Precalculus Examples

$x^{3} + x^{2} - 26 x + 24 \geq 0$

Step 1

Convert the inequality to an equation.

$x^{3} + x^{2} - 26 x + 24 = 0$

Step 2

Factor the left side of the equation.

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Step 2.1

Factor $x^{3} + x^{2} - 26 x + 24$ using the rational roots test.

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Step 2.1.1

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.

$p = \pm 1, \pm 24, \pm 2, \pm 12, \pm 3, \pm 8, \pm 4, \pm 6$

$q = \pm 1$

Step 2.1.2

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

$\pm 1, \pm 24, \pm 2, \pm 12, \pm 3, \pm 8, \pm 4, \pm 6$

Step 2.1.3

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

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Step 2.1.3.1

Substitute $1$ into the polynomial.

$1^{3} + 1^{2} - 26 \cdot 1 + 24$

Step 2.1.3.2

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

$1 + 1^{2} - 26 \cdot 1 + 24$

Step 2.1.3.3

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

$1 + 1 - 26 \cdot 1 + 24$

Step 2.1.3.4

Add $1$ and $1$ .

$2 - 26 \cdot 1 + 24$

Step 2.1.3.5

Multiply $- 26$ by $1$ .

$2 - 26 + 24$

Step 2.1.3.6

Subtract $26$ from $2$ .

$- 24 + 24$

Step 2.1.3.7

Add $- 24$ and $24$ .

$0$

Step 2.1.4

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.

$\frac{x^{3} + x^{2} - 26 x + 24}{x - 1}$

Step 2.1.5

Divide $x^{3} + x^{2} - 26 x + 24$ by $x - 1$ .

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Step 2.1.5.1

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

$x$

$1$

$x^{3}$

$x^{2}$

$26 x$

$24$

Step 2.1.5.2

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

					$x^{2}$
	$x$	-	$1$		$x^{3}$	+	$x^{2}$	-	$26 x$	+	$24$

Step 2.1.5.3

Multiply the new quotient term by the divisor.

				$x^{2}$
$x$	-	$1$		$x^{3}$	+	$x^{2}$	-	$26 x$	+	$24$
			+	$x^{3}$	-	$x^{2}$

Step 2.1.5.4

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

				$x^{2}$
$x$	-	$1$		$x^{3}$	+	$x^{2}$	-	$26 x$	+	$24$
			-	$x^{3}$	+	$x^{2}$

Step 2.1.5.5

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

				$x^{2}$
$x$	-	$1$		$x^{3}$	+	$x^{2}$	-	$26 x$	+	$24$
			-	$x^{3}$	+	$x^{2}$
					+	$2 x^{2}$

Step 2.1.5.6

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

				$x^{2}$
$x$	-	$1$		$x^{3}$	+	$x^{2}$	-	$26 x$	+	$24$
			-	$x^{3}$	+	$x^{2}$
					+	$2 x^{2}$	-	$26 x$

Step 2.1.5.7

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

				$x^{2}$	+	$2 x$
$x$	-	$1$		$x^{3}$	+	$x^{2}$	-	$26 x$	+	$24$
			-	$x^{3}$	+	$x^{2}$
					+	$2 x^{2}$	-	$26 x$

Step 2.1.5.8

Multiply the new quotient term by the divisor.

				$x^{2}$	+	$2 x$
$x$	-	$1$		$x^{3}$	+	$x^{2}$	-	$26 x$	+	$24$
			-	$x^{3}$	+	$x^{2}$
					+	$2 x^{2}$	-	$26 x$
					+	$2 x^{2}$	-	$2 x$

Step 2.1.5.9

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

				$x^{2}$	+	$2 x$
$x$	-	$1$		$x^{3}$	+	$x^{2}$	-	$26 x$	+	$24$
			-	$x^{3}$	+	$x^{2}$
					+	$2 x^{2}$	-	$26 x$
					-	$2 x^{2}$	+	$2 x$

Step 2.1.5.10

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

				$x^{2}$	+	$2 x$
$x$	-	$1$		$x^{3}$	+	$x^{2}$	-	$26 x$	+	$24$
			-	$x^{3}$	+	$x^{2}$
					+	$2 x^{2}$	-	$26 x$
					-	$2 x^{2}$	+	$2 x$
							-	$24 x$

Step 2.1.5.11

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

				$x^{2}$	+	$2 x$
$x$	-	$1$		$x^{3}$	+	$x^{2}$	-	$26 x$	+	$24$
			-	$x^{3}$	+	$x^{2}$
					+	$2 x^{2}$	-	$26 x$
					-	$2 x^{2}$	+	$2 x$
							-	$24 x$	+	$24$

Step 2.1.5.12

Divide the highest order term in the dividend $- 24 x$ by the highest order term in divisor $x$ .

				$x^{2}$	+	$2 x$	-	$24$
$x$	-	$1$		$x^{3}$	+	$x^{2}$	-	$26 x$	+	$24$
			-	$x^{3}$	+	$x^{2}$
					+	$2 x^{2}$	-	$26 x$
					-	$2 x^{2}$	+	$2 x$
							-	$24 x$	+	$24$

Step 2.1.5.13

Multiply the new quotient term by the divisor.

				$x^{2}$	+	$2 x$	-	$24$
$x$	-	$1$		$x^{3}$	+	$x^{2}$	-	$26 x$	+	$24$
			-	$x^{3}$	+	$x^{2}$
					+	$2 x^{2}$	-	$26 x$
					-	$2 x^{2}$	+	$2 x$
							-	$24 x$	+	$24$
							-	$24 x$	+	$24$

Step 2.1.5.14

The expression needs to be subtracted from the dividend, so change all the signs in $- 24 x + 24$

				$x^{2}$	+	$2 x$	-	$24$
$x$	-	$1$		$x^{3}$	+	$x^{2}$	-	$26 x$	+	$24$
			-	$x^{3}$	+	$x^{2}$
					+	$2 x^{2}$	-	$26 x$
					-	$2 x^{2}$	+	$2 x$
							-	$24 x$	+	$24$
							+	$24 x$	-	$24$

Step 2.1.5.15

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

				$x^{2}$	+	$2 x$	-	$24$
$x$	-	$1$		$x^{3}$	+	$x^{2}$	-	$26 x$	+	$24$
			-	$x^{3}$	+	$x^{2}$
					+	$2 x^{2}$	-	$26 x$
					-	$2 x^{2}$	+	$2 x$
							-	$24 x$	+	$24$
							+	$24 x$	-	$24$
										$0$

Step 2.1.5.16

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

$x^{2} + 2 x - 24$

Step 2.1.6

Write $x^{3} + x^{2} - 26 x + 24$ as a set of factors.

$(x - 1) (x^{2} + 2 x - 24) = 0$

Step 2.2

Factor $x^{2} + 2 x - 24$ using the AC method.

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Step 2.2.1

Factor $x^{2} + 2 x - 24$ using the AC method.

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Step 2.2.1.1

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 $- 24$ and whose sum is $2$ .

$- 4, 6$

Step 2.2.1.2

Write the factored form using these integers.

$(x - 1) ((x - 4) (x + 6)) = 0$

Step 2.2.2

Remove unnecessary parentheses.

$(x - 1) (x - 4) (x + 6) = 0$

Step 3

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

$x - 1 = 0$

$x - 4 = 0$

$x + 6 = 0$

Step 4

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

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Step 4.1

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

$x - 1 = 0$

Step 4.2

Add $1$ to both sides of the equation.

$x = 1$

Step 5

Set $x - 4$ equal to $0$ and solve for $x$ .

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Step 5.1

Set $x - 4$ equal to $0$ .

$x - 4 = 0$

Step 5.2

Add $4$ to both sides of the equation.

$x = 4$

Step 6

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

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Step 6.1

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

$x + 6 = 0$

Step 6.2

Subtract $6$ from both sides of the equation.

$x = - 6$

Step 7

The final solution is all the values that make $(x - 1) (x - 4) (x + 6) = 0$ true.

$x = 1, 4, - 6$

Step 8

Use each root to create test intervals.

$x < - 6$

$- 6 < x < 1$

$1 < x < 4$

$x > 4$

Step 9

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

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Step 9.1

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

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Step 9.1.1

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

$x = - 8$

Step 9.1.2

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

${(- 8)}^{3} + {(- 8)}^{2} - 26 \cdot - 8 + 24 \geq 0$

Step 9.1.3

The left side $- 216$ is less than the right side $0$ , which means that the given statement is false.

False

Step 9.2

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

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Step 9.2.1

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

$x = 0$

Step 9.2.2

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

${(0)}^{3} + {(0)}^{2} - 26 \cdot 0 + 24 \geq 0$

Step 9.2.3

The left side $24$ is greater than the right side $0$ , which means that the given statement is always true.

True

Step 9.3

Test a value on the interval $1 < x < 4$ to see if it makes the inequality true.

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Step 9.3.1

Choose a value on the interval $1 < x < 4$ and see if this value makes the original inequality true.

$x = 2$

Step 9.3.2

Replace $x$ with $2$ in the original inequality.

${(2)}^{3} + {(2)}^{2} - 26 \cdot 2 + 24 \geq 0$

Step 9.3.3

The left side $- 16$ is less than the right side $0$ , which means that the given statement is false.

False

Step 9.4

Test a value on the interval $x > 4$ to see if it makes the inequality true.

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Step 9.4.1

Choose a value on the interval $x > 4$ and see if this value makes the original inequality true.

$x = 6$

Step 9.4.2

Replace $x$ with $6$ in the original inequality.

${(6)}^{3} + {(6)}^{2} - 26 \cdot 6 + 24 \geq 0$

Step 9.4.3

The left side $120$ is greater than the right side $0$ , which means that the given statement is always true.

True

Step 9.5

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

$x < - 6$ False

$- 6 < x < 1$ True

$1 < x < 4$ False

$x > 4$ True

$x < - 6$ False

$- 6 < x < 1$ True

$1 < x < 4$ False

$x > 4$ True

Step 10

The solution consists of all of the true intervals.

$- 6 \leq x \leq 1$ or $x \geq 4$

Step 11

Convert the inequality to interval notation.

$[- 6, 1] \cup [4, \infty)$

Step 12