Find the Roots (Zeros) x^3-x^2+5x-5=0

$x^{3} - x^{2} + 5 x - 5 = 0$

Step 1

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.

$(x^{3} - x^{2}) + 5 x - 5 = 0$

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

$x^{2} (x - 1) + 5 (x - 1) = 0$

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

$(x - 1) (x^{2} + 5) = 0$

Step 2

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

$x - 1 = 0$

$x^{2} + 5 = 0$

Step 3

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

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

$x - 1 = 0$

Add $1$ to both sides of the equation.

$x = 1$

Step 4

Set $x^{2} + 5$ equal to $0$ and solve for $x$ .

Set $x^{2} + 5$ equal to $0$ .

$x^{2} + 5 = 0$

Solve $x^{2} + 5 = 0$ for $x$ .

Subtract $5$ from both sides of the equation.

$x^{2} = - 5$

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

$x = \pm \sqrt{- 5}$

Simplify $\pm \sqrt{- 5}$ .

Rewrite $- 5$ as $- 1 (5)$ .

$x = \pm \sqrt{- 1 (5)}$

Rewrite $\sqrt{- 1 (5)}$ as $\sqrt{- 1} \cdot \sqrt{5}$ .

$x = \pm \sqrt{- 1} \cdot \sqrt{5}$

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

$x = \pm i \sqrt{5}$

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.

$x = i \sqrt{5}$

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

$x = - i \sqrt{5}$

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

$x = i \sqrt{5}, - i \sqrt{5}$

Step 5

The final solution is all the values that make $(x - 1) (x^{2} + 5) = 0$ true.

$x = 1, i \sqrt{5}, - i \sqrt{5}$

Step 6

Find the Roots (Zeros) x^3-x^2+5x-5=0

Related Questions

Support LCMGCF.com by disabling your adblocker.