Precalculus Examples

$\frac{x^{5} - 4 x^{4} + 2 x^{3} - 7 x^{2} + 2 x + 12}{x^{3} - 4 x^{2} + 2 x - 8}$

Step 1

Divide using long polynomial division.

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Step 1.1

Set up the polynomials to be divided. If there is not a term for every exponent, insert one with a value of $0$ .

$x^{3}$

$4 x^{2}$

$2 x$

$8$

$x^{5}$

$4 x^{4}$

$2 x^{3}$

$7 x^{2}$

$2 x$

$12$

Step 1.2

Divide the highest order term in the dividend $x^{5}$ by the highest order term in divisor $x^{3}$ .

$x^{2}$

$x^{3}$

$4 x^{2}$

$2 x$

$8$

$x^{5}$

$4 x^{4}$

$2 x^{3}$

$7 x^{2}$

$2 x$

$12$

Step 1.3

Multiply the new quotient term by the divisor.

$x^{2}$

$x^{3}$

$4 x^{2}$

$2 x$

$8$

$x^{5}$

$4 x^{4}$

$2 x^{3}$

$7 x^{2}$

$2 x$

$12$

$x^{5}$

$4 x^{4}$

$2 x^{3}$

$8 x^{2}$

Step 1.4

The expression needs to be subtracted from the dividend, so change all the signs in $x^{5} - 4 x^{4} + 2 x^{3} - 8 x^{2}$

$x^{2}$

$x^{3}$

$4 x^{2}$

$2 x$

$8$

$x^{5}$

$4 x^{4}$

$2 x^{3}$

$7 x^{2}$

$2 x$

$12$

$x^{5}$

$4 x^{4}$

$2 x^{3}$

$8 x^{2}$

Step 1.5

After changing the signs, add the last dividend from the multiplied polynomial to find the new dividend.

$x^{2}$

$x^{3}$

$4 x^{2}$

$2 x$

$8$

$x^{5}$

$4 x^{4}$

$2 x^{3}$

$7 x^{2}$

$2 x$

$12$

$x^{5}$

$4 x^{4}$

$2 x^{3}$

$8 x^{2}$

$x^{2}$

Step 1.6

Pull the next terms from the original dividend down into the current dividend.

$x^{2}$

$x^{3}$

$4 x^{2}$

$2 x$

$8$

$x^{5}$

$4 x^{4}$

$2 x^{3}$

$7 x^{2}$

$2 x$

$12$

$x^{5}$

$4 x^{4}$

$2 x^{3}$

$8 x^{2}$

$x^{2}$

$2 x$

$12$

Step 1.7

The final answer is the quotient plus the remainder over the divisor.

$x^{2} + \frac{x^{2} + 2 x + 12}{x^{3} - 4 x^{2} + 2 x - 8}$

Step 2

Decompose the fraction and multiply through by the common denominator.

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Step 2.1

Factor the fraction.

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Step 2.1.1

Factor out the greatest common factor from each group.

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Step 2.1.1.1

Group the first two terms and the last two terms.

$\frac{x^{2} + 2 x + 12}{(x^{3} - 4 x^{2}) + 2 x - 8}$

Step 2.1.1.2

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

$\frac{x^{2} + 2 x + 12}{x^{2} (x - 4) + 2 (x - 4)}$

Step 2.1.2

Factor the polynomial by factoring out the greatest common factor, $x - 4$ .

$\frac{x^{2} + 2 x + 12}{(x - 4) (x^{2} + 2)}$

Step 2.2

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 $A$ .

$\frac{A}{x - 4}$

Step 2.3

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.

$\frac{A}{x - 4} + \frac{B x + C}{x^{2} + 2}$

Step 2.4

Multiply each fraction in the equation by the denominator of the original expression. In this case, the denominator is $(x - 4) (x^{2} + 2)$ .

$\frac{(x^{2} + 2 x + 12) (x - 4) (x^{2} + 2)}{(x - 4) (x^{2} + 2)} = \frac{(A) (x - 4) (x^{2} + 2)}{x - 4} + \frac{(B x + C) (x - 4) (x^{2} + 2)}{x^{2} + 2}$

Step 2.5

Cancel the common factor of $x - 4$ .

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Step 2.5.1

Cancel the common factor.

$\frac{(x^{2} + 2 x + 12) (x - 4) (x^{2} + 2)}{(x - 4) (x^{2} + 2)} = \frac{(A) (x - 4) (x^{2} + 2)}{x - 4} + \frac{(B x + C) (x - 4) (x^{2} + 2)}{x^{2} + 2}$

Step 2.5.2

Rewrite the expression.

$\frac{(x^{2} + 2 x + 12) (x^{2} + 2)}{x^{2} + 2} = \frac{(A) (x - 4) (x^{2} + 2)}{x - 4} + \frac{(B x + C) (x - 4) (x^{2} + 2)}{x^{2} + 2}$

Step 2.6

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

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Step 2.6.1

Cancel the common factor.

$\frac{(x^{2} + 2 x + 12) (x^{2} + 2)}{x^{2} + 2} = \frac{(A) (x - 4) (x^{2} + 2)}{x - 4} + \frac{(B x + C) (x - 4) (x^{2} + 2)}{x^{2} + 2}$

Step 2.6.2

Divide $x^{2} + 2 x + 12$ by $1$ .

$x^{2} + 2 x + 12 = \frac{(A) (x - 4) (x^{2} + 2)}{x - 4} + \frac{(B x + C) (x - 4) (x^{2} + 2)}{x^{2} + 2}$

Step 2.7

Simplify each term.

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Step 2.7.1

Cancel the common factor of $x - 4$ .

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Step 2.7.1.1

Cancel the common factor.

$x^{2} + 2 x + 12 = \frac{A (x - 4) (x^{2} + 2)}{x - 4} + \frac{(B x + C) (x - 4) (x^{2} + 2)}{x^{2} + 2}$

Step 2.7.1.2

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

$x^{2} + 2 x + 12 = (A) (x^{2} + 2) + \frac{(B x + C) (x - 4) (x^{2} + 2)}{x^{2} + 2}$

Step 2.7.2

Apply the distributive property.

$x^{2} + 2 x + 12 = A x^{2} + A \cdot 2 + \frac{(B x + C) (x - 4) (x^{2} + 2)}{x^{2} + 2}$

Step 2.7.3

Move $2$ to the left of $A$ .

$x^{2} + 2 x + 12 = A x^{2} + 2 \cdot A + \frac{(B x + C) (x - 4) (x^{2} + 2)}{x^{2} + 2}$

Step 2.7.4

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

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Step 2.7.4.1

Cancel the common factor.

$x^{2} + 2 x + 12 = A x^{2} + 2 A + \frac{(B x + C) (x - 4) (x^{2} + 2)}{x^{2} + 2}$

Step 2.7.4.2

Divide $(B x + C) (x - 4)$ by $1$ .

$x^{2} + 2 x + 12 = A x^{2} + 2 A + (B x + C) (x - 4)$

Step 2.7.5

Expand $(B x + C) (x - 4)$ using the FOIL Method.

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Step 2.7.5.1

Apply the distributive property.

$x^{2} + 2 x + 12 = A x^{2} + 2 A + B x (x - 4) + C (x - 4)$

Step 2.7.5.2

Apply the distributive property.

$x^{2} + 2 x + 12 = A x^{2} + 2 A + B x \cdot x + B x \cdot - 4 + C (x - 4)$

Step 2.7.5.3

Apply the distributive property.

$x^{2} + 2 x + 12 = A x^{2} + 2 A + B x \cdot x + B x \cdot - 4 + C x + C \cdot - 4$

Step 2.7.6

Simplify each term.

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Step 2.7.6.1

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

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Step 2.7.6.1.1

Move $x$ .

$x^{2} + 2 x + 12 = A x^{2} + 2 A + B (x \cdot x) + B x \cdot - 4 + C x + C \cdot - 4$

Step 2.7.6.1.2

Multiply $x$ by $x$ .

$x^{2} + 2 x + 12 = A x^{2} + 2 A + B x^{2} + B x \cdot - 4 + C x + C \cdot - 4$

Step 2.7.6.2

Move $- 4$ to the left of $B x$ .

$x^{2} + 2 x + 12 = A x^{2} + 2 A + B x^{2} - 4 \cdot (B x) + C x + C \cdot - 4$

Step 2.7.6.3

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

$x^{2} + 2 x + 12 = A x^{2} + 2 A + B x^{2} - 4 B x + C x - 4 C$

Step 2.8

Simplify the expression.

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Step 2.8.1

Move $B$ .

$x^{2} + 2 x + 12 = A x^{2} + 2 A + B x^{2} - 4 x B + C x - 4 C$

Step 2.8.2

Move $2 A$ .

$x^{2} + 2 x + 12 = A x^{2} + B x^{2} - 4 x B + C x + 2 A - 4 C$

Step 3

Create equations for the partial fraction variables and use them to set up a system of equations.

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Step 3.1

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.

$1 = A + B$

Step 3.2

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.

$2 = - 4 B + C$

Step 3.3

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.

$12 = 2 A - 4 C$

Step 3.4

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

$1 = A + B$

$2 = - 4 B + C$

$12 = 2 A - 4 C$

$1 = A + B$

$2 = - 4 B + C$

$12 = 2 A - 4 C$

Step 4

Solve the system of equations.

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Step 4.1

Solve for $A$ in $1 = A + B$ .

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Step 4.1.1

Rewrite the equation as $A + B = 1$ .

$A + B = 1$

$2 = - 4 B + C$

$12 = 2 A - 4 C$

Step 4.1.2

Subtract $B$ from both sides of the equation.

$A = 1 - B$

$2 = - 4 B + C$

$12 = 2 A - 4 C$

$A = 1 - B$

$2 = - 4 B + C$

$12 = 2 A - 4 C$

Step 4.2

Replace all occurrences of $A$ with $1 - B$ in each equation.

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Step 4.2.1

Replace all occurrences of $A$ in $12 = 2 A - 4 C$ with $1 - B$ .

$12 = 2 (1 - B) - 4 C$

$A = 1 - B$

$2 = - 4 B + C$

Step 4.2.2

Simplify the right side.

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Step 4.2.2.1

Simplify each term.

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Step 4.2.2.1.1

Apply the distributive property.

$12 = 2 \cdot 1 + 2 (- B) - 4 C$

$A = 1 - B$

$2 = - 4 B + C$

Step 4.2.2.1.2

Multiply $2$ by $1$ .

$12 = 2 + 2 (- B) - 4 C$

$A = 1 - B$

$2 = - 4 B + C$

Step 4.2.2.1.3

Multiply $- 1$ by $2$ .

$12 = 2 - 2 B - 4 C$

$A = 1 - B$

$2 = - 4 B + C$

$12 = 2 - 2 B - 4 C$

$A = 1 - B$

$2 = - 4 B + C$

$12 = 2 - 2 B - 4 C$

$A = 1 - B$

$2 = - 4 B + C$

$12 = 2 - 2 B - 4 C$

$A = 1 - B$

$2 = - 4 B + C$

Step 4.3

Reorder $1$ and $- B$ .

$A = - B + 1$

$12 = 2 - 2 B - 4 C$

$2 = - 4 B + C$

Step 4.4

Solve for $C$ in $2 = - 4 B + C$ .

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Step 4.4.1

Rewrite the equation as $- 4 B + C = 2$ .

$- 4 B + C = 2$

$A = - B + 1$

$12 = 2 - 2 B - 4 C$

Step 4.4.2

Add $4 B$ to both sides of the equation.

$C = 2 + 4 B$

$A = - B + 1$

$12 = 2 - 2 B - 4 C$

$C = 2 + 4 B$

$A = - B + 1$

$12 = 2 - 2 B - 4 C$

Step 4.5

Replace all occurrences of $C$ with $2 + 4 B$ in each equation.

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Step 4.5.1

Replace all occurrences of $C$ in $12 = 2 - 2 B - 4 C$ with $2 + 4 B$ .

$12 = 2 - 2 B - 4 (2 + 4 B)$

$C = 2 + 4 B$

$A = - B + 1$

Step 4.5.2

Simplify the right side.

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Step 4.5.2.1

Simplify $2 - 2 B - 4 (2 + 4 B)$ .

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Step 4.5.2.1.1

Simplify each term.

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Step 4.5.2.1.1.1

Apply the distributive property.

$12 = 2 - 2 B - 4 \cdot 2 - 4 (4 B)$

$C = 2 + 4 B$

$A = - B + 1$

Step 4.5.2.1.1.2

Multiply $- 4$ by $2$ .

$12 = 2 - 2 B - 8 - 4 (4 B)$

$C = 2 + 4 B$

$A = - B + 1$

Step 4.5.2.1.1.3

Multiply $4$ by $- 4$ .

$12 = 2 - 2 B - 8 - 16 B$

$C = 2 + 4 B$

$A = - B + 1$

$12 = 2 - 2 B - 8 - 16 B$

$C = 2 + 4 B$

$A = - B + 1$

Step 4.5.2.1.2

Simplify by adding terms.

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Step 4.5.2.1.2.1

Subtract $8$ from $2$ .

$12 = - 2 B - 6 - 16 B$

$C = 2 + 4 B$

$A = - B + 1$

Step 4.5.2.1.2.2

Subtract $16 B$ from $- 2 B$ .

$12 = - 18 B - 6$

$C = 2 + 4 B$

$A = - B + 1$

$12 = - 18 B - 6$

$C = 2 + 4 B$

$A = - B + 1$

$12 = - 18 B - 6$

$C = 2 + 4 B$

$A = - B + 1$

$12 = - 18 B - 6$

$C = 2 + 4 B$

$A = - B + 1$

$12 = - 18 B - 6$

$C = 2 + 4 B$

$A = - B + 1$

Step 4.6

Solve for $B$ in $12 = - 18 B - 6$ .

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Step 4.6.1

Rewrite the equation as $- 18 B - 6 = 12$ .

$- 18 B - 6 = 12$

$C = 2 + 4 B$

$A = - B + 1$

Step 4.6.2

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

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Step 4.6.2.1

Add $6$ to both sides of the equation.

$- 18 B = 12 + 6$

$C = 2 + 4 B$

$A = - B + 1$

Step 4.6.2.2

Add $12$ and $6$ .

$- 18 B = 18$

$C = 2 + 4 B$

$A = - B + 1$

$- 18 B = 18$

$C = 2 + 4 B$

$A = - B + 1$

Step 4.6.3

Divide each term in $- 18 B = 18$ by $- 18$ and simplify.

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Step 4.6.3.1

Divide each term in $- 18 B = 18$ by $- 18$ .

$\frac{- 18 B}{- 18} = \frac{18}{- 18}$

$C = 2 + 4 B$

$A = - B + 1$

Step 4.6.3.2

Simplify the left side.

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Step 4.6.3.2.1

Cancel the common factor of $- 18$ .

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Step 4.6.3.2.1.1

Cancel the common factor.

$\frac{- 18 B}{- 18} = \frac{18}{- 18}$

$C = 2 + 4 B$

$A = - B + 1$

Step 4.6.3.2.1.2

Divide $B$ by $1$ .

$B = \frac{18}{- 18}$

$C = 2 + 4 B$

$A = - B + 1$

$B = \frac{18}{- 18}$

$C = 2 + 4 B$

$A = - B + 1$

$B = \frac{18}{- 18}$

$C = 2 + 4 B$

$A = - B + 1$

Step 4.6.3.3

Simplify the right side.

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Step 4.6.3.3.1

Divide $18$ by $- 18$ .

$B = - 1$

$C = 2 + 4 B$

$A = - B + 1$

$B = - 1$

$C = 2 + 4 B$

$A = - B + 1$

$B = - 1$

$C = 2 + 4 B$

$A = - B + 1$

$B = - 1$

$C = 2 + 4 B$

$A = - B + 1$

Step 4.7

Replace all occurrences of $B$ with $- 1$ in each equation.

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Step 4.7.1

Replace all occurrences of $B$ in $C = 2 + 4 B$ with $- 1$ .

$C = 2 + 4 (- 1)$

$B = - 1$

$A = - B + 1$

Step 4.7.2

Simplify the right side.

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Step 4.7.2.1

Simplify $2 + 4 (- 1)$ .

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Step 4.7.2.1.1

Multiply $4$ by $- 1$ .

$C = 2 - 4$

$B = - 1$

$A = - B + 1$

Step 4.7.2.1.2

Subtract $4$ from $2$ .

$C = - 2$

$B = - 1$

$A = - B + 1$

$C = - 2$

$B = - 1$

$A = - B + 1$

$C = - 2$

$B = - 1$

$A = - B + 1$

Step 4.7.3

Replace all occurrences of $B$ in $A = - B + 1$ with $- 1$ .

$A = - (- 1) + 1$

$C = - 2$

$B = - 1$

Step 4.7.4

Simplify the right side.

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Step 4.7.4.1

Simplify $- (- 1) + 1$ .

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Step 4.7.4.1.1

Multiply $- 1$ by $- 1$ .

$A = 1 + 1$

$C = - 2$

$B = - 1$

Step 4.7.4.1.2

Add $1$ and $1$ .

$A = 2$

$C = - 2$

$B = - 1$

$A = 2$

$C = - 2$

$B = - 1$

$A = 2$

$C = - 2$

$B = - 1$

$A = 2$

$C = - 2$

$B = - 1$

Step 4.8

List all of the solutions.

$A = 2, C = - 2, B = - 1$

Step 5

Replace each of the partial fraction coefficients in $x^{2} + \frac{A}{x - 4} + \frac{B x + C}{x^{2} + 2}$ with the values found for $A$ , $B$ , and $C$ .

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

Step 6

Rewrite $- 1 x$ as $- x$ .

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