Solve by Factoring 12x^2-1=-x

$12 x^{2} - 1 = - x$

Step 1

Add $x$ to both sides of the equation.

$12 x^{2} - 1 + x = 0$

Step 2

Factor by grouping.

Reorder terms.

$12 x^{2} + x - 1 = 0$

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 $a \cdot c = 12 \cdot - 1 = - 12$ and whose sum is $b = 1$ .

Multiply by $1$ .

$12 x^{2} + 1 x - 1 = 0$

Rewrite $1$ as $- 3$ plus $4$

$12 x^{2} + (- 3 + 4) x - 1 = 0$

Apply the distributive property.

$12 x^{2} - 3 x + 4 x - 1 = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(12 x^{2} - 3 x) + 4 x - 1 = 0$

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

$3 x (4 x - 1) + 1 (4 x - 1) = 0$

Factor the polynomial by factoring out the greatest common factor, $4 x - 1$ .

$(4 x - 1) (3 x + 1) = 0$

Step 3

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

$4 x - 1 = 0$

$3 x + 1 = 0$

Step 4

Set $4 x - 1$ equal to $0$ and solve for $x$ .

Set $4 x - 1$ equal to $0$ .

$4 x - 1 = 0$

Solve $4 x - 1 = 0$ for $x$ .

Add $1$ to both sides of the equation.

$4 x = 1$

Divide each term in $4 x = 1$ by $4$ and simplify.

Divide each term in $4 x = 1$ by $4$ .

$\frac{4 x}{4} = \frac{1}{4}$

Simplify the left side.

Cancel the common factor of $4$ .

Cancel the common factor.

$\frac{4 x}{4} = \frac{1}{4}$

Divide $x$ by $1$ .

$x = \frac{1}{4}$

Step 5

Set $3 x + 1$ equal to $0$ and solve for $x$ .

Set $3 x + 1$ equal to $0$ .

$3 x + 1 = 0$

Solve $3 x + 1 = 0$ for $x$ .

Subtract $1$ from both sides of the equation.

$3 x = - 1$

Divide each term in $3 x = - 1$ by $3$ and simplify.

Divide each term in $3 x = - 1$ by $3$ .

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

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

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

Divide $x$ by $1$ .

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

Simplify the right side.

Move the negative in front of the fraction.

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

Step 6

The final solution is all the values that make $(4 x - 1) (3 x + 1) = 0$ true.

$x = \frac{1}{4}, - \frac{1}{3}$

Solve by Factoring 12x^2-1=-x

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