Algebra Examples

f (x) = \frac{x^{3} + 4 x^{2} + x - 6}{x^{2} + 5 x + 6}

$f (x) = \frac{x^{3} + 4 x^{2} + x - 6}{x^{2} + 5 x + 6}$

Step 1

Factor

x^{3} + 4 x^{2} + x - 6

$x^{3} + 4 x^{2} + x - 6$ .

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Step 1.1

Factor

x^{3} + 4 x^{2} + x - 6

$x^{3} + 4 x^{2} + x - 6$ using the rational roots test.

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Step 1.1.1

If a polynomial function has integer coefficients, then every rational zero will have the form

\frac{p}{q}

$\frac{p}{q}$ where

p

$p$ is a factor of the constant and

q

$q$ is a factor of the leading coefficient.

p = \pm 1, \pm 6, \pm 2, \pm 3

$p = \pm 1, \pm 6, \pm 2, \pm 3$

q = \pm 1

$q = \pm 1$

Step 1.1.2

Find every combination of

\pm \frac{p}{q}

$\pm \frac{p}{q}$ . These are the possible roots of the polynomial function.

\pm 1, \pm 6, \pm 2, \pm 3

$\pm 1, \pm 6, \pm 2, \pm 3$

Step 1.1.3

Substitute

1

$1$ and simplify the expression. In this case, the expression is equal to

0

$0$ so

1

$1$ is a root of the polynomial.

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Step 1.1.3.1

Substitute

1

$1$ into the polynomial.

1^{3} + 4 \cdot 1^{2} + 1 - 6

$1^{3} + 4 \cdot 1^{2} + 1 - 6$

Step 1.1.3.2

Raise

1

$1$ to the power of

3

$3$ .

1 + 4 \cdot 1^{2} + 1 - 6

$1 + 4 \cdot 1^{2} + 1 - 6$

Step 1.1.3.3

Raise

1

$1$ to the power of

2

$2$ .

1 + 4 \cdot 1 + 1 - 6

$1 + 4 \cdot 1 + 1 - 6$

Step 1.1.3.4

Multiply

4

$4$ by

1

$1$ .

1 + 4 + 1 - 6

$1 + 4 + 1 - 6$

Step 1.1.3.5

Add

1

$1$ and

4

$4$ .

5 + 1 - 6

$5 + 1 - 6$

Step 1.1.3.6

Add

5

$5$ and

1

$1$ .

6 - 6

$6 - 6$

Step 1.1.3.7

Subtract

6

$6$ from

6

$6$ .

0

$0$

0

$0$

Step 1.1.4

Since

1

$1$ is a known root, divide the polynomial by

x - 1

$x - 1$ to find the quotient polynomial. This polynomial can then be used to find the remaining roots.

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

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

Step 1.1.5

Divide

x^{3} + 4 x^{2} + x - 6

$x^{3} + 4 x^{2} + x - 6$ by

x - 1

$x - 1$ .

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Step 1.1.5.1

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

0

$0$ .

x

$x$

1

$1$

x^{3}

$x^{3}$

4 x^{2}

$4 x^{2}$

x

$x$

6

$6$

Step 1.1.5.2

Divide the highest order term in the dividend

x^{3}

$x^{3}$ by the highest order term in divisor

x

$x$ .

					$x^{2}$ $x^{2}$
	$x$ $x$	-	$1$ $1$		$x^{3}$ $x^{3}$	+	$4 x^{2}$ $4 x^{2}$	+	$x$ $x$	-	$6$ $6$

Step 1.1.5.3

Multiply the new quotient term by the divisor.

				$x^{2}$ $x^{2}$
$x$ $x$	-	$1$ $1$		$x^{3}$ $x^{3}$	+	$4 x^{2}$ $4 x^{2}$	+	$x$ $x$	-	$6$ $6$
			+	$x^{3}$ $x^{3}$	-	$x^{2}$ $x^{2}$

Step 1.1.5.4

The expression needs to be subtracted from the dividend, so change all the signs in

x^{3} - x^{2}

$x^{3} - x^{2}$

				$x^{2}$ $x^{2}$
$x$ $x$	-	$1$ $1$		$x^{3}$ $x^{3}$	+	$4 x^{2}$ $4 x^{2}$	+	$x$ $x$	-	$6$ $6$
			-	$x^{3}$ $x^{3}$	+	$x^{2}$ $x^{2}$

Step 1.1.5.5

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

				$x^{2}$ $x^{2}$
$x$ $x$	-	$1$ $1$		$x^{3}$ $x^{3}$	+	$4 x^{2}$ $4 x^{2}$	+	$x$ $x$	-	$6$ $6$
			-	$x^{3}$ $x^{3}$	+	$x^{2}$ $x^{2}$
					+	$5 x^{2}$ $5 x^{2}$

Step 1.1.5.6

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

				$x^{2}$ $x^{2}$
$x$ $x$	-	$1$ $1$		$x^{3}$ $x^{3}$	+	$4 x^{2}$ $4 x^{2}$	+	$x$ $x$	-	$6$ $6$
			-	$x^{3}$ $x^{3}$	+	$x^{2}$ $x^{2}$
					+	$5 x^{2}$ $5 x^{2}$	+	$x$ $x$

Step 1.1.5.7

Divide the highest order term in the dividend

5 x^{2}

$5 x^{2}$ by the highest order term in divisor

x

$x$ .

				$x^{2}$ $x^{2}$	+	$5 x$ $5 x$
$x$ $x$	-	$1$ $1$		$x^{3}$ $x^{3}$	+	$4 x^{2}$ $4 x^{2}$	+	$x$ $x$	-	$6$ $6$
			-	$x^{3}$ $x^{3}$	+	$x^{2}$ $x^{2}$
					+	$5 x^{2}$ $5 x^{2}$	+	$x$ $x$

Step 1.1.5.8

Multiply the new quotient term by the divisor.

				$x^{2}$ $x^{2}$	+	$5 x$ $5 x$
$x$ $x$	-	$1$ $1$		$x^{3}$ $x^{3}$	+	$4 x^{2}$ $4 x^{2}$	+	$x$ $x$	-	$6$ $6$
			-	$x^{3}$ $x^{3}$	+	$x^{2}$ $x^{2}$
					+	$5 x^{2}$ $5 x^{2}$	+	$x$ $x$
					+	$5 x^{2}$ $5 x^{2}$	-	$5 x$ $5 x$

Step 1.1.5.9

The expression needs to be subtracted from the dividend, so change all the signs in

5 x^{2} - 5 x

$5 x^{2} - 5 x$

				$x^{2}$ $x^{2}$	+	$5 x$ $5 x$
$x$ $x$	-	$1$ $1$		$x^{3}$ $x^{3}$	+	$4 x^{2}$ $4 x^{2}$	+	$x$ $x$	-	$6$ $6$
			-	$x^{3}$ $x^{3}$	+	$x^{2}$ $x^{2}$
					+	$5 x^{2}$ $5 x^{2}$	+	$x$ $x$
					-	$5 x^{2}$ $5 x^{2}$	+	$5 x$ $5 x$

Step 1.1.5.10

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

				$x^{2}$ $x^{2}$	+	$5 x$ $5 x$
$x$ $x$	-	$1$ $1$		$x^{3}$ $x^{3}$	+	$4 x^{2}$ $4 x^{2}$	+	$x$ $x$	-	$6$ $6$
			-	$x^{3}$ $x^{3}$	+	$x^{2}$ $x^{2}$
					+	$5 x^{2}$ $5 x^{2}$	+	$x$ $x$
					-	$5 x^{2}$ $5 x^{2}$	+	$5 x$ $5 x$
							+	$6 x$ $6 x$

Step 1.1.5.11

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

				$x^{2}$ $x^{2}$	+	$5 x$ $5 x$
$x$ $x$	-	$1$ $1$		$x^{3}$ $x^{3}$	+	$4 x^{2}$ $4 x^{2}$	+	$x$ $x$	-	$6$ $6$
			-	$x^{3}$ $x^{3}$	+	$x^{2}$ $x^{2}$
					+	$5 x^{2}$ $5 x^{2}$	+	$x$ $x$
					-	$5 x^{2}$ $5 x^{2}$	+	$5 x$ $5 x$
							+	$6 x$ $6 x$	-	$6$ $6$

Step 1.1.5.12

Divide the highest order term in the dividend

6 x

$6 x$ by the highest order term in divisor

x

$x$ .

				$x^{2}$ $x^{2}$	+	$5 x$ $5 x$	+	$6$ $6$
$x$ $x$	-	$1$ $1$		$x^{3}$ $x^{3}$	+	$4 x^{2}$ $4 x^{2}$	+	$x$ $x$	-	$6$ $6$
			-	$x^{3}$ $x^{3}$	+	$x^{2}$ $x^{2}$
					+	$5 x^{2}$ $5 x^{2}$	+	$x$ $x$
					-	$5 x^{2}$ $5 x^{2}$	+	$5 x$ $5 x$
							+	$6 x$ $6 x$	-	$6$ $6$

Step 1.1.5.13

Multiply the new quotient term by the divisor.

				$x^{2}$ $x^{2}$	+	$5 x$ $5 x$	+	$6$ $6$
$x$ $x$	-	$1$ $1$		$x^{3}$ $x^{3}$	+	$4 x^{2}$ $4 x^{2}$	+	$x$ $x$	-	$6$ $6$
			-	$x^{3}$ $x^{3}$	+	$x^{2}$ $x^{2}$
					+	$5 x^{2}$ $5 x^{2}$	+	$x$ $x$
					-	$5 x^{2}$ $5 x^{2}$	+	$5 x$ $5 x$
							+	$6 x$ $6 x$	-	$6$ $6$
							+	$6 x$ $6 x$	-	$6$ $6$

Step 1.1.5.14

The expression needs to be subtracted from the dividend, so change all the signs in

6 x - 6

$6 x - 6$

				$x^{2}$ $x^{2}$	+	$5 x$ $5 x$	+	$6$ $6$
$x$ $x$	-	$1$ $1$		$x^{3}$ $x^{3}$	+	$4 x^{2}$ $4 x^{2}$	+	$x$ $x$	-	$6$ $6$
			-	$x^{3}$ $x^{3}$	+	$x^{2}$ $x^{2}$
					+	$5 x^{2}$ $5 x^{2}$	+	$x$ $x$
					-	$5 x^{2}$ $5 x^{2}$	+	$5 x$ $5 x$
							+	$6 x$ $6 x$	-	$6$ $6$
							-	$6 x$ $6 x$	+	$6$ $6$

Step 1.1.5.15

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

				$x^{2}$ $x^{2}$	+	$5 x$ $5 x$	+	$6$ $6$
$x$ $x$	-	$1$ $1$		$x^{3}$ $x^{3}$	+	$4 x^{2}$ $4 x^{2}$	+	$x$ $x$	-	$6$ $6$
			-	$x^{3}$ $x^{3}$	+	$x^{2}$ $x^{2}$
					+	$5 x^{2}$ $5 x^{2}$	+	$x$ $x$
					-	$5 x^{2}$ $5 x^{2}$	+	$5 x$ $5 x$
							+	$6 x$ $6 x$	-	$6$ $6$
							-	$6 x$ $6 x$	+	$6$ $6$
										$0$ $0$

Step 1.1.5.16

Since the remander is

0

$0$ , the final answer is the quotient.

x^{2} + 5 x + 6

$x^{2} + 5 x + 6$

x^{2} + 5 x + 6

$x^{2} + 5 x + 6$

Step 1.1.6

Write

x^{3} + 4 x^{2} + x - 6

$x^{3} + 4 x^{2} + x - 6$ as a set of factors.

f (x) = \frac{(x - 1) (x^{2} + 5 x + 6)}{x^{2} + 5 x + 6}

$f (x) = \frac{(x - 1) (x^{2} + 5 x + 6)}{x^{2} + 5 x + 6}$

Step 1.2

Factor

$x^{2} + 5 x + 6$ using the AC method.

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Step 1.2.1

Factor

$x^{2} + 5 x + 6$ using the AC method.

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Step 1.2.1.1

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

$6$ and whose sum is

$5$ .

$2, 3$

Step 1.2.1.2

Write the factored form using these integers.

$f (x) = \frac{(x - 1) ((x + 2) (x + 3))}{x^{2} + 5 x + 6}$

Step 1.2.2

Remove unnecessary parentheses.

$f (x) = \frac{(x - 1) (x + 2) (x + 3)}{x^{2} + 5 x + 6}$

Step 2

Factor

$x^{2} + 5 x + 6$ using the AC method.

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Step 2.1

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

$6$ and whose sum is

$5$ .

$2, 3$

Step 2.2

Write the factored form using these integers.

$f (x) = \frac{(x - 1) (x + 2) (x + 3)}{(x + 2) (x + 3)}$

Step 3

Cancel the common factor of

$x + 2$ .

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Step 3.1

Cancel the common factor.

$f (x) = \frac{(x - 1) (x + 2) (x + 3)}{(x + 2) (x + 3)}$

Step 3.2

Rewrite the expression.

$f (x) = \frac{(x - 1) (x + 3)}{x + 3}$

Step 4

Cancel the common factor of

$x + 3$ .

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Step 4.1

Cancel the common factor.

$f (x) = \frac{(x - 1) (x + 3)}{x + 3}$

Step 4.2

Divide

$x - 1$ by

$1$ .

$f (x) = x - 1$

Step 5

To find the holes in the graph, look at the denominator factors that were cancelled.

$x + 2, x + 3$

Step 6

To find the coordinates of the holes, set each factor that was cancelled equal to

$0$ , solve, and substitute back in to

$x - 1$ .

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Step 6.1

Set

$x + 2$ equal to

$0$ .

$x + 2 = 0$

Step 6.2

Subtract

$2$ from both sides of the equation.

$x = - 2$

Step 6.3

Substitute

$- 2$ for

$x$ in

$x - 1$ and simplify.

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Step 6.3.1

Substitute

$- 2$ for

$x$ to find the

$y$ coordinate of the hole.

$- 2 - 1$

Step 6.3.2

Subtract

$1$ from

$- 2$ .

$- 3$

Step 6.4

Set

$x + 3$ equal to

$0$ .

$x + 3 = 0$

Step 6.5

Subtract

$3$ from both sides of the equation.

$x = - 3$

Step 6.6

Substitute

$- 3$ for

$x$ in

$x - 1$ and simplify.

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Step 6.6.1

Substitute

$- 3$ for

$x$ to find the

$y$ coordinate of the hole.

$- 3 - 1$

Step 6.6.2

Subtract

$1$ from

$- 3$ .

$- 4$

Step 6.7

The holes in the graph are the points where any of the cancelled factors are equal to

$0$ .

$(- 2, - 3), (- 3, - 4)$

Step 7

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