Find the Roots (Zeros) x^4+x^3-6x^2

$x^{4} + x^{3} - 6 x^{2}$

Set $x^{4} + x^{3} - 6 x^{2}$ equal to $0$ .

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

Solve for $x$ .

Factor the left side of the equation.

Factor $x^{2}$ out of $x^{4} + x^{3} - 6 x^{2}$ .

Factor $x^{2}$ out of $x^{4}$ .

$x^{2} x^{2} + x^{3} - 6 x^{2} = 0$

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

$x^{2} x^{2} + x^{2} x - 6 x^{2} = 0$

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

$x^{2} x^{2} + x^{2} x + x^{2} \cdot - 6 = 0$

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

$x^{2} (x^{2} + x) + x^{2} \cdot - 6 = 0$

Factor $x^{2}$ out of $x^{2} (x^{2} + x) + x^{2} \cdot - 6$ .

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

Factor.

Factor $x^{2} + x - 6$ 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 $- 6$ and whose sum is $1$ .

$- 2, 3$

Write the factored form using these integers.

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

Remove unnecessary parentheses.

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

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

$x^{2} = 0$

$x - 2 = 0$

$x + 3 = 0$

Set $x^{2}$ equal to $0$ and solve for $x$ .

Set $x^{2}$ equal to $0$ .

$x^{2} = 0$

Solve $x^{2} = 0$ for $x$ .

Take the square root of both sides of the equation to eliminate the exponent on the left side.

$x = \pm \sqrt{0}$

Simplify $\pm \sqrt{0}$ .

Rewrite $0$ as $0^{2}$ .

$x = \pm \sqrt{0^{2}}$

Pull terms out from under the radical, assuming positive real numbers.

$x = \pm 0$

Plus or minus $0$ is $0$ .

$x = 0$

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

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

$x - 2 = 0$

Add $2$ to both sides of the equation.

$x = 2$

Set $x + 3$ equal to $0$ and solve for $x$ .

Set $x + 3$ equal to $0$ .

$x + 3 = 0$

Subtract $3$ from both sides of the equation.

$x = - 3$

The final solution is all the values that make $x^{2} (x - 2) (x + 3) = 0$ true.

$x = 0, 2, - 3$

Find the Roots (Zeros) x^4+x^3-6x^2

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