Precalculus Examples

$f (x) = - 2 x^{3} + 6 x^{2} - \frac{9}{2} x$

Step 1

Set $- 2 x^{3} + 6 x^{2} - \frac{9}{2} x$ equal to $0$ .

$- 2 x^{3} + 6 x^{2} - \frac{9}{2} x = 0$

Step 2

Solve for $x$ .

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Step 2.1

Simplify each term.

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Step 2.1.1

Combine $x$ and $\frac{9}{2}$ .

$- 2 x^{3} + 6 x^{2} - \frac{x \cdot 9}{2} = 0$

Step 2.1.2

Move $9$ to the left of $x$ .

$- 2 x^{3} + 6 x^{2} - \frac{9 x}{2} = 0$

Step 2.2

Factor $- x$ out of $- 2 x^{3} + 6 x^{2} - \frac{9 x}{2}$ .

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Step 2.2.1

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

$- x (2 x^{2}) + 6 x^{2} - \frac{9 x}{2} = 0$

Step 2.2.2

Factor $- x$ out of $6 x^{2}$ .

$- x (2 x^{2}) - x (- 6 x) - \frac{9 x}{2} = 0$

Step 2.2.3

Factor $- x$ out of $- \frac{9 x}{2}$ .

$- x (2 x^{2}) - x (- 6 x) - x \frac{9}{2} = 0$

Step 2.2.4

Factor $- x$ out of $- x (2 x^{2}) - x (- 6 x)$ .

$- x (2 x^{2} - 6 x) - x \frac{9}{2} = 0$

Step 2.2.5

Factor $- x$ out of $- x (2 x^{2} - 6 x) - x (\frac{9}{2})$ .

$- x (2 x^{2} - 6 x + \frac{9}{2}) = 0$

Step 2.3

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

$x = 0$

$2 x^{2} - 6 x + \frac{9}{2} = 0$

Step 2.4

Set $x$ equal to $0$ .

$x = 0$

Step 2.5

Set $2 x^{2} - 6 x + \frac{9}{2}$ equal to $0$ and solve for $x$ .

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Step 2.5.1

Set $2 x^{2} - 6 x + \frac{9}{2}$ equal to $0$ .

$2 x^{2} - 6 x + \frac{9}{2} = 0$

Step 2.5.2

Solve $2 x^{2} - 6 x + \frac{9}{2} = 0$ for $x$ .

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Step 2.5.2.1

Multiply through by the least common denominator $2$ , then simplify.

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Step 2.5.2.1.1

Apply the distributive property.

$2 (2 x^{2}) + 2 (- 6 x) + 2 (\frac{9}{2}) = 0$

Step 2.5.2.1.2

Simplify.

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Step 2.5.2.1.2.1

Multiply $2$ by $2$ .

$4 x^{2} + 2 (- 6 x) + 2 (\frac{9}{2}) = 0$

Step 2.5.2.1.2.2

Multiply $- 6$ by $2$ .

$4 x^{2} - 12 x + 2 (\frac{9}{2}) = 0$

Step 2.5.2.1.2.3

Cancel the common factor of $2$ .

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Step 2.5.2.1.2.3.1

Cancel the common factor.

$4 x^{2} - 12 x + 2 (\frac{9}{2}) = 0$

Step 2.5.2.1.2.3.2

Rewrite the expression.

$4 x^{2} - 12 x + 9 = 0$

Step 2.5.2.2

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Step 2.5.2.3

Substitute the values $a = 4$ , $b = - 12$ , and $c = 9$ into the quadratic formula and solve for $x$ .

$\frac{12 \pm \sqrt{{(- 12)}^{2} - 4 \cdot (4 \cdot 9)}}{2 \cdot 4}$

Step 2.5.2.4

Simplify.

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Step 2.5.2.4.1

Simplify the numerator.

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Step 2.5.2.4.1.1

Raise $- 12$ to the power of $2$ .

$x = \frac{12 \pm \sqrt{144 - 4 \cdot 4 \cdot 9}}{2 \cdot 4}$

Step 2.5.2.4.1.2

Multiply $- 4 \cdot 4 \cdot 9$ .

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Step 2.5.2.4.1.2.1

Multiply $- 4$ by $4$ .

$x = \frac{12 \pm \sqrt{144 - 16 \cdot 9}}{2 \cdot 4}$

Step 2.5.2.4.1.2.2

Multiply $- 16$ by $9$ .

$x = \frac{12 \pm \sqrt{144 - 144}}{2 \cdot 4}$

Step 2.5.2.4.1.3

Subtract $144$ from $144$ .

$x = \frac{12 \pm \sqrt{0}}{2 \cdot 4}$

Step 2.5.2.4.1.4

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

$x = \frac{12 \pm \sqrt{0^{2}}}{2 \cdot 4}$

Step 2.5.2.4.1.5

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

$x = \frac{12 \pm 0}{2 \cdot 4}$

Step 2.5.2.4.1.6

$12$ plus or minus $0$ is $12$ .

$x = \frac{12}{2 \cdot 4}$

Step 2.5.2.4.2

Multiply $2$ by $4$ .

$x = \frac{12}{8}$

Step 2.5.2.4.3

Cancel the common factor of $12$ and $8$ .

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Step 2.5.2.4.3.1

Factor $4$ out of $12$ .

$x = \frac{4 (3)}{8}$

Step 2.5.2.4.3.2

Cancel the common factors.

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Step 2.5.2.4.3.2.1

Factor $4$ out of $8$ .

$x = \frac{4 \cdot 3}{4 \cdot 2}$

Step 2.5.2.4.3.2.2

Cancel the common factor.

$x = \frac{4 \cdot 3}{4 \cdot 2}$

Step 2.5.2.4.3.2.3

Rewrite the expression.

$x = \frac{3}{2}$

Step 2.5.2.5

The final answer is the combination of both solutions.

$x = \frac{3}{2}$ Double roots

Step 2.6

The final solution is all the values that make $- x (2 x^{2} - 6 x + \frac{9}{2}) = 0$ true.

$x = 0, \frac{3}{2}$

Step 3