Find the Roots (Zeros) 2x^5+24x=14x^3

$2 x^{5} + 24 x = 14 x^{3}$

Step 1

Subtract $14 x^{3}$ from both sides of the equation.

$2 x^{5} + 24 x - 14 x^{3} = 0$

Step 2

Factor the left side of the equation.

Factor $2 x$ out of $2 x^{5} + 24 x - 14 x^{3}$ .

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

$2 x (x^{4}) + 24 x - 14 x^{3} = 0$

Factor $2 x$ out of $24 x$ .

$2 x (x^{4}) + 2 x (12) - 14 x^{3} = 0$

Factor $2 x$ out of $- 14 x^{3}$ .

$2 x (x^{4}) + 2 x (12) + 2 x (- 7 x^{2}) = 0$

Factor $2 x$ out of $2 x (x^{4}) + 2 x (12)$ .

$2 x (x^{4} + 12) + 2 x (- 7 x^{2}) = 0$

Factor $2 x$ out of $2 x (x^{4} + 12) + 2 x (- 7 x^{2})$ .

$2 x (x^{4} + 12 - 7 x^{2}) = 0$

Factor $x^{4} + 12 - 7 x^{2}$ using the AC method.

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $12$ and whose sum is $- 7$ .

$- 4, - 3$

Write the factored form using these integers.

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

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

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

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 $b = 2$ .

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

Remove unnecessary parentheses.

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

Step 3

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

$x = 0$

$x + 2 = 0$

$x - 2 = 0$

$x^{2} - 3 = 0$

Step 4

Set $x$ equal to $0$ .

$x = 0$

Step 5

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.

$x = - 2$

Step 6

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.

$x = 2$

Step 7

Set $x^{2} - 3$ equal to $0$ and solve for $x$ .

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

$x^{2} - 3 = 0$

Solve $x^{2} - 3 = 0$ for $x$ .

Add $3$ to both sides of the equation.

$x^{2} = 3$

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

$x = \pm \sqrt{3}$

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.

$x = \sqrt{3}$

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

$x = - \sqrt{3}$

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

$x = \sqrt{3}, - \sqrt{3}$

Step 8

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

$x = 0, - 2, 2, \sqrt{3}, - \sqrt{3}$

Step 9

The result can be shown in multiple forms.

Exact Form:

$x = 0, - 2, 2, \sqrt{3}, - \sqrt{3}$

Decimal Form:

$x = 0, - 2, 2, 1.73205080 \dots, - 1.73205080 \dots$

Step 10

Find the Roots (Zeros) 2x^5+24x=14x^3

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