Solve by Factoring 6x^3+54x^2+120x=0

$6 x^{3} + 54 x^{2} + 120 x = 0$

Step 1

Factor $6 x$ out of $6 x^{3} + 54 x^{2} + 120 x$ .

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

$6 x (x^{2}) + 54 x^{2} + 120 x = 0$

Factor $6 x$ out of $54 x^{2}$ .

$6 x (x^{2}) + 6 x (9 x) + 120 x = 0$

Factor $6 x$ out of $120 x$ .

$6 x (x^{2}) + 6 x (9 x) + 6 x (20) = 0$

Factor $6 x$ out of $6 x (x^{2}) + 6 x (9 x)$ .

$6 x (x^{2} + 9 x) + 6 x (20) = 0$

Factor $6 x$ out of $6 x (x^{2} + 9 x) + 6 x (20)$ .

$6 x (x^{2} + 9 x + 20) = 0$

Step 2

Factor.

Factor $x^{2} + 9 x + 20$ 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 $20$ and whose sum is $9$ .

$4, 5$

Write the factored form using these integers.

$6 x ((x + 4) (x + 5)) = 0$

Remove unnecessary parentheses.

$6 x (x + 4) (x + 5) = 0$

Step 3

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

$x = 0$

$x + 4 = 0$

$x + 5 = 0$

Step 4

Set $x$ equal to $0$ .

$x = 0$

Step 5

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.

$x = - 4$

Step 6

Set $x + 5$ equal to $0$ and solve for $x$ .

Set $x + 5$ equal to $0$ .

$x + 5 = 0$

Subtract $5$ from both sides of the equation.

$x = - 5$

Step 7

The final solution is all the values that make $6 x (x + 4) (x + 5) = 0$ true.

$x = 0, - 4, - 5$

Solve by Factoring 6x^3+54x^2+120x=0

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