Solve for x x^3-5x^2=-6x

$x^{3} - 5 x^{2} = - 6 x$

Add $6 x$ to both sides of the equation.

$x^{3} - 5 x^{2} + 6 x = 0$

Factor the left side of the equation.

Factor $x$ out of $x^{3} - 5 x^{2} + 6 x$ .

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

$x \cdot x^{2} - 5 x^{2} + 6 x = 0$

Factor $x$ out of $- 5 x^{2}$ .

$x \cdot x^{2} + x (- 5 x) + 6 x = 0$

Factor $x$ out of $6 x$ .

$x \cdot x^{2} + x (- 5 x) + x \cdot 6 = 0$

Factor $x$ out of $x \cdot x^{2} + x (- 5 x)$ .

$x (x^{2} - 5 x) + x \cdot 6 = 0$

Factor $x$ out of $x (x^{2} - 5 x) + x \cdot 6$ .

$x (x^{2} - 5 x + 6) = 0$

Factor.

Factor $x^{2} - 5 x + 6$ 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 $- 5$ .

$- 3, - 2$

Write the factored form using these integers.

$x ((x - 3) (x - 2)) = 0$

Remove unnecessary parentheses.

$x (x - 3) (x - 2) = 0$

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

$x = 0$

$x - 3 = 0$

$x - 2 = 0$

Set $x$ equal to $0$ .

$x = 0$

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

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

$x - 3 = 0$

Add $3$ to both sides of the equation.

$x = 3$

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.

$x = 2$

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

$x = 0, 3, 2$

Solve for x x^3-5x^2=-6x

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