Solve for x x^3+2x^2-x=2

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

Step 1

Subtract $2$ from both sides of the equation.

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

Step 2

Factor the left side of the equation.

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

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

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

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

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

$(x + 2) (x^{2} - 1) = 0$

Rewrite $1$ as $1^{2}$ .

$(x + 2) (x^{2} - 1^{2}) = 0$

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 $b = 1$ .

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

Remove unnecessary parentheses.

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

Step 3

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

$x + 2 = 0$

$x + 1 = 0$

$x - 1 = 0$

Step 4

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 5

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

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

$x + 1 = 0$

Subtract $1$ from both sides of the equation.

$x = - 1$

Step 6

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 7

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

$x = - 2, - 1, 1$

Solve for x x^3+2x^2-x=2

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