Find the Roots (Zeros) x^3-3x^2+4x-12=0

$x^{3} - 3 x^{2} + 4 x - 12 = 0$

Factor the left side of the equation.

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(x^{3} - 3 x^{2}) + 4 x - 12 = 0$

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

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

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

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

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

$x - 3 = 0$

$x^{2} + 4 = 0$

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

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

$x - 3 = 0$

Add $3$ to both sides of the equation.

$x = 3$

Set $x^{2} + 4$ equal to $0$ and solve for $x$ .

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

$x^{2} + 4 = 0$

Solve $x^{2} + 4 = 0$ for $x$ .

Subtract $4$ from both sides of the equation.

$x^{2} = - 4$

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

$x = \pm \sqrt{- 4}$

Simplify $\pm \sqrt{- 4}$ .

Rewrite $- 4$ as $- 1 (4)$ .

$x = \pm \sqrt{- 1 (4)}$

Rewrite $\sqrt{- 1 (4)}$ as $\sqrt{- 1} \cdot \sqrt{4}$ .

$x = \pm \sqrt{- 1} \cdot \sqrt{4}$

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

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

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

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

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

$x = \pm i \cdot 2$

Move $2$ to the left of $i$ .

$x = \pm 2 i$

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

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

$x = 2 i$

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

$x = - 2 i$

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

$x = 2 i, - 2 i$

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

$x = 3, 2 i, - 2 i$

Find the Roots (Zeros) x^3-3x^2+4x-12=0

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