Precalculus Examples

$12 x^{3} - 20 x^{2} < 0$

Convert the inequality to an equation.

$12 x^{3} - 20 x^{2} = 0$

Factor $4 x^{2}$ out of $12 x^{3} - 20 x^{2}$ .

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Factor $4 x^{2}$ out of $12 x^{3}$ .

$4 x^{2} (3 x) - 20 x^{2} = 0$

Factor $4 x^{2}$ out of $- 20 x^{2}$ .

$4 x^{2} (3 x) + 4 x^{2} (- 5) = 0$

Factor $4 x^{2}$ out of $4 x^{2} (3 x) + 4 x^{2} (- 5)$ .

$4 x^{2} (3 x - 5) = 0$

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

$x^{2} = 0$

$3 x - 5 = 0$

Set $x^{2}$ equal to $0$ and solve for $x$ .

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Set $x^{2}$ equal to $0$ .

$x^{2} = 0$

Solve $x^{2} = 0$ for $x$ .

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Take the square root of both sides of the equation to eliminate the exponent on the left side.

$x = \pm \sqrt{0}$

Simplify $\pm \sqrt{0}$ .

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Rewrite $0$ as $0^{2}$ .

$x = \pm \sqrt{0^{2}}$

Pull terms out from under the radical, assuming positive real numbers.

$x = \pm 0$

Plus or minus $0$ is $0$ .

$x = 0$

Set $3 x - 5$ equal to $0$ and solve for $x$ .

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Set $3 x - 5$ equal to $0$ .

$3 x - 5 = 0$

Solve $3 x - 5 = 0$ for $x$ .

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Add $5$ to both sides of the equation.

$3 x = 5$

Divide each term in $3 x = 5$ by $3$ and simplify.

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Divide each term in $3 x = 5$ by $3$ .

$\frac{3 x}{3} = \frac{5}{3}$

Simplify the left side.

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Cancel the common factor of $3$ .

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Cancel the common factor.

$\frac{3 x}{3} = \frac{5}{3}$

Divide $x$ by $1$ .

$x = \frac{5}{3}$

The final solution is all the values that make $4 x^{2} (3 x - 5) = 0$ true.

$x = 0, \frac{5}{3}$

Use each root to create test intervals.

$x < 0$

$0 < x < \frac{5}{3}$

$x > \frac{5}{3}$

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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Test a value on the interval $x < 0$ to see if it makes the inequality true.

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Choose a value on the interval $x < 0$ and see if this value makes the original inequality true.

$x = - 2$

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

$12 {(- 2)}^{3} - 20 {(- 2)}^{2} < 0$

The left side $- 176$ is less than the right side $0$ , which means that the given statement is always true.

True

Test a value on the interval $0 < x < \frac{5}{3}$ to see if it makes the inequality true.

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Choose a value on the interval $0 < x < \frac{5}{3}$ and see if this value makes the original inequality true.

$x = 1$

Replace $x$ with $1$ in the original inequality.

$12 {(1)}^{3} - 20 {(1)}^{2} < 0$

The left side $- 8$ is less than the right side $0$ , which means that the given statement is always true.

True

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

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

$x = 4$

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

$12 {(4)}^{3} - 20 {(4)}^{2} < 0$

The left side $448$ is not less than the right side $0$ , which means that the given statement is false.

False

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

$x < 0$ True

$0 < x < \frac{5}{3}$ True

$x > \frac{5}{3}$ False

$x < 0$ True

$0 < x < \frac{5}{3}$ True

$x > \frac{5}{3}$ False

The solution consists of all of the true intervals.

$x < 0$ or $0 < x < \frac{5}{3}$

Convert the inequality to interval notation.

$(- \infty, 0) \cup (0, \frac{5}{3})$