Solve for x -x^3+6x^2+x-6=0

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

Step 1

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.

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

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

$x^{2} (- x + 6) - 1 (- x + 6) = 0$

Factor the polynomial by factoring out the greatest common factor, $- x + 6$ .

$(- x + 6) (x^{2} - 1) = 0$

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

$(- x + 6) (x^{2} - 1^{2}) = 0$

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 $b = 1$ .

$(- x + 6) ((x + 1) (x - 1)) = 0$

Remove unnecessary parentheses.

$(- x + 6) (x + 1) (x - 1) = 0$

Step 2

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

$- x + 6 = 0$

$x + 1 = 0$

$x - 1 = 0$

Step 3

Set $- x + 6$ equal to $0$ and solve for $x$ .

Set $- x + 6$ equal to $0$ .

$- x + 6 = 0$

Solve $- x + 6 = 0$ for $x$ .

Subtract $6$ from both sides of the equation.

$- x = - 6$

Divide each term in $- x = - 6$ by $- 1$ and simplify.

Divide each term in $- x = - 6$ by $- 1$ .

$\frac{- x}{- 1} = \frac{- 6}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{x}{1} = \frac{- 6}{- 1}$

Divide $x$ by $1$ .

$x = \frac{- 6}{- 1}$

Simplify the right side.

Divide $- 6$ by $- 1$ .

$x = 6$

Step 4

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 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

The final solution is all the values that make $(- x + 6) (x + 1) (x - 1) = 0$ true.

$x = 6, - 1, 1$

Solve for x -x^3+6x^2+x-6=0

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