Solve by Factoring n^3-9n^2-n+9=0

$n^{3} - 9 n^{2} - n + 9 = 0$

Step 1

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(n^{3} - 9 n^{2}) - n + 9 = 0$

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

$n^{2} (n - 9) - (n - 9) = 0$

Step 2

Factor the polynomial by factoring out the greatest common factor, $n - 9$ .

$(n - 9) (n^{2} - 1) = 0$

Step 3

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

$(n - 9) (n^{2} - 1^{2}) = 0$

Step 4

Factor.

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = n$ and $b = 1$ .

$(n - 9) ((n + 1) (n - 1)) = 0$

Remove unnecessary parentheses.

$(n - 9) (n + 1) (n - 1) = 0$

Step 5

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

$n - 9 = 0$

$n + 1 = 0$

$n - 1 = 0$

Step 6

Set $n - 9$ equal to $0$ and solve for $n$ .

Set $n - 9$ equal to $0$ .

$n - 9 = 0$

Add $9$ to both sides of the equation.

$n = 9$

Step 7

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

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

$n + 1 = 0$

Subtract $1$ from both sides of the equation.

$n = - 1$

Step 8

Set $n - 1$ equal to $0$ and solve for $n$ .

Set $n - 1$ equal to $0$ .

$n - 1 = 0$

Add $1$ to both sides of the equation.

$n = 1$

Step 9

The final solution is all the values that make $(n - 9) (n + 1) (n - 1) = 0$ true.

$n = 9, - 1, 1$