Solve for x -2x^6+9x^5+8x^4-36x^3=0

$- 2 x^{6} + 9 x^{5} + 8 x^{4} - 36 x^{3} = 0$

Step 1

Factor the left side of the equation.

Factor $x^{3}$ out of $- 2 x^{6} + 9 x^{5} + 8 x^{4} - 36 x^{3}$ .

Factor $x^{3}$ out of $- 2 x^{6}$ .

$x^{3} (- 2 x^{3}) + 9 x^{5} + 8 x^{4} - 36 x^{3} = 0$

Factor $x^{3}$ out of $9 x^{5}$ .

$x^{3} (- 2 x^{3}) + x^{3} (9 x^{2}) + 8 x^{4} - 36 x^{3} = 0$

Factor $x^{3}$ out of $8 x^{4}$ .

$x^{3} (- 2 x^{3}) + x^{3} (9 x^{2}) + x^{3} (8 x) - 36 x^{3} = 0$

Factor $x^{3}$ out of $- 36 x^{3}$ .

$x^{3} (- 2 x^{3}) + x^{3} (9 x^{2}) + x^{3} (8 x) + x^{3} \cdot - 36 = 0$

Factor $x^{3}$ out of $x^{3} (- 2 x^{3}) + x^{3} (9 x^{2})$ .

$x^{3} (- 2 x^{3} + 9 x^{2}) + x^{3} (8 x) + x^{3} \cdot - 36 = 0$

Factor $x^{3}$ out of $x^{3} (- 2 x^{3} + 9 x^{2}) + x^{3} (8 x)$ .

$x^{3} (- 2 x^{3} + 9 x^{2} + 8 x) + x^{3} \cdot - 36 = 0$

Factor $x^{3}$ out of $x^{3} (- 2 x^{3} + 9 x^{2} + 8 x) + x^{3} \cdot - 36$ .

$x^{3} (- 2 x^{3} + 9 x^{2} + 8 x - 36) = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$x^{3} ((- 2 x^{3} + 9 x^{2}) + 8 x - 36) = 0$

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

$x^{3} (x^{2} (- 2 x + 9) - 4 (- 2 x + 9)) = 0$

Factor the polynomial by factoring out the greatest common factor, $- 2 x + 9$ .

$x^{3} ((- 2 x + 9) (x^{2} - 4)) = 0$

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

$x^{3} ((- 2 x + 9) (x^{2} - 2^{2})) = 0$

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 $b = 2$ .

$x^{3} ((- 2 x + 9) ((x + 2) (x - 2))) = 0$

Remove unnecessary parentheses.

$x^{3} ((- 2 x + 9) (x + 2) (x - 2)) = 0$

Remove unnecessary parentheses.

$x^{3} (- 2 x + 9) (x + 2) (x - 2) = 0$

Step 2

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

$x^{3} = 0$

$- 2 x + 9 = 0$

$x + 2 = 0$

$x - 2 = 0$

Step 3

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

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

$x^{3} = 0$

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

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

$x = \sqrt[3]{0}$

Simplify $\sqrt[3]{0}$ .

Rewrite $0$ as $0^{3}$ .

$x = \sqrt[3]{0^{3}}$

Pull terms out from under the radical, assuming real numbers.

$x = 0$

Step 4

Set $- 2 x + 9$ equal to $0$ and solve for $x$ .

Set $- 2 x + 9$ equal to $0$ .

$- 2 x + 9 = 0$

Solve $- 2 x + 9 = 0$ for $x$ .

Subtract $9$ from both sides of the equation.

$- 2 x = - 9$

Divide each term in $- 2 x = - 9$ by $- 2$ and simplify.

Divide each term in $- 2 x = - 9$ by $- 2$ .

$\frac{- 2 x}{- 2} = \frac{- 9}{- 2}$

Simplify the left side.

Cancel the common factor of $- 2$ .

Cancel the common factor.

$\frac{- 2 x}{- 2} = \frac{- 9}{- 2}$

Divide $x$ by $1$ .

$x = \frac{- 9}{- 2}$

Simplify the right side.

Dividing two negative values results in a positive value.

$x = \frac{9}{2}$

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

The final solution is all the values that make $x^{3} (- 2 x + 9) (x + 2) (x - 2) = 0$ true.

$x = 0, \frac{9}{2}, - 2, 2$

Step 8

The result can be shown in multiple forms.

Exact Form:

$x = 0, \frac{9}{2}, - 2, 2$

Decimal Form:

$x = 0, 4.5, - 2, 2$

Mixed Number Form:

$x = 0, 4 \frac{1}{2}, - 2, 2$

Solve for x -2x^6+9x^5+8x^4-36x^3=0

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