Find the Roots (Zeros) 8x^3+12x^2+6x+1=0

$8 x^{3} + 12 x^{2} + 6 x + 1 = 0$

Factor the left side of the equation.

Regroup terms.

$8 x^{3} + 1 + 12 x^{2} + 6 x = 0$

Rewrite $8 x^{3}$ as ${(2 x)}^{3}$ .

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

Rewrite $1$ as $1^{3}$ .

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

Since both terms are perfect cubes, factor using the sum of cubes formula, $a^{3} + b^{3} = (a + b) (a^{2} - a b + b^{2})$ where $a = 2 x$ and $b = 1$ .

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

Simplify.

Apply the product rule to $2 x$ .

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

Raise $2$ to the power of $2$ .

$(2 x + 1) (4 x^{2} - 2 x \cdot 1 + 1^{2}) + 12 x^{2} + 6 x = 0$

Multiply $2$ by $- 1$ .

$(2 x + 1) (4 x^{2} - 2 x \cdot 1 + 1^{2}) + 12 x^{2} + 6 x = 0$

Multiply $- 2$ by $1$ .

$(2 x + 1) (4 x^{2} - 2 x + 1^{2}) + 12 x^{2} + 6 x = 0$

One to any power is one.

$(2 x + 1) (4 x^{2} - 2 x + 1) + 12 x^{2} + 6 x = 0$

Factor $6 x$ out of $12 x^{2} + 6 x$ .

Factor $6 x$ out of $12 x^{2}$ .

$(2 x + 1) (4 x^{2} - 2 x + 1) + 6 x (2 x) + 6 x = 0$

Factor $6 x$ out of $6 x$ .

$(2 x + 1) (4 x^{2} - 2 x + 1) + 6 x (2 x) + 6 x (1) = 0$

Factor $6 x$ out of $6 x (2 x) + 6 x (1)$ .

$(2 x + 1) (4 x^{2} - 2 x + 1) + 6 x (2 x + 1) = 0$

Factor $2 x + 1$ out of $(2 x + 1) (4 x^{2} - 2 x + 1) + 6 x (2 x + 1)$ .

Factor $2 x + 1$ out of $6 x (2 x + 1)$ .

$(2 x + 1) (4 x^{2} - 2 x + 1) + (2 x + 1) (6 x) = 0$

Factor $2 x + 1$ out of $(2 x + 1) (4 x^{2} - 2 x + 1) + (2 x + 1) (6 x)$ .

$(2 x + 1) (4 x^{2} - 2 x + 1 + 6 x) = 0$

Add $- 2 x$ and $6 x$ .

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

Factor using the perfect square rule.

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

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

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

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

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$4 x = 2 \cdot (2 x) \cdot 1$

Rewrite the polynomial.

$(2 x + 1) ({(2 x)}^{2} + 2 \cdot (2 x) \cdot 1 + 1^{2}) = 0$

Factor using the perfect square trinomial rule $a^{2} + 2 a b + b^{2} = {(a + b)}^{2}$ , where $a = 2 x$ and $b = 1$ .

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

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

$2 x + 1 = 0$

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

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.

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

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

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

Find the Roots (Zeros) 8x^3+12x^2+6x+1=0

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