Find the Roots (Zeros) 2x^4+8x^3+10x^2+40x

$2 x^{4} + 8 x^{3} + 10 x^{2} + 40 x$

Step 1

Set $2 x^{4} + 8 x^{3} + 10 x^{2} + 40 x$ equal to $0$ .

$2 x^{4} + 8 x^{3} + 10 x^{2} + 40 x = 0$

Step 2

Solve for $x$ .

Factor the left side of the equation.

Factor $2 x$ out of $2 x^{4} + 8 x^{3} + 10 x^{2} + 40 x$ .

Factor $2 x$ out of $2 x^{4}$ .

$2 x (x^{3}) + 8 x^{3} + 10 x^{2} + 40 x = 0$

Factor $2 x$ out of $8 x^{3}$ .

$2 x (x^{3}) + 2 x (4 x^{2}) + 10 x^{2} + 40 x = 0$

Factor $2 x$ out of $10 x^{2}$ .

$2 x (x^{3}) + 2 x (4 x^{2}) + 2 x (5 x) + 40 x = 0$

Factor $2 x$ out of $40 x$ .

$2 x (x^{3}) + 2 x (4 x^{2}) + 2 x (5 x) + 2 x (20) = 0$

Factor $2 x$ out of $2 x (x^{3}) + 2 x (4 x^{2})$ .

$2 x (x^{3} + 4 x^{2}) + 2 x (5 x) + 2 x (20) = 0$

Factor $2 x$ out of $2 x (x^{3} + 4 x^{2}) + 2 x (5 x)$ .

$2 x (x^{3} + 4 x^{2} + 5 x) + 2 x (20) = 0$

Factor $2 x$ out of $2 x (x^{3} + 4 x^{2} + 5 x) + 2 x (20)$ .

$2 x (x^{3} + 4 x^{2} + 5 x + 20) = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$2 x ((x^{3} + 4 x^{2}) + 5 x + 20) = 0$

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

$2 x (x^{2} (x + 4) + 5 (x + 4)) = 0$

Factor.

Factor the polynomial by factoring out the greatest common factor, $x + 4$ .

$2 x ((x + 4) (x^{2} + 5)) = 0$

Remove unnecessary parentheses.

$2 x (x + 4) (x^{2} + 5) = 0$

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

$x = 0$

$x + 4 = 0$

$x^{2} + 5 = 0$

Set $x$ equal to $0$ .

$x = 0$

Set $x + 4$ equal to $0$ and solve for $x$ .

Set $x + 4$ equal to $0$ .

$x + 4 = 0$

Subtract $4$ from both sides of the equation.

$x = - 4$

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 square 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.

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

The final solution is all the values that make $2 x (x + 4) (x^{2} + 5) = 0$ true.

$x = 0, - 4, i \sqrt{5}, - i \sqrt{5}$

Step 3

Find the Roots (Zeros) 2x^4+8x^3+10x^2+40x

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