Find the Roots (Zeros) 4x^3+x^2-12x-3=0

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

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.

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

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

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

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

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

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

$4 x + 1 = 0$

$x^{2} - 3 = 0$

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 $x$ .

Subtract $1$ from 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 $1$ .

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

Simplify the right side.

Move the negative in front of the fraction.

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

Set $x^{2} - 3$ equal to $0$ and solve for $x$ .

Set $x^{2} - 3$ equal to $0$ .

$x^{2} - 3 = 0$

Solve $x^{2} - 3 = 0$ for $x$ .

Add $3$ to both sides of the equation.

$x^{2} = 3$

Take the square root of both sides of the equation to eliminate the exponent on the left side.

$x = \pm \sqrt{3}$

The complete solution is the result of both the positive and negative portions of the solution.

First, use the positive value of the $\pm$ to find the first solution.

$x = \sqrt{3}$

Next, use the negative value of the $\pm$ to find the second solution.

$x = - \sqrt{3}$

The complete solution is the result of both the positive and negative portions of the solution.

$x = \sqrt{3}, - \sqrt{3}$

The final solution is all the values that make $(4 x + 1) (x^{2} - 3) = 0$ true.

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

The result can be shown in multiple forms.

Exact Form:

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

Decimal Form:

$x = - 0.25, 1.73205080 \dots, - 1.73205080 \dots$

Find the Roots (Zeros) 4x^3+x^2-12x-3=0

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