Solve by Factoring 2x^3-x^2-18x+9=0

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

Step 1

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

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

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

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

Step 2

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

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

Step 3

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

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

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 $a = x$ and $b = 3$ .

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

Remove unnecessary parentheses.

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

Step 5

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

$2 x - 1 = 0$

$x + 3 = 0$

$x - 3 = 0$

Step 6

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

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

$2 x - 1 = 0$

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

Add $1$ to both sides of the equation.

$2 x = 1$

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

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

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

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

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

Divide $x$ by $1$ .

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

Step 7

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.

$x = - 3$

Step 8

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.

$x = 3$

Step 9

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

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

Solve by Factoring 2x^3-x^2-18x+9=0

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