Precalculus Examples

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

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 $0$ .

$x^{2}$

$3 x$

$2$

$2 x^{3}$

$x^{2}$

$x$

$5$

Step 1.2

Divide the highest order term in the dividend $2 x^{3}$ by the highest order term in divisor $x^{2}$ .

$2 x$

$x^{2}$

$3 x$

$2$

$2 x^{3}$

$x^{2}$

$x$

$5$

Step 1.3

Multiply the new quotient term by the divisor.

$2 x$

$x^{2}$

$3 x$

$2$

$2 x^{3}$

$x^{2}$

$x$

$5$

$2 x^{3}$

$6 x^{2}$

$4 x$

Step 1.4

The expression needs to be subtracted from the dividend, so change all the signs in $2 x^{3} + 6 x^{2} + 4 x$

$2 x$

$x^{2}$

$3 x$

$2$

$2 x^{3}$

$x^{2}$

$x$

$5$

$2 x^{3}$

$6 x^{2}$

$4 x$

Step 1.5

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

$2 x$

$x^{2}$

$3 x$

$2$

$2 x^{3}$

$x^{2}$

$x$

$5$

$2 x^{3}$

$6 x^{2}$

$4 x$

$7 x^{2}$

$3 x$

Step 1.6

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

$2 x$

$x^{2}$

$3 x$

$2$

$2 x^{3}$

$x^{2}$

$x$

$5$

$2 x^{3}$

$6 x^{2}$

$4 x$

$7 x^{2}$

$3 x$

$5$

Step 1.7

Divide the highest order term in the dividend $- 7 x^{2}$ by the highest order term in divisor $x^{2}$ .

						$2 x$	-	$7$
$x^{2}$	+	$3 x$	+	$2$		$2 x^{3}$	-	$x^{2}$	+	$x$	+	$5$
					-	$2 x^{3}$	-	$6 x^{2}$	-	$4 x$
							-	$7 x^{2}$	-	$3 x$	+	$5$

Step 1.8

Multiply the new quotient term by the divisor.

						$2 x$	-	$7$
$x^{2}$	+	$3 x$	+	$2$		$2 x^{3}$	-	$x^{2}$	+	$x$	+	$5$
					-	$2 x^{3}$	-	$6 x^{2}$	-	$4 x$
							-	$7 x^{2}$	-	$3 x$	+	$5$
							-	$7 x^{2}$	-	$21 x$	-	$14$

Step 1.9

The expression needs to be subtracted from the dividend, so change all the signs in $- 7 x^{2} - 21 x - 14$

$2 x$

$7$

$x^{2}$

$3 x$

$2$

$2 x^{3}$

$x^{2}$

$x$

$5$

$2 x^{3}$

$6 x^{2}$

$4 x$

$7 x^{2}$

$3 x$

$5$

$7 x^{2}$

$21 x$

$14$

Step 1.10

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

$2 x$

$7$

$x^{2}$

$3 x$

$2$

$2 x^{3}$

$x^{2}$

$x$

$5$

$2 x^{3}$

$6 x^{2}$

$4 x$

$7 x^{2}$

$3 x$

$5$

$7 x^{2}$

$21 x$

$14$

$18 x$

$19$

Step 1.11

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

$2 x - 7 + \frac{18 x + 19}{x^{2} + 3 x + 2}$

Step 2

Decompose the fraction and multiply through by the common denominator.

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Step 2.1

Factor $x^{2} + 3 x + 2$ using the AC method.

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Step 2.1.1

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $2$ and whose sum is $3$ .

$1, 2$

Step 2.1.2

Write the factored form using these integers.

$\frac{18 x + 19}{(x + 1) (x + 2)}$

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 $A$ .

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

Step 2.3

$\frac{A}{x + 1} + \frac{B}{x + 2}$

Step 2.4

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

$\frac{(18 x + 19) (x + 1) (x + 2)}{(x + 1) (x + 2)} = \frac{(A) (x + 1) (x + 2)}{x + 1} + \frac{(B) (x + 1) (x + 2)}{x + 2}$

Step 2.5

Cancel the common factor of $x + 1$ .

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Step 2.5.1

Cancel the common factor.

$\frac{(18 x + 19) (x + 1) (x + 2)}{(x + 1) (x + 2)} = \frac{(A) (x + 1) (x + 2)}{x + 1} + \frac{(B) (x + 1) (x + 2)}{x + 2}$

Step 2.5.2

Rewrite the expression.

$\frac{(18 x + 19) (x + 2)}{x + 2} = \frac{(A) (x + 1) (x + 2)}{x + 1} + \frac{(B) (x + 1) (x + 2)}{x + 2}$

Step 2.6

Cancel the common factor of $x + 2$ .

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Step 2.6.1

Cancel the common factor.

$\frac{(18 x + 19) (x + 2)}{x + 2} = \frac{(A) (x + 1) (x + 2)}{x + 1} + \frac{(B) (x + 1) (x + 2)}{x + 2}$

Step 2.6.2

Divide $18 x + 19$ by $1$ .

$18 x + 19 = \frac{(A) (x + 1) (x + 2)}{x + 1} + \frac{(B) (x + 1) (x + 2)}{x + 2}$

Step 2.7

Simplify each term.

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Step 2.7.1

Cancel the common factor of $x + 1$ .

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Step 2.7.1.1

Cancel the common factor.

$18 x + 19 = \frac{A (x + 1) (x + 2)}{x + 1} + \frac{(B) (x + 1) (x + 2)}{x + 2}$

Step 2.7.1.2

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

$18 x + 19 = (A) (x + 2) + \frac{(B) (x + 1) (x + 2)}{x + 2}$

Step 2.7.2

Apply the distributive property.

$18 x + 19 = A x + A \cdot 2 + \frac{(B) (x + 1) (x + 2)}{x + 2}$

Step 2.7.3

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

$18 x + 19 = A x + 2 \cdot A + \frac{(B) (x + 1) (x + 2)}{x + 2}$

Step 2.7.4

Cancel the common factor of $x + 2$ .

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Step 2.7.4.1

Cancel the common factor.

$18 x + 19 = A x + 2 A + \frac{(B) (x + 1) (x + 2)}{x + 2}$

Step 2.7.4.2

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

$18 x + 19 = A x + 2 A + (B) (x + 1)$

Step 2.7.5

Apply the distributive property.

$18 x + 19 = A x + 2 A + B x + B \cdot 1$

Step 2.7.6

Multiply $B$ by $1$ .

$18 x + 19 = A x + 2 A + B x + B$

Step 2.8

Move $2 A$ .

$18 x + 19 = A x + B x + 2 A + B$

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 $x$ from each side of the equation. For the equation to be equal, the equivalent coefficients on each side of the equation must be equal.

$18 = A + B$

Step 3.2

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.

$19 = 2 A + B$

Step 3.3

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

$18 = A + B$

$19 = 2 A + B$

$18 = A + B$

$19 = 2 A + B$

Step 4

Solve the system of equations.

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Step 4.1

Solve for $A$ in $18 = A + B$ .

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Step 4.1.1

Rewrite the equation as $A + B = 18$ .

$A + B = 18$

$19 = 2 A + B$

Step 4.1.2

Subtract $B$ from both sides of the equation.

$A = 18 - B$

$19 = 2 A + B$

$A = 18 - B$

$19 = 2 A + B$

Step 4.2

Replace all occurrences of $A$ with $18 - B$ in each equation.

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Step 4.2.1

Replace all occurrences of $A$ in $19 = 2 A + B$ with $18 - B$ .

$19 = 2 (18 - B) + B$

$A = 18 - B$

Step 4.2.2

Simplify the right side.

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Step 4.2.2.1

Simplify $2 (18 - B) + B$ .

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Step 4.2.2.1.1

Simplify each term.

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Step 4.2.2.1.1.1

Apply the distributive property.

$19 = 2 \cdot 18 + 2 (- B) + B$

$A = 18 - B$

Step 4.2.2.1.1.2

Multiply $2$ by $18$ .

$19 = 36 + 2 (- B) + B$

$A = 18 - B$

Step 4.2.2.1.1.3

Multiply $- 1$ by $2$ .

$19 = 36 - 2 B + B$

$A = 18 - B$

$19 = 36 - 2 B + B$

$A = 18 - B$

Step 4.2.2.1.2

Add $- 2 B$ and $B$ .

$19 = 36 - B$

$A = 18 - B$

$19 = 36 - B$

$A = 18 - B$

$19 = 36 - B$

$A = 18 - B$

$19 = 36 - B$

$A = 18 - B$

Step 4.3

Solve for $B$ in $19 = 36 - B$ .

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Step 4.3.1

Rewrite the equation as $36 - B = 19$ .

$36 - B = 19$

$A = 18 - B$

Step 4.3.2

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

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Step 4.3.2.1

Subtract $36$ from both sides of the equation.

$- B = 19 - 36$

$A = 18 - B$

Step 4.3.2.2

Subtract $36$ from $19$ .

$- B = - 17$

$A = 18 - B$

$- B = - 17$

$A = 18 - B$

Step 4.3.3

Divide each term in $- B = - 17$ by $- 1$ and simplify.

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Step 4.3.3.1

Divide each term in $- B = - 17$ by $- 1$ .

$\frac{- B}{- 1} = \frac{- 17}{- 1}$

$A = 18 - B$

Step 4.3.3.2

Simplify the left side.

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Step 4.3.3.2.1

Dividing two negative values results in a positive value.

$\frac{B}{1} = \frac{- 17}{- 1}$

$A = 18 - B$

Step 4.3.3.2.2

Divide $B$ by $1$ .

$B = \frac{- 17}{- 1}$

$A = 18 - B$

$B = \frac{- 17}{- 1}$

$A = 18 - B$

Step 4.3.3.3

Simplify the right side.

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Step 4.3.3.3.1

Divide $- 17$ by $- 1$ .

$B = 17$

$A = 18 - B$

$B = 17$

$A = 18 - B$

$B = 17$

$A = 18 - B$

$B = 17$

$A = 18 - B$

Step 4.4

Replace all occurrences of $B$ with $17$ in each equation.

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Step 4.4.1

Replace all occurrences of $B$ in $A = 18 - B$ with $17$ .

$A = 18 - (17)$

$B = 17$

Step 4.4.2

Simplify the right side.

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Step 4.4.2.1

Simplify $18 - (17)$ .

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Step 4.4.2.1.1

Multiply $- 1$ by $17$ .

$A = 18 - 17$

$B = 17$

Step 4.4.2.1.2

Subtract $17$ from $18$ .

$A = 1$

$B = 17$

$A = 1$

$B = 17$

$A = 1$

$B = 17$

$A = 1$

$B = 17$

Step 4.5

List all of the solutions.

$A = 1, B = 17$

Step 5

Replace each of the partial fraction coefficients in $2 x - 7 + \frac{A}{x + 1} + \frac{B}{x + 2}$ with the values found for $A$ and $B$ .

$2 x - 7 + \frac{1}{x + 1} + \frac{17}{x + 2}$