 3)}^{2})}{{(x - 3)}^{2}} + \frac{(C) (x {(x - 3)}^{2})}{x - 3}$

Step 1.8.3

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

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Step 1.8.3.1

Apply the distributive property.

$- 4 x + 27 = A (x (x - 3) - 3 (x - 3)) + \frac{(B) (x {(x - 3)}^{2})}{{(x - 3)}^{2}} + \frac{(C) (x {(x - 3)}^{2})}{x - 3}$

Step 1.8.3.2

Apply the distributive property.

$- 4 x + 27 = A (x \cdot x + x \cdot - 3 - 3 (x - 3)) + \frac{(B) (x {(x - 3)}^{2})}{{(x - 3)}^{2}} + \frac{(C) (x {(x - 3)}^{2})}{x - 3}$

Step 1.8.3.3

Apply the distributive property.

$- 4 x + 27 = A (x \cdot x + x \cdot - 3 - 3 x - 3 \cdot - 3) + \frac{(B) (x {(x - 3)}^{2})}{{(x - 3)}^{2}} + \frac{(C) (x {(x - 3)}^{2})}{x - 3}$

Step 1.8.4

Simplify and combine like terms.

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Step 1.8.4.1

Simplify each term.

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Step 1.8.4.1.1

Multiply $x$ by $x$ .

$- 4 x + 27 = A (x^{2} + x \cdot - 3 - 3 x - 3 \cdot - 3) + \frac{(B) (x {(x - 3)}^{2})}{{(x - 3)}^{2}} + \frac{(C) (x {(x - 3)}^{2})}{x - 3}$

Step 1.8.4.1.2

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

$- 4 x + 27 = A (x^{2} - 3 \cdot x - 3 x - 3 \cdot - 3) + \frac{(B) (x {(x - 3)}^{2})}{{(x - 3)}^{2}} + \frac{(C) (x {(x - 3)}^{2})}{x - 3}$

Step 1.8.4.1.3

Multiply $- 3$ by $- 3$ .

$- 4 x + 27 = A (x^{2} - 3 x - 3 x + 9) + \frac{(B) (x {(x - 3)}^{2})}{{(x - 3)}^{2}} + \frac{(C) (x {(x - 3)}^{2})}{x - 3}$

Step 1.8.4.2

Subtract $3 x$ from $- 3 x$ .

$- 4 x + 27 = A (x^{2} - 6 x + 9) + \frac{(B) (x {(x - 3)}^{2})}{{(x - 3)}^{2}} + \frac{(C) (x {(x - 3)}^{2})}{x - 3}$

Step 1.8.5

Apply the distributive property.

$- 4 x + 27 = A x^{2} + A (- 6 x) + A \cdot 9 + \frac{(B) (x {(x - 3)}^{2})}{{(x - 3)}^{2}} + \frac{(C) (x {(x - 3)}^{2})}{x - 3}$

Step 1.8.6

Simplify.

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Step 1.8.6.1

Rewrite using the commutative property of multiplication.

$- 4 x + 27 = A x^{2} - 6 A x + A \cdot 9 + \frac{(B) (x {(x - 3)}^{2})}{{(x - 3)}^{2}} + \frac{(C) (x {(x - 3)}^{2})}{x - 3}$

Step 1.8.6.2

Move $9$ to the left of $A$ .

$- 4 x + 27 = A x^{2} - 6 A x + 9 \cdot A + \frac{(B) (x {(x - 3)}^{2})}{{(x - 3)}^{2}} + \frac{(C) (x {(x - 3)}^{2})}{x - 3}$

Step 1.8.7

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

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Step 1.8.7.1

Cancel the common factor.

$- 4 x + 27 = A x^{2} - 6 A x + 9 A + \frac{B (x {(x - 3)}^{2})}{{(x - 3)}^{2}} + \frac{(C) (x {(x - 3)}^{2})}{x - 3}$

Step 1.8.7.2

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

$- 4 x + 27 = A x^{2} - 6 A x + 9 A + (B) (x) + \frac{(C) (x {(x - 3)}^{2})}{x - 3}$

Step 1.8.8

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

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Step 1.8.8.1

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

$- 4 x + 27 = A x^{2} - 6 A x + 9 A + B x + \frac{(x - 3) (C (x (x - 3)))}{x - 3}$

Step 1.8.8.2

Cancel the common factors.

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Step 1.8.8.2.1

Multiply by $1$ .

$- 4 x + 27 = A x^{2} - 6 A x + 9 A + B x + \frac{(x - 3) (C (x (x - 3)))}{(x - 3) \cdot 1}$

Step 1.8.8.2.2

Cancel the common factor.

$- 4 x + 27 = A x^{2} - 6 A x + 9 A + B x + \frac{(x - 3) (C (x (x - 3)))}{(x - 3) \cdot 1}$

Step 1.8.8.2.3

Rewrite the expression.

$- 4 x + 27 = A x^{2} - 6 A x + 9 A + B x + \frac{C (x (x - 3))}{1}$

Step 1.8.8.2.4

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

$- 4 x + 27 = A x^{2} - 6 A x + 9 A + B x + C (x (x - 3))$

Step 1.8.9

Apply the distributive property.

$- 4 x + 27 = A x^{2} - 6 A x + 9 A + B x + C (x \cdot x + x \cdot - 3)$

Step 1.8.10

Multiply $x$ by $x$ .

$- 4 x + 27 = A x^{2} - 6 A x + 9 A + B x + C (x^{2} + x \cdot - 3)$

Step 1.8.11

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

$- 4 x + 27 = A x^{2} - 6 A x + 9 A + B x + C (x^{2} - 3 \cdot x)$

Step 1.8.12

Apply the distributive property.

$- 4 x + 27 = A x^{2} - 6 A x + 9 A + B x + C x^{2} + C (- 3 x)$

Step 1.8.13

Rewrite using the commutative property of multiplication.

$- 4 x + 27 = A x^{2} - 6 A x + 9 A + B x + C x^{2} - 3 C x$

Step 1.9

Simplify the expression.

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Step 1.9.1

Move $A$ .

$- 4 x + 27 = A x^{2} - 6 x A + 9 A + B x + C x^{2} - 3 C x$

Step 1.9.2

Reorder $B$ and $x$ .

$- 4 x + 27 = A x^{2} - 6 x A + 9 A + B x + C x^{2} - 3 C x$

Step 1.9.3

Move $9 A$ .

$- 4 x + 27 = A x^{2} - 6 x A + B x + C x^{2} - 3 C x + 9 A$

Step 1.9.4

Move $B x$ .

$- 4 x + 27 = A x^{2} - 6 x A + C x^{2} + B x - 3 C x + 9 A$

Step 1.9.5

Move $- 6 x A$ .

$- 4 x + 27 = A x^{2} + C x^{2} - 6 x A + B x - 3 C x + 9 A$

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.

$0 = A + C$

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.

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

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.

$27 = 9 A$

Step 2.4

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

$0 = A + C$

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

$27 = 9 A$

$0 = A + C$

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

$27 = 9 A$

Step 3

Solve the system of equations.

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Step 3.1

Solve for $A$ in $27 = 9 A$ .

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Step 3.1.1

Rewrite the equation as $9 A = 27$ .

$9 A = 27$

$0 = A + C$

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

Step 3.1.2

Divide each term in $9 A = 27$ by $9$ and simplify.

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Step 3.1.2.1

Divide each term in $9 A = 27$ by $9$ .

$\frac{9 A}{9} = \frac{27}{9}$

$0 = A + C$

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

Step 3.1.2.2

Simplify the left side.

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Step 3.1.2.2.1

Cancel the common factor of $9$ .

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Step 3.1.2.2.1.1

Cancel the common factor.

$\frac{9 A}{9} = \frac{27}{9}$

$0 = A + C$

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

Step 3.1.2.2.1.2

Divide $A$ by $1$ .

$A = \frac{27}{9}$

$0 = A + C$

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

$A = \frac{27}{9}$

$0 = A + C$

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

$A = \frac{27}{9}$

$0 = A + C$

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

Step 3.1.2.3

Simplify the right side.

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Step 3.1.2.3.1

Divide $27$ by $9$ .

$A = 3$

$0 = A + C$

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

$A = 3$

$0 = A + C$

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

$A = 3$

$0 = A + C$

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

$A = 3$

$0 = A + C$

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

Step 3.2

Replace all occurrences of $A$ with $3$ in each equation.

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Step 3.2.1

Replace all occurrences of $A$ in $0 = A + C$ with $3$ .

$0 = (3) + C$

$A = 3$

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

Step 3.2.2

Simplify the right side.

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Step 3.2.2.1

Remove parentheses.

$0 = 3 + C$

$A = 3$

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

$0 = 3 + C$

$A = 3$

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

Step 3.2.3

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

$- 4 = - 6 \cdot 3 + B - 3 C$

$0 = 3 + C$

$A = 3$

Step 3.2.4

Simplify the right side.

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Step 3.2.4.1

Multiply $- 6$ by $3$ .

$- 4 = - 18 + B - 3 C$

$0 = 3 + C$

$A = 3$

$- 4 = - 18 + B - 3 C$

$0 = 3 + C$

$A = 3$

$- 4 = - 18 + B - 3 C$

$0 = 3 + C$

$A = 3$

Step 3.3

Solve for $C$ in $0 = 3 + C$ .

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Step 3.3.1

Rewrite the equation as $3 + C = 0$ .

$3 + C = 0$

$- 4 = - 18 + B - 3 C$

$A = 3$

Step 3.3.2

Subtract $3$ from both sides of the equation.

$C = - 3$

$- 4 = - 18 + B - 3 C$

$A = 3$

$C = - 3$

$- 4 = - 18 + B - 3 C$

$A = 3$

Step 3.4

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

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Step 3.4.1

Replace all occurrences of $C$ in $- 4 = - 18 + B - 3 C$ with $- 3$ .

$- 4 = - 18 + B - 3 \cdot - 3$

$C = - 3$

$A = 3$

Step 3.4.2

Simplify the right side.

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Step 3.4.2.1

Simplify $- 18 + B - 3 \cdot - 3$ .

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Step 3.4.2.1.1

Multiply $- 3$ by $- 3$ .

$- 4 = - 18 + B + 9$

$C = - 3$

$A = 3$

Step 3.4.2.1.2

Add $- 18$ and $9$ .

$- 4 = B - 9$

$C = - 3$

$A = 3$

$- 4 = B - 9$

$C = - 3$

$A = 3$

$- 4 = B - 9$

$C = - 3$

$A = 3$

$- 4 = B - 9$

$C = - 3$

$A = 3$

Step 3.5

Solve for $B$ in $- 4 = B - 9$ .

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Step 3.5.1

Rewrite the equation as $B - 9 = - 4$ .

$B - 9 = - 4$

$C = - 3$

$A = 3$

Step 3.5.2

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

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Step 3.5.2.1

Add $9$ to both sides of the equation.

$B = - 4 + 9$

$C = - 3$

$A = 3$

Step 3.5.2.2

Add $- 4$ and $9$ .

$B = 5$

$C = - 3$

$A = 3$

$B = 5$

$C = - 3$

$A = 3$

$B = 5$

$C = - 3$

$A = 3$

Step 3.6

Solve the system of equations.

$B = 5 C = - 3 A = 3$

Step 3.7

List all of the solutions.

$B = 5, C = - 3, A = 3$

Step 4

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

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