Algebra Examples

$y = x^{3} - 3 x^{2} + 16 x - 48$

Step 1

Set $x^{3} - 3 x^{2} + 16 x - 48$ equal to $0$ .

$x^{3} - 3 x^{2} + 16 x - 48 = 0$

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 out the greatest common factor from each group.

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Step 2.1.1.1

Group the first two terms and the last two terms.

$(x^{3} - 3 x^{2}) + 16 x - 48 = 0$

Step 2.1.1.2

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

$x^{2} (x - 3) + 16 (x - 3) = 0$

Step 2.1.2

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

$(x - 3) (x^{2} + 16) = 0$

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 $0$ .

$x - 3 = 0$

$x^{2} + 16 = 0$

Step 2.3

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

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Step 2.3.1

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

$x - 3 = 0$

Step 2.3.2

Add $3$ to both sides of the equation.

$x = 3$

Step 2.4

Set $x^{2} + 16$ equal to $0$ and solve for $x$ .

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Step 2.4.1

Set $x^{2} + 16$ equal to $0$ .

$x^{2} + 16 = 0$

Step 2.4.2

Solve $x^{2} + 16 = 0$ for $x$ .

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Step 2.4.2.1

Subtract $16$ from both sides of the equation.

$x^{2} = - 16$

Step 2.4.2.2

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

$x = \pm \sqrt{- 16}$

Step 2.4.2.3

Simplify $\pm \sqrt{- 16}$ .

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Step 2.4.2.3.1

Rewrite $- 16$ as $- 1 (16)$ .

$x = \pm \sqrt{- 1 (16)}$

Step 2.4.2.3.2

Rewrite $\sqrt{- 1 (16)}$ as $\sqrt{- 1} \cdot \sqrt{16}$ .

$x = \pm \sqrt{- 1} \cdot \sqrt{16}$

Step 2.4.2.3.3

Rewrite $\sqrt{- 1}$ as $i$ .

$x = \pm i \cdot \sqrt{16}$

Step 2.4.2.3.4

Rewrite $16$ as $4^{2}$ .

$x = \pm i \cdot \sqrt{4^{2}}$

Step 2.4.2.3.5

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

$x = \pm i \cdot 4$

Step 2.4.2.3.6

Move $4$ to the left of $i$ .

$x = \pm 4 i$

Step 2.4.2.4

The complete solution is the result of both the positive and negative portions of the solution.

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Step 2.4.2.4.1

First, use the positive value of the $\pm$ to find the first solution.

$x = 4 i$

Step 2.4.2.4.2

Next, use the negative value of the $\pm$ to find the second solution.

$x = - 4 i$

Step 2.4.2.4.3

The complete solution is the result of both the positive and negative portions of the solution.

$x = 4 i, - 4 i$

Step 2.5

The final solution is all the values that make $(x - 3) (x^{2} + 16) = 0$ true.

$x = 3, 4 i, - 4 i$

Step 3