Find the Roots (Zeros) f(x)=x^3-3x^2+x-3

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

Step 1

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

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

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.

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

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

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

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

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

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

$x - 3 = 0$

$x^{2} + 1 = 0$

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 $x^{2} + 1$ equal to $0$ and solve for $x$ .

Set $x^{2} + 1$ equal to $0$ .

$x^{2} + 1 = 0$

Solve $x^{2} + 1 = 0$ for $x$ .

Subtract $1$ from both sides of the equation.

$x^{2} = - 1$

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

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

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

$x = \pm i$

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.

$x = i$

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

$x = - i$

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

$x = i, - i$

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

$x = 3, i, - i$

Step 3