Precalculus Examples

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

Step 1

Set $- 2 x^{4} - x^{3} + 3 x^{2}$ equal to $0$ .

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

Step 2

Solve for $x$ .

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Step 2.1

Factor the left side of the equation.

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Step 2.1.1

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

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Step 2.1.1.1

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

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

Step 2.1.1.2

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

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

Step 2.1.1.3

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

$- x^{2} (2 x^{2}) - x^{2} x - x^{2} \cdot - 3 = 0$

Step 2.1.1.4

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

$- x^{2} (2 x^{2} + x) - x^{2} \cdot - 3 = 0$

Step 2.1.1.5

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

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

Step 2.1.2

Factor.

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Step 2.1.2.1

Factor by grouping.

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Step 2.1.2.1.1

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 $a \cdot c = 2 \cdot - 3 = - 6$ and whose sum is $b = 1$ .

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Step 2.1.2.1.1.1

Multiply by $1$ .

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

Step 2.1.2.1.1.2

Rewrite $1$ as $- 2$ plus $3$

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

Step 2.1.2.1.1.3

Apply the distributive property.

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

Step 2.1.2.1.2

Factor out the greatest common factor from each group.

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Step 2.1.2.1.2.1

Group the first two terms and the last two terms.

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

Step 2.1.2.1.2.2

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

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

Step 2.1.2.1.3

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

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

Step 2.1.2.2

Remove unnecessary parentheses.

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

Step 2.2

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

$x^{2} = 0$

$x - 1 = 0$

$2 x + 3 = 0$

Step 2.3

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

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Step 2.3.1

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

$x^{2} = 0$

Step 2.3.2

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

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Step 2.3.2.1

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

$x = \pm \sqrt{0}$

Step 2.3.2.2

Simplify $\pm \sqrt{0}$ .

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Step 2.3.2.2.1

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

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

Step 2.3.2.2.2

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

$x = \pm 0$

Step 2.3.2.2.3

Plus or minus $0$ is $0$ .

$x = 0$

Step 2.4

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

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Step 2.4.1

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

$x - 1 = 0$

Step 2.4.2

Add $1$ to both sides of the equation.

$x = 1$

Step 2.5

Set $2 x + 3$ equal to $0$ and solve for $x$ .

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Step 2.5.1

Set $2 x + 3$ equal to $0$ .

$2 x + 3 = 0$

Step 2.5.2

Solve $2 x + 3 = 0$ for $x$ .

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Step 2.5.2.1

Subtract $3$ from both sides of the equation.

$2 x = - 3$

Step 2.5.2.2

Divide each term in $2 x = - 3$ by $2$ and simplify.

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Step 2.5.2.2.1

Divide each term in $2 x = - 3$ by $2$ .

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

Step 2.5.2.2.2

Simplify the left side.

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Step 2.5.2.2.2.1

Cancel the common factor of $2$ .

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Step 2.5.2.2.2.1.1

Cancel the common factor.

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

Step 2.5.2.2.2.1.2

Divide $x$ by $1$ .

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

Step 2.5.2.2.3

Simplify the right side.

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Step 2.5.2.2.3.1

Move the negative in front of the fraction.

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

Step 2.6

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

$x = 0, 1, - \frac{3}{2}$

Step 3