Find the Roots (Zeros) y=2x^3+x^2-x

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

Set $2 x^{3} + x^{2} - x$ equal to $0$ .

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

Solve for $x$ .

Factor the left side of the equation.

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

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

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

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

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

Factor $x$ out of $- x$ .

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

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

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

Factor $x$ out of $x (2 x^{2} + x) + x \cdot - 1$ .

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

Factor.

Factor by grouping.

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = 2 \cdot - 1 = - 2$ and whose sum is $b = 1$ .

Multiply by $1$ .

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

Rewrite $1$ as $- 1$ plus $2$

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

Apply the distributive property.

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

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

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

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

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

Factor the polynomial by factoring out the greatest common factor, $2 x - 1$ .

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

Remove unnecessary parentheses.

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

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

$x = 0$

$2 x - 1 = 0$

$x + 1 = 0$

Set $x$ equal to $0$ .

$x = 0$

Set $2 x - 1$ equal to $0$ and solve for $x$ .

Set $2 x - 1$ equal to $0$ .

$2 x - 1 = 0$

Solve $2 x - 1 = 0$ for $x$ .

Add $1$ to both sides of the equation.

$2 x = 1$

Divide each term in $2 x = 1$ by $2$ and simplify.

Divide each term in $2 x = 1$ by $2$ .

$\frac{2 x}{2} = \frac{1}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 x}{2} = \frac{1}{2}$

Divide $x$ by $1$ .

$x = \frac{1}{2}$

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.

$x = - 1$

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

$x = 0, \frac{1}{2}, - 1$

Find the Roots (Zeros) y=2x^3+x^2-x

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