Solve for m 2m^3-m^2-18m+9=0

$2 m^{3} - m^{2} - 18 m + 9 = 0$

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.

$(2 m^{3} - m^{2}) - 18 m + 9 = 0$

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

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

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

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

Rewrite $9$ as $3^{2}$ .

$(2 m - 1) (m^{2} - 3^{2}) = 0$

Factor.

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = m$ and $b = 3$ .

$(2 m - 1) ((m + 3) (m - 3)) = 0$

Remove unnecessary parentheses.

$(2 m - 1) (m + 3) (m - 3) = 0$

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

$2 m - 1 = 0$

$m + 3 = 0$

$m - 3 = 0$

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

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

$2 m - 1 = 0$

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

Add $1$ to both sides of the equation.

$2 m = 1$

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

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

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

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

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

Divide $m$ by $1$ .

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

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

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

$m + 3 = 0$

Subtract $3$ from both sides of the equation.

$m = - 3$

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

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

$m - 3 = 0$

Add $3$ to both sides of the equation.

$m = 3$

The final solution is all the values that make $(2 m - 1) (m + 3) (m - 3) = 0$ true.

$m = \frac{1}{2}, - 3, 3$

The result can be shown in multiple forms.

Exact Form:

$m = \frac{1}{2}, - 3, 3$

Decimal Form:

$m = 0.5, - 3, 3$

Solve for m 2m^3-m^2-18m+9=0

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