Find the Roots (Zeros) 2x^5-3x^3-20x=0

$2 x^{5} - 3 x^{3} - 20 x = 0$

Factor the left side of the equation.

Factor $x$ out of $2 x^{5} - 3 x^{3} - 20 x$ .

Factor $x$ out of $2 x^{5}$ .

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

Factor $x$ out of $- 3 x^{3}$ .

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

Factor $x$ out of $- 20 x$ .

$x (2 x^{4}) + x (- 3 x^{2}) + x \cdot - 20 = 0$

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

$x (2 x^{4} - 3 x^{2}) + x \cdot - 20 = 0$

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

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

Rewrite $x^{4}$ as ${(x^{2})}^{2}$ .

$x (2 {(x^{2})}^{2} - 3 x^{2} - 20) = 0$

Let $u = x^{2}$ . Substitute $u$ for all occurrences of $x^{2}$ .

$x (2 u^{2} - 3 u - 20) = 0$

Factor by grouping.

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = 2 \cdot - 20 = - 40$ and whose sum is $b = - 3$ .

Factor $- 3$ out of $- 3 u$ .

$x (2 u^{2} - 3 u - 20) = 0$

Rewrite $- 3$ as $5$ plus $- 8$

$x (2 u^{2} + (5 - 8) u - 20) = 0$

Apply the distributive property.

$x (2 u^{2} + 5 u - 8 u - 20) = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$x ((2 u^{2} + 5 u) - 8 u - 20) = 0$

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

$x (u (2 u + 5) - 4 (2 u + 5)) = 0$

Factor the polynomial by factoring out the greatest common factor, $2 u + 5$ .

$x ((2 u + 5) (u - 4)) = 0$

Replace all occurrences of $u$ with $x^{2}$ .

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

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

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

Factor.

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = x$ and $b = 2$ .

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

Remove unnecessary parentheses.

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

Remove unnecessary parentheses.

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

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

$x = 0$

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

$x + 2 = 0$

$x - 2 = 0$

Set $x$ equal to $0$ .

$x = 0$

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

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

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

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

Subtract $5$ from both sides of the equation.

$2 x^{2} = - 5$

Divide each term in $2 x^{2} = - 5$ by $2$ and simplify.

Divide each term in $2 x^{2} = - 5$ by $2$ .

$\frac{2 x^{2}}{2} = \frac{- 5}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 x^{2}}{2} = \frac{- 5}{2}$

Divide $x^{2}$ by $1$ .

$x^{2} = \frac{- 5}{2}$

Simplify the right side.

Move the negative in front of the fraction.

$x^{2} = - \frac{5}{2}$

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

$x = \pm \sqrt{- \frac{5}{2}}$

Simplify $\pm \sqrt{- \frac{5}{2}}$ .

Rewrite $- 1$ as $i^{2}$ .

$x = \pm \sqrt{i^{2} \frac{5}{2}}$

Pull terms out from under the radical.

$x = \pm i \sqrt{\frac{5}{2}}$

Rewrite $\sqrt{\frac{5}{2}}$ as $\frac{\sqrt{5}}{\sqrt{2}}$ .

$x = \pm i \frac{\sqrt{5}}{\sqrt{2}}$

Multiply $\frac{\sqrt{5}}{\sqrt{2}}$ by $\frac{\sqrt{2}}{\sqrt{2}}$ .

$x = \pm i (\frac{\sqrt{5}}{\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}})$

Combine and simplify the denominator.

Multiply $\frac{\sqrt{5}}{\sqrt{2}}$ by $\frac{\sqrt{2}}{\sqrt{2}}$ .

$x = \pm i \frac{\sqrt{5} \sqrt{2}}{\sqrt{2} \sqrt{2}}$

Raise $\sqrt{2}$ to the power of $1$ .

$x = \pm i \frac{\sqrt{5} \sqrt{2}}{{\sqrt{2}}^{1} \sqrt{2}}$

Raise $\sqrt{2}$ to the power of $1$ .

$x = \pm i \frac{\sqrt{5} \sqrt{2}}{{\sqrt{2}}^{1} {\sqrt{2}}^{1}}$

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$x = \pm i \frac{\sqrt{5} \sqrt{2}}{{\sqrt{2}}^{1 + 1}}$

Add $1$ and $1$ .

$x = \pm i \frac{\sqrt{5} \sqrt{2}}{{\sqrt{2}}^{2}}$

Rewrite ${\sqrt{2}}^{2}$ as $2$ .

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{2}$ as $2^{\frac{1}{2}}$ .

$x = \pm i \frac{\sqrt{5} \sqrt{2}}{{(2^{\frac{1}{2}})}^{2}}$

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$x = \pm i \frac{\sqrt{5} \sqrt{2}}{2^{\frac{1}{2} \cdot 2}}$

Combine $\frac{1}{2}$ and $2$ .

$x = \pm i \frac{\sqrt{5} \sqrt{2}}{2^{\frac{2}{2}}}$

Cancel the common factor of $2$ .

Cancel the common factor.

$x = \pm i \frac{\sqrt{5} \sqrt{2}}{2^{\frac{2}{2}}}$

Rewrite the expression.

$x = \pm i \frac{\sqrt{5} \sqrt{2}}{2^{1}}$

Evaluate the exponent.

$x = \pm i \frac{\sqrt{5} \sqrt{2}}{2}$

Simplify the numerator.

Combine using the product rule for radicals.

$x = \pm i \frac{\sqrt{5 \cdot 2}}{2}$

Multiply $5$ by $2$ .

$x = \pm i \frac{\sqrt{10}}{2}$

Combine $i$ and $\frac{\sqrt{10}}{2}$ .

$x = \pm \frac{i \sqrt{10}}{2}$

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.

$x = \frac{i \sqrt{10}}{2}$

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

$x = - \frac{i \sqrt{10}}{2}$

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

$x = \frac{i \sqrt{10}}{2}, - \frac{i \sqrt{10}}{2}$

Set $x + 2$ equal to $0$ and solve for $x$ .

Set $x + 2$ equal to $0$ .

$x + 2 = 0$

Subtract $2$ from both sides of the equation.

$x = - 2$

Set $x - 2$ equal to $0$ and solve for $x$ .

Set $x - 2$ equal to $0$ .

$x - 2 = 0$

Add $2$ to both sides of the equation.

$x = 2$

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

$x = 0, \frac{i \sqrt{10}}{2}, - \frac{i \sqrt{10}}{2}, - 2, 2$

Find the Roots (Zeros) 2x^5-3x^3-20x=0

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