Solve by Factoring 2x^3+2x^2-4x=0

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

Step 1

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

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

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

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

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

Factor $2 x$ out of $- 4 x$ .

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

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

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

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

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

Step 2

Factor.

Factor $x^{2} + x - 2$ 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 $- 2$ and whose sum is $1$ .

$- 1, 2$

Write the factored form using these integers.

$2 x ((x - 1) (x + 2)) = 0$

Remove unnecessary parentheses.

$2 x (x - 1) (x + 2) = 0$

Step 3

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

$x = 0$

$x - 1 = 0$

$x + 2 = 0$

Step 4

Set $x$ equal to $0$ .

$x = 0$

Step 5

Set $x - 1$ equal to $0$ and solve for $x$ .

Set $x - 1$ equal to $0$ .

$x - 1 = 0$

Add $1$ to both sides of the equation.

$x = 1$

Step 6

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 7

The final solution is all the values that make $2 x (x - 1) (x + 2) = 0$ true.

$x = 0, 1, - 2$

Solve by Factoring 2x^3+2x^2-4x=0

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