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

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

Step 1

Set $4 x^{3} + 12 x^{2} - x - 3$ equal to $0$ .

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

Step 2

Solve for $x$ .

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.

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

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

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

Factor the polynomial by factoring out the greatest common factor, $x + 3$ .

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

Rewrite $4 x^{2}$ as ${(2 x)}^{2}$ .

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

Rewrite $1$ as $1^{2}$ .

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

Factor.

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = 2 x$ and $b = 1$ .

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

Remove unnecessary parentheses.

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

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

$x + 3 = 0$

$2 x + 1 = 0$

$2 x - 1 = 0$

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.

$x = - 3$

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 $2 x - 1$ 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 $1$ .

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

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

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

Step 3

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

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