Solve for x (3x-1)^2+12x-4=0

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

Factor the left side of the equation.

Let $u = x$ . Substitute $u$ for all occurrences of $x$ .

$9 u^{2} + 6 u - 3 = 0$

Factor $3$ out of $9 u^{2} + 6 u - 3$ .

Factor $3$ out of $9 u^{2}$ .

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

Factor $3$ out of $6 u$ .

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

Factor $3$ out of $- 3$ .

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

Factor $3$ out of $3 (3 u^{2}) + 3 (2 u)$ .

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

Factor $3$ out of $3 (3 u^{2} + 2 u) + 3 (- 1)$ .

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

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 $a \cdot c = 3 \cdot - 1 = - 3$ and whose sum is $b = 2$ .

Factor $2$ out of $2 u$ .

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

Rewrite $2$ as $- 1$ plus $3$

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

Apply the distributive property.

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

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

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

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

$3 (u (3 u - 1) + 1 (3 u - 1)) = 0$

Factor the polynomial by factoring out the greatest common factor, $3 u - 1$ .

$3 ((3 u - 1) (u + 1)) = 0$

Remove unnecessary parentheses.

$3 (3 u - 1) (u + 1) = 0$

Replace all occurrences of $u$ with $x$ .

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

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

$3 x - 1 = 0$

$x + 1 = 0$

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 $x$ .

Add $1$ to 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 $1$ .

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

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

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

$x + 1 = 0$

Subtract $1$ from both sides of the equation.

$x = - 1$

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

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

The result can be shown in multiple forms.

Exact Form:

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

Decimal Form:

$x = 0.\overline{3}, - 1$

Solve for x (3x-1)^2+12x-4=0

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