Find the Roots (Zeros) x^3-5x^2+7x-35=0

$x^{3} - 5 x^{2} + 7 x - 35 = 0$

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.

$(x^{3} - 5 x^{2}) + 7 x - 35 = 0$

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

$x^{2} (x - 5) + 7 (x - 5) = 0$

Factor the polynomial by factoring out the greatest common factor, $x - 5$ .

$(x - 5) (x^{2} + 7) = 0$

Step 2

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

$x - 5 = 0$

$x^{2} + 7 = 0$

Step 3

Set $x - 5$ equal to $0$ and solve for $x$ .

Set $x - 5$ equal to $0$ .

$x - 5 = 0$

Add $5$ to both sides of the equation.

$x = 5$

Step 4

Set $x^{2} + 7$ equal to $0$ and solve for $x$ .

Set $x^{2} + 7$ equal to $0$ .

$x^{2} + 7 = 0$

Solve $x^{2} + 7 = 0$ for $x$ .

Subtract $7$ from both sides of the equation.

$x^{2} = - 7$

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

$x = \pm \sqrt{- 7}$

Simplify $\pm \sqrt{- 7}$ .

Rewrite $- 7$ as $- 1 (7)$ .

$x = \pm \sqrt{- 1 (7)}$

Rewrite $\sqrt{- 1 (7)}$ as $\sqrt{- 1} \cdot \sqrt{7}$ .

$x = \pm \sqrt{- 1} \cdot \sqrt{7}$

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

$x = \pm i \sqrt{7}$

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.

$x = i \sqrt{7}$

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

$x = - i \sqrt{7}$

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

$x = i \sqrt{7}, - i \sqrt{7}$

Step 5

The final solution is all the values that make $(x - 5) (x^{2} + 7) = 0$ true.

$x = 5, i \sqrt{7}, - i \sqrt{7}$

Step 6

Find the Roots (Zeros) x^3-5x^2+7x-35=0

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