Find the Roots (Zeros) 6x^2-33x-18

$6 x^{2} - 33 x - 18$

Step 1

Set $6 x^{2} - 33 x - 18$ equal to $0$ .

$6 x^{2} - 33 x - 18 = 0$

Step 2

Solve for $x$ .

Factor the left side of the equation.

Factor $3$ out of $6 x^{2} - 33 x - 18$ .

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

$3 (2 x^{2}) - 33 x - 18 = 0$

Factor $3$ out of $- 33 x$ .

$3 (2 x^{2}) + 3 (- 11 x) - 18 = 0$

Factor $3$ out of $- 18$ .

$3 (2 x^{2}) + 3 (- 11 x) + 3 (- 6) = 0$

Factor $3$ out of $3 (2 x^{2}) + 3 (- 11 x)$ .

$3 (2 x^{2} - 11 x) + 3 (- 6) = 0$

Factor $3$ out of $3 (2 x^{2} - 11 x) + 3 (- 6)$ .

$3 (2 x^{2} - 11 x - 6) = 0$

Factor.

Factor by grouping.

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = 2 \cdot - 6 = - 12$ and whose sum is $b = - 11$ .

Factor $- 11$ out of $- 11 x$ .

$3 (2 x^{2} - 11 x - 6) = 0$

Rewrite $- 11$ as $1$ plus $- 12$

$3 (2 x^{2} + (1 - 12) x - 6) = 0$

Apply the distributive property.

$3 (2 x^{2} + 1 x - 12 x - 6) = 0$

Multiply $x$ by $1$ .

$3 (2 x^{2} + x - 12 x - 6) = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$3 ((2 x^{2} + x) - 12 x - 6) = 0$

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

$3 (x (2 x + 1) - 6 (2 x + 1)) = 0$

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

$3 ((2 x + 1) (x - 6)) = 0$

Remove unnecessary parentheses.

$3 (2 x + 1) (x - 6) = 0$

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

$2 x + 1 = 0$

$x - 6 = 0$

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 $x$ .

Subtract $1$ from 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 $1$ .

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

Simplify the right side.

Move the negative in front of the fraction.

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

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

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

$x - 6 = 0$

Add $6$ to both sides of the equation.

$x = 6$

The final solution is all the values that make $3 (2 x + 1) (x - 6) = 0$ true.

$x = - \frac{1}{2}, 6$

Step 3

Find the Roots (Zeros) 6x^2-33x-18

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