Find the Roots (Zeros) x^5-2x^4-24x^3=0

$x^{5} - 2 x^{4} - 24 x^{3} = 0$

Factor the left side of the equation.

Factor $x^{3}$ out of $x^{5} - 2 x^{4} - 24 x^{3}$ .

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

$x^{3} x^{2} - 2 x^{4} - 24 x^{3} = 0$

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

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

Factor $x^{3}$ out of $- 24 x^{3}$ .

$x^{3} x^{2} + x^{3} (- 2 x) + x^{3} \cdot - 24 = 0$

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

$x^{3} (x^{2} - 2 x) + x^{3} \cdot - 24 = 0$

Factor $x^{3}$ out of $x^{3} (x^{2} - 2 x) + x^{3} \cdot - 24$ .

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

Factor.

Factor $x^{2} - 2 x - 24$ 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 $- 24$ and whose sum is $- 2$ .

$- 6, 4$

Write the factored form using these integers.

$x^{3} ((x - 6) (x + 4)) = 0$

Remove unnecessary parentheses.

$x^{3} (x - 6) (x + 4) = 0$

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

$x^{3} = 0$

$x - 6 = 0$

$x + 4 = 0$

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.

$x = 0$

Set $x - 6$ equal to $0$ and solve for $x$ .

Set $x - 6$ equal to $0$ .

$x - 6 = 0$

Add $6$ to both sides of the equation.

$x = 6$

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.

$x = - 4$

The final solution is all the values that make $x^{3} (x - 6) (x + 4) = 0$ true.

$x = 0, 6, - 4$

Find the Roots (Zeros) x^5-2x^4-24x^3=0

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