Precalculus Examples

$- x^{3} - 5 x^{2} > - 8 x - 12$

Step 1

Add $8 x$ to both sides of the inequality.

$- x^{3} - 5 x^{2} + 8 x > - 12$

Step 2

Convert the inequality to an equation.

$- x^{3} - 5 x^{2} + 8 x = - 12$

Step 3

Add $12$ to both sides of the equation.

$- x^{3} - 5 x^{2} + 8 x + 12 = 0$

Step 4

Factor the left side of the equation.

Tap for more steps...

Step 4.1

Factor $- x^{3} - 5 x^{2} + 8 x + 12$ using the rational roots test.

Tap for more steps...

Step 4.1.1

If a polynomial function has integer coefficients, then every rational zero will have the form $\frac{p}{q}$ where $p$ is a factor of the constant and $q$ is a factor of the leading coefficient.

$p = \pm 1, \pm 12, \pm 2, \pm 6, \pm 3, \pm 4$

$q = \pm 1$

Step 4.1.2

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

$\pm 1, \pm 12, \pm 2, \pm 6, \pm 3, \pm 4$

Step 4.1.3

Substitute $- 1$ and simplify the expression. In this case, the expression is equal to $0$ so $- 1$ is a root of the polynomial.

Tap for more steps...

Step 4.1.3.1

Substitute $- 1$ into the polynomial.

$- {(- 1)}^{3} - 5 {(- 1)}^{2} + 8 \cdot - 1 + 12$

Step 4.1.3.2

Raise $- 1$ to the power of $3$ .

$- - 1 - 5 {(- 1)}^{2} + 8 \cdot - 1 + 12$

Step 4.1.3.3

Multiply $- 1$ by $- 1$ .

$1 - 5 {(- 1)}^{2} + 8 \cdot - 1 + 12$

Step 4.1.3.4

Raise $- 1$ to the power of $2$ .

$1 - 5 \cdot 1 + 8 \cdot - 1 + 12$

Step 4.1.3.5

Multiply $- 5$ by $1$ .

$1 - 5 + 8 \cdot - 1 + 12$

Step 4.1.3.6

Subtract $5$ from $1$ .

$- 4 + 8 \cdot - 1 + 12$

Step 4.1.3.7

Multiply $8$ by $- 1$ .

$- 4 - 8 + 12$

Step 4.1.3.8

Subtract $8$ from $- 4$ .

$- 12 + 12$

Step 4.1.3.9

Add $- 12$ and $12$ .

$0$

Step 4.1.4

Since $- 1$ is a known root, divide the polynomial by $x + 1$ to find the quotient polynomial. This polynomial can then be used to find the remaining roots.

$\frac{- x^{3} - 5 x^{2} + 8 x + 12}{x + 1}$

Step 4.1.5

Divide $- x^{3} - 5 x^{2} + 8 x + 12$ by $x + 1$ .

Tap for more steps...

Step 4.1.5.1

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

$x$

$1$

$x^{3}$

$5 x^{2}$

$8 x$

$12$

Step 4.1.5.2

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

				-	$x^{2}$
	$x$	+	$1$	-	$x^{3}$	-	$5 x^{2}$	+	$8 x$	+	$12$

Step 4.1.5.3

Multiply the new quotient term by the divisor.

			-	$x^{2}$
$x$	+	$1$	-	$x^{3}$	-	$5 x^{2}$	+	$8 x$	+	$12$
			-	$x^{3}$	-	$x^{2}$

Step 4.1.5.4

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

			-	$x^{2}$
$x$	+	$1$	-	$x^{3}$	-	$5 x^{2}$	+	$8 x$	+	$12$
			+	$x^{3}$	+	$x^{2}$

Step 4.1.5.5

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

			-	$x^{2}$
$x$	+	$1$	-	$x^{3}$	-	$5 x^{2}$	+	$8 x$	+	$12$
			+	$x^{3}$	+	$x^{2}$
					-	$4 x^{2}$

Step 4.1.5.6

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

			-	$x^{2}$
$x$	+	$1$	-	$x^{3}$	-	$5 x^{2}$	+	$8 x$	+	$12$
			+	$x^{3}$	+	$x^{2}$
					-	$4 x^{2}$	+	$8 x$

Step 4.1.5.7

Divide the highest order term in the dividend $- 4 x^{2}$ by the highest order term in divisor $x$ .

			-	$x^{2}$	-	$4 x$
$x$	+	$1$	-	$x^{3}$	-	$5 x^{2}$	+	$8 x$	+	$12$
			+	$x^{3}$	+	$x^{2}$
					-	$4 x^{2}$	+	$8 x$

Step 4.1.5.8

Multiply the new quotient term by the divisor.

			-	$x^{2}$	-	$4 x$
$x$	+	$1$	-	$x^{3}$	-	$5 x^{2}$	+	$8 x$	+	$12$
			+	$x^{3}$	+	$x^{2}$
					-	$4 x^{2}$	+	$8 x$
					-	$4 x^{2}$	-	$4 x$

Step 4.1.5.9

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

			-	$x^{2}$	-	$4 x$
$x$	+	$1$	-	$x^{3}$	-	$5 x^{2}$	+	$8 x$	+	$12$
			+	$x^{3}$	+	$x^{2}$
					-	$4 x^{2}$	+	$8 x$
					+	$4 x^{2}$	+	$4 x$

Step 4.1.5.10

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

			-	$x^{2}$	-	$4 x$
$x$	+	$1$	-	$x^{3}$	-	$5 x^{2}$	+	$8 x$	+	$12$
			+	$x^{3}$	+	$x^{2}$
					-	$4 x^{2}$	+	$8 x$
					+	$4 x^{2}$	+	$4 x$
							+	$12 x$

Step 4.1.5.11

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

			-	$x^{2}$	-	$4 x$
$x$	+	$1$	-	$x^{3}$	-	$5 x^{2}$	+	$8 x$	+	$12$
			+	$x^{3}$	+	$x^{2}$
					-	$4 x^{2}$	+	$8 x$
					+	$4 x^{2}$	+	$4 x$
							+	$12 x$	+	$12$

Step 4.1.5.12

Divide the highest order term in the dividend $12 x$ by the highest order term in divisor $x$ .

			-	$x^{2}$	-	$4 x$	+	$12$
$x$	+	$1$	-	$x^{3}$	-	$5 x^{2}$	+	$8 x$	+	$12$
			+	$x^{3}$	+	$x^{2}$
					-	$4 x^{2}$	+	$8 x$
					+	$4 x^{2}$	+	$4 x$
							+	$12 x$	+	$12$

Step 4.1.5.13

Multiply the new quotient term by the divisor.

			-	$x^{2}$	-	$4 x$	+	$12$
$x$	+	$1$	-	$x^{3}$	-	$5 x^{2}$	+	$8 x$	+	$12$
			+	$x^{3}$	+	$x^{2}$
					-	$4 x^{2}$	+	$8 x$
					+	$4 x^{2}$	+	$4 x$
							+	$12 x$	+	$12$
							+	$12 x$	+	$12$

Step 4.1.5.14

The expression needs to be subtracted from the dividend, so change all the signs in $12 x + 12$

			-	$x^{2}$	-	$4 x$	+	$12$
$x$	+	$1$	-	$x^{3}$	-	$5 x^{2}$	+	$8 x$	+	$12$
			+	$x^{3}$	+	$x^{2}$
					-	$4 x^{2}$	+	$8 x$
					+	$4 x^{2}$	+	$4 x$
							+	$12 x$	+	$12$
							-	$12 x$	-	$12$

Step 4.1.5.15

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

			-	$x^{2}$	-	$4 x$	+	$12$
$x$	+	$1$	-	$x^{3}$	-	$5 x^{2}$	+	$8 x$	+	$12$
			+	$x^{3}$	+	$x^{2}$
					-	$4 x^{2}$	+	$8 x$
					+	$4 x^{2}$	+	$4 x$
							+	$12 x$	+	$12$
							-	$12 x$	-	$12$
										$0$

Step 4.1.5.16

Since the remander is $0$ , the final answer is the quotient.

$- x^{2} - 4 x + 12$

Step 4.1.6

Write $- x^{3} - 5 x^{2} + 8 x + 12$ as a set of factors.

$(x + 1) (- x^{2} - 4 x + 12) = 0$

Step 4.2

Factor by grouping.

Tap for more steps...

Step 4.2.1

Factor by grouping.

Tap for more steps...

Step 4.2.1.1

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = - 1 \cdot 12 = - 12$ and whose sum is $b = - 4$ .

Tap for more steps...

Step 4.2.1.1.1

Factor $- 4$ out of $- 4 x$ .

$(x + 1) (- x^{2} - 4 x + 12) = 0$

Step 4.2.1.1.2

Rewrite $- 4$ as $2$ plus $- 6$

$(x + 1) (- x^{2} + (2 - 6) x + 12) = 0$

Step 4.2.1.1.3

Apply the distributive property.

$(x + 1) (- x^{2} + 2 x - 6 x + 12) = 0$

Step 4.2.1.2

Factor out the greatest common factor from each group.

Tap for more steps...

Step 4.2.1.2.1

Group the first two terms and the last two terms.

$(x + 1) ((- x^{2} + 2 x) - 6 x + 12) = 0$

Step 4.2.1.2.2

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

$(x + 1) (x (- x + 2) + 6 (- x + 2)) = 0$

Step 4.2.1.3

Factor the polynomial by factoring out the greatest common factor, $- x + 2$ .

$(x + 1) ((- x + 2) (x + 6)) = 0$

Step 4.2.2

Remove unnecessary parentheses.

$(x + 1) (- x + 2) (x + 6) = 0$

Step 5

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$x + 1 = 0$

$- x + 2 = 0$

$x + 6 = 0$

Step 6

Set $x + 1$ equal to $0$ and solve for $x$ .

Tap for more steps...

Step 6.1

Set $x + 1$ equal to $0$ .

$x + 1 = 0$

Step 6.2

Subtract $1$ from both sides of the equation.

$x = - 1$

Step 7

Set $- x + 2$ equal to $0$ and solve for $x$ .

Tap for more steps...

Step 7.1

Set $- x + 2$ equal to $0$ .

$- x + 2 = 0$

Step 7.2

Solve $- x + 2 = 0$ for $x$ .

Tap for more steps...

Step 7.2.1

Subtract $2$ from both sides of the equation.

$- x = - 2$

Step 7.2.2

Divide each term in $- x = - 2$ by $- 1$ and simplify.

Tap for more steps...

Step 7.2.2.1

Divide each term in $- x = - 2$ by $- 1$ .

$\frac{- x}{- 1} = \frac{- 2}{- 1}$

Step 7.2.2.2

Simplify the left side.

Tap for more steps...

Step 7.2.2.2.1

Dividing two negative values results in a positive value.

$\frac{x}{1} = \frac{- 2}{- 1}$

Step 7.2.2.2.2

Divide $x$ by $1$ .

$x = \frac{- 2}{- 1}$

Step 7.2.2.3

Simplify the right side.

Tap for more steps...

Step 7.2.2.3.1

Divide $- 2$ by $- 1$ .

$x = 2$

Step 8

Set $x + 6$ equal to $0$ and solve for $x$ .

Tap for more steps...

Step 8.1

Set $x + 6$ equal to $0$ .

$x + 6 = 0$

Step 8.2

Subtract $6$ from both sides of the equation.

$x = - 6$

Step 9

The final solution is all the values that make $(x + 1) (- x + 2) (x + 6) = 0$ true.

$x = - 1, 2, - 6$

Step 10

Use each root to create test intervals.

$x < - 6$

$- 6 < x < - 1$

$- 1 < x < 2$

$x > 2$

Step 11

Choose a test value from each interval and plug this value into the original inequality to determine which intervals satisfy the inequality.

Tap for more steps...

Step 11.1

Test a value on the interval $x < - 6$ to see if it makes the inequality true.

Tap for more steps...

Step 11.1.1

Choose a value on the interval $x < - 6$ and see if this value makes the original inequality true.

$x = - 8$

Step 11.1.2

Replace $x$ with $- 8$ in the original inequality.

$- {(- 8)}^{3} - 5 {(- 8)}^{2} > - 8 \cdot - 8 - 12$

Step 11.1.3

The left side $192$ is greater than the right side $52$ , which means that the given statement is always true.

True

Step 11.2

Test a value on the interval $- 6 < x < - 1$ to see if it makes the inequality true.

Tap for more steps...

Step 11.2.1

Choose a value on the interval $- 6 < x < - 1$ and see if this value makes the original inequality true.

$x = - 4$

Step 11.2.2

Replace $x$ with $- 4$ in the original inequality.

$- {(- 4)}^{3} - 5 {(- 4)}^{2} > - 8 \cdot - 4 - 12$

Step 11.2.3

The left side $- 16$ is not greater than the right side $20$ , which means that the given statement is false.

False

Step 11.3

Test a value on the interval $- 1 < x < 2$ to see if it makes the inequality true.

Tap for more steps...

Step 11.3.1

Choose a value on the interval $- 1 < x < 2$ and see if this value makes the original inequality true.

$x = 0$

Step 11.3.2

Replace $x$ with $0$ in the original inequality.

$- {(0)}^{3} - 5 {(0)}^{2} > - 8 \cdot 0 - 12$

Step 11.3.3

The left side $0$ is greater than the right side $- 12$ , which means that the given statement is always true.

True

Step 11.4

Test a value on the interval $x > 2$ to see if it makes the inequality true.

Tap for more steps...

Step 11.4.1

Choose a value on the interval $x > 2$ and see if this value makes the original inequality true.

$x = 4$

Step 11.4.2

Replace $x$ with $4$ in the original inequality.

$- {(4)}^{3} - 5 {(4)}^{2} > - 8 \cdot 4 - 12$

Step 11.4.3

The left side $- 144$ is not greater than the right side $- 44$ , which means that the given statement is false.

False

Step 11.5

Compare the intervals to determine which ones satisfy the original inequality.

$x < - 6$ True

$- 6 < x < - 1$ False

$- 1 < x < 2$ True

$x > 2$ False

$x < - 6$ True

$- 6 < x < - 1$ False

$- 1 < x < 2$ True

$x > 2$ False

Step 12

The solution consists of all of the true intervals.

$x < - 6$ or $- 1 < x < 2$

Step 13

Convert the inequality to interval notation.

$(- \infty, - 6) \cup (- 1, 2)$

Step 14