Precalculus Examples

$6 x^{3} + 25 x^{2} - 24 x + 5 > 0$

Step 1

Convert the inequality to an equation.

$6 x^{3} + 25 x^{2} - 24 x + 5 = 0$

Step 2

Factor the left side of the equation.

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Step 2.1

Factor $6 x^{3} + 25 x^{2} - 24 x + 5$ 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.

$p = \pm 1, \pm 5$

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

Step 2.1.2

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

$\pm 1, \pm 0.1\overline{6}, \pm 0.5, \pm 0.\overline{3}, \pm 5, \pm 0.8\overline{3}, \pm 2.5, \pm 1.\overline{6}$

Step 2.1.3

Substitute $0.5$ and simplify the expression. In this case, the expression is equal to $0$ so $0.5$ is a root of the polynomial.

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Step 2.1.3.1

Substitute $0.5$ into the polynomial.

$6 \cdot {0.5}^{3} + 25 \cdot {0.5}^{2} - 24 \cdot 0.5 + 5$

Step 2.1.3.2

Raise $0.5$ to the power of $3$ .

$6 \cdot 0.125 + 25 \cdot {0.5}^{2} - 24 \cdot 0.5 + 5$

Step 2.1.3.3

Multiply $6$ by $0.125$ .

$0.75 + 25 \cdot {0.5}^{2} - 24 \cdot 0.5 + 5$

Step 2.1.3.4

Raise $0.5$ to the power of $2$ .

$0.75 + 25 \cdot 0.25 - 24 \cdot 0.5 + 5$

Step 2.1.3.5

Multiply $25$ by $0.25$ .

$0.75 + 6.25 - 24 \cdot 0.5 + 5$

Step 2.1.3.6

Add $0.75$ and $6.25$ .

$7 - 24 \cdot 0.5 + 5$

Step 2.1.3.7

Multiply $- 24$ by $0.5$ .

$7 - 12 + 5$

Step 2.1.3.8

Subtract $12$ from $7$ .

$- 5 + 5$

Step 2.1.3.9

Add $- 5$ and $5$ .

$0$

Step 2.1.4

Since $0.5$ is a known root, divide the polynomial by $2 x - 1$ to find the quotient polynomial. This polynomial can then be used to find the remaining roots.

$\frac{6 x^{3} + 25 x^{2} - 24 x + 5}{2 x - 1}$

Step 2.1.5

Divide $6 x^{3} + 25 x^{2} - 24 x + 5$ by $2 x - 1$ .

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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 $0$ .

$2 x$

$1$

$6 x^{3}$

$25 x^{2}$

$24 x$

$5$

Step 2.1.5.2

Divide the highest order term in the dividend $6 x^{3}$ by the highest order term in divisor $2 x$ .

					$3 x^{2}$
	$2 x$	-	$1$		$6 x^{3}$	+	$25 x^{2}$	-	$24 x$	+	$5$

Step 2.1.5.3

Multiply the new quotient term by the divisor.

				$3 x^{2}$
$2 x$	-	$1$		$6 x^{3}$	+	$25 x^{2}$	-	$24 x$	+	$5$
			+	$6 x^{3}$	-	$3 x^{2}$

Step 2.1.5.4

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

				$3 x^{2}$
$2 x$	-	$1$		$6 x^{3}$	+	$25 x^{2}$	-	$24 x$	+	$5$
			-	$6 x^{3}$	+	$3 x^{2}$

Step 2.1.5.5

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

				$3 x^{2}$
$2 x$	-	$1$		$6 x^{3}$	+	$25 x^{2}$	-	$24 x$	+	$5$
			-	$6 x^{3}$	+	$3 x^{2}$
					+	$28 x^{2}$

Step 2.1.5.6

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

				$3 x^{2}$
$2 x$	-	$1$		$6 x^{3}$	+	$25 x^{2}$	-	$24 x$	+	$5$
			-	$6 x^{3}$	+	$3 x^{2}$
					+	$28 x^{2}$	-	$24 x$

Step 2.1.5.7

Divide the highest order term in the dividend $28 x^{2}$ by the highest order term in divisor $2 x$ .

				$3 x^{2}$	+	$14 x$
$2 x$	-	$1$		$6 x^{3}$	+	$25 x^{2}$	-	$24 x$	+	$5$
			-	$6 x^{3}$	+	$3 x^{2}$
					+	$28 x^{2}$	-	$24 x$

Step 2.1.5.8

Multiply the new quotient term by the divisor.

				$3 x^{2}$	+	$14 x$
$2 x$	-	$1$		$6 x^{3}$	+	$25 x^{2}$	-	$24 x$	+	$5$
			-	$6 x^{3}$	+	$3 x^{2}$
					+	$28 x^{2}$	-	$24 x$
					+	$28 x^{2}$	-	$14 x$

Step 2.1.5.9

The expression needs to be subtracted from the dividend, so change all the signs in $28 x^{2} - 14 x$

				$3 x^{2}$	+	$14 x$
$2 x$	-	$1$		$6 x^{3}$	+	$25 x^{2}$	-	$24 x$	+	$5$
			-	$6 x^{3}$	+	$3 x^{2}$
					+	$28 x^{2}$	-	$24 x$
					-	$28 x^{2}$	+	$14 x$

Step 2.1.5.10

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

				$3 x^{2}$	+	$14 x$
$2 x$	-	$1$		$6 x^{3}$	+	$25 x^{2}$	-	$24 x$	+	$5$
			-	$6 x^{3}$	+	$3 x^{2}$
					+	$28 x^{2}$	-	$24 x$
					-	$28 x^{2}$	+	$14 x$
							-	$10 x$

Step 2.1.5.11

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

				$3 x^{2}$	+	$14 x$
$2 x$	-	$1$		$6 x^{3}$	+	$25 x^{2}$	-	$24 x$	+	$5$
			-	$6 x^{3}$	+	$3 x^{2}$
					+	$28 x^{2}$	-	$24 x$
					-	$28 x^{2}$	+	$14 x$
							-	$10 x$	+	$5$

Step 2.1.5.12

Divide the highest order term in the dividend $- 10 x$ by the highest order term in divisor $2 x$ .

				$3 x^{2}$	+	$14 x$	-	$5$
$2 x$	-	$1$		$6 x^{3}$	+	$25 x^{2}$	-	$24 x$	+	$5$
			-	$6 x^{3}$	+	$3 x^{2}$
					+	$28 x^{2}$	-	$24 x$
					-	$28 x^{2}$	+	$14 x$
							-	$10 x$	+	$5$

Step 2.1.5.13

Multiply the new quotient term by the divisor.

				$3 x^{2}$	+	$14 x$	-	$5$
$2 x$	-	$1$		$6 x^{3}$	+	$25 x^{2}$	-	$24 x$	+	$5$
			-	$6 x^{3}$	+	$3 x^{2}$
					+	$28 x^{2}$	-	$24 x$
					-	$28 x^{2}$	+	$14 x$
							-	$10 x$	+	$5$
							-	$10 x$	+	$5$

Step 2.1.5.14

The expression needs to be subtracted from the dividend, so change all the signs in $- 10 x + 5$

				$3 x^{2}$	+	$14 x$	-	$5$
$2 x$	-	$1$		$6 x^{3}$	+	$25 x^{2}$	-	$24 x$	+	$5$
			-	$6 x^{3}$	+	$3 x^{2}$
					+	$28 x^{2}$	-	$24 x$
					-	$28 x^{2}$	+	$14 x$
							-	$10 x$	+	$5$
							+	$10 x$	-	$5$

Step 2.1.5.15

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

				$3 x^{2}$	+	$14 x$	-	$5$
$2 x$	-	$1$		$6 x^{3}$	+	$25 x^{2}$	-	$24 x$	+	$5$
			-	$6 x^{3}$	+	$3 x^{2}$
					+	$28 x^{2}$	-	$24 x$
					-	$28 x^{2}$	+	$14 x$
							-	$10 x$	+	$5$
							+	$10 x$	-	$5$
										$0$

Step 2.1.5.16

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

$3 x^{2} + 14 x - 5$

Step 2.1.6

Write $6 x^{3} + 25 x^{2} - 24 x + 5$ as a set of factors.

$(2 x - 1) (3 x^{2} + 14 x - 5) = 0$

Step 2.2

Factor by grouping.

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Step 2.2.1

Factor by grouping.

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Step 2.2.1.1

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 $a \cdot c = 3 \cdot - 5 = - 15$ and whose sum is $b = 14$ .

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Step 2.2.1.1.1

Factor $14$ out of $14 x$ .

$(2 x - 1) (3 x^{2} + 14 (x) - 5) = 0$

Step 2.2.1.1.2

Rewrite $14$ as $- 1$ plus $15$

$(2 x - 1) (3 x^{2} + (- 1 + 15) x - 5) = 0$

Step 2.2.1.1.3

Apply the distributive property.

$(2 x - 1) (3 x^{2} - 1 x + 15 x - 5) = 0$

Step 2.2.1.2

Factor out the greatest common factor from each group.

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Step 2.2.1.2.1

Group the first two terms and the last two terms.

$(2 x - 1) ((3 x^{2} - 1 x) + 15 x - 5) = 0$

Step 2.2.1.2.2

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

$(2 x - 1) (x (3 x - 1) + 5 (3 x - 1)) = 0$

Step 2.2.1.3

Factor the polynomial by factoring out the greatest common factor, $3 x - 1$ .

$(2 x - 1) ((3 x - 1) (x + 5)) = 0$

Step 2.2.2

Remove unnecessary parentheses.

$(2 x - 1) (3 x - 1) (x + 5) = 0$

Step 3

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

$2 x - 1 = 0$

$3 x - 1 = 0$

$x + 5 = 0$

Step 4

Set $2 x - 1$ equal to $0$ and solve for $x$ .

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Step 4.1

Set $2 x - 1$ equal to $0$ .

$2 x - 1 = 0$

Step 4.2

Solve $2 x - 1 = 0$ for $x$ .

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Step 4.2.1

Add $1$ to both sides of the equation.

$2 x = 1$

Step 4.2.2

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

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Step 4.2.2.1

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

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

Step 4.2.2.2

Simplify the left side.

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Step 4.2.2.2.1

Cancel the common factor of $2$ .

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Step 4.2.2.2.1.1

Cancel the common factor.

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

Step 4.2.2.2.1.2

Divide $x$ by $1$ .

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

Step 5

Set $3 x - 1$ equal to $0$ and solve for $x$ .

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Step 5.1

Set $3 x - 1$ equal to $0$ .

$3 x - 1 = 0$

Step 5.2

Solve $3 x - 1 = 0$ for $x$ .

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Step 5.2.1

Add $1$ to both sides of the equation.

$3 x = 1$

Step 5.2.2

Divide each term in $3 x = 1$ by $3$ and simplify.

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Step 5.2.2.1

Divide each term in $3 x = 1$ by $3$ .

$\frac{3 x}{3} = \frac{1}{3}$

Step 5.2.2.2

Simplify the left side.

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Step 5.2.2.2.1

Cancel the common factor of $3$ .

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Step 5.2.2.2.1.1

Cancel the common factor.

$\frac{3 x}{3} = \frac{1}{3}$

Step 5.2.2.2.1.2

Divide $x$ by $1$ .

$x = \frac{1}{3}$

Step 6

Set $x + 5$ equal to $0$ and solve for $x$ .

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Step 6.1

Set $x + 5$ equal to $0$ .

$x + 5 = 0$

Step 6.2

Subtract $5$ from both sides of the equation.

$x = - 5$

Step 7

The final solution is all the values that make $(2 x - 1) (3 x - 1) (x + 5) = 0$ true.

$x = \frac{1}{2}, \frac{1}{3}, - 5$

Step 8

Use each root to create test intervals.

$x < - 5$

$- 5 < x < \frac{1}{3}$

$\frac{1}{3} < x < \frac{1}{2}$

$x > \frac{1}{2}$

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 $x < - 5$ to see if it makes the inequality true.

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Step 9.1.1

Choose a value on the interval $x < - 5$ and see if this value makes the original inequality true.

$x = - 8$

Step 9.1.2

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

$6 {(- 8)}^{3} + 25 {(- 8)}^{2} - 24 \cdot - 8 + 5 > 0$

Step 9.1.3

The left side $- 1275$ is not greater than the right side $0$ , which means that the given statement is false.

False

Step 9.2

Test a value on the interval $- 5 < x < \frac{1}{3}$ to see if it makes the inequality true.

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Step 9.2.1

Choose a value on the interval $- 5 < x < \frac{1}{3}$ and see if this value makes the original inequality true.

$x = 0$

Step 9.2.2

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

$6 {(0)}^{3} + 25 {(0)}^{2} - 24 \cdot 0 + 5 > 0$

Step 9.2.3

The left side $5$ 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 $\frac{1}{3} < x < \frac{1}{2}$ to see if it makes the inequality true.

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Step 9.3.1

Choose a value on the interval $\frac{1}{3} < x < \frac{1}{2}$ and see if this value makes the original inequality true.

$x = 0.42$

Step 9.3.2

Replace $x$ with $0.42$ in the original inequality.

$6 {(0.42)}^{3} + 25 {(0.42)}^{2} - 24 \cdot 0.42 + 5 > 0$

Step 9.3.3

The left side $- 0.225472$ is not greater than the right side $0$ , which means that the given statement is false.

False

Step 9.4

Test a value on the interval $x > \frac{1}{2}$ to see if it makes the inequality true.

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Step 9.4.1

Choose a value on the interval $x > \frac{1}{2}$ and see if this value makes the original inequality true.

$x = 3$

Step 9.4.2

Replace $x$ with $3$ in the original inequality.

$6 {(3)}^{3} + 25 {(3)}^{2} - 24 \cdot 3 + 5 > 0$

Step 9.4.3

The left side $320$ 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.

$x < - 5$ False

$- 5 < x < \frac{1}{3}$ True

$\frac{1}{3} < x < \frac{1}{2}$ False

$x > \frac{1}{2}$ True

$x < - 5$ False

$- 5 < x < \frac{1}{3}$ True

$\frac{1}{3} < x < \frac{1}{2}$ False

$x > \frac{1}{2}$ True

Step 10

The solution consists of all of the true intervals.

$- 5 < x < \frac{1}{3}$ or $x > \frac{1}{2}$

Step 11

Convert the inequality to interval notation.

$(- 5, \frac{1}{3}) \cup (\frac{1}{2}, \infty)$

Step 12