Find the Roots (Zeros) y=x^3-3x^2+16x-48

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

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 $x$ .

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.

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

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

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

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

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

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

$x - 3 = 0$

$x^{2} + 16 = 0$

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

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

$x^{2} + 16 = 0$

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

Subtract $16$ from both sides of the equation.

$x^{2} = - 16$

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

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

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

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

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

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

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

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

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

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

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

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

$x = \pm i \cdot 4$

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

$x = \pm 4 i$

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.

$x = 4 i$

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

$x = - 4 i$

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

$x = 4 i, - 4 i$

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

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

Step 3

Find the Roots (Zeros) y=x^3-3x^2+16x-48

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