Solve by Factoring 3x^5+6x^4-72x^3=0

$3 x^{5} + 6 x^{4} - 72 x^{3} = 0$

Step 1

Factor $3 x^{3}$ out of $3 x^{5} + 6 x^{4} - 72 x^{3}$ .

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

$3 x^{3} (x^{2}) + 6 x^{4} - 72 x^{3} = 0$

Factor $3 x^{3}$ out of $6 x^{4}$ .

$3 x^{3} (x^{2}) + 3 x^{3} (2 x) - 72 x^{3} = 0$

Factor $3 x^{3}$ out of $- 72 x^{3}$ .

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

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

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

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

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

Step 2

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 $2$ .

$- 4, 6$

Write the factored form using these integers.

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

Remove unnecessary parentheses.

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

Step 3

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

$x^{3} = 0$

$x - 4 = 0$

$x + 6 = 0$

Step 4

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 specified 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 5

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

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

$x - 4 = 0$

Add $4$ to both sides of the equation.

$x = 4$

Step 6

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

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

$x + 6 = 0$

Subtract $6$ from both sides of the equation.

$x = - 6$

Step 7

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

$x = 0, 4, - 6$

Solve by Factoring 3x^5+6x^4-72x^3=0

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