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

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

Step 1

Set $8 x^{3} - 12 x^{2} + 6 x - 1$ equal to $0$ .

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

Step 2

Solve for $x$ .

Factor the left side of the equation.

Factor $8 x^{3} - 12 x^{2} + 6 x - 1$ using the rational roots test.

If a polynomial function has integer coefficients, then every rational zero will have the form $\frac{p}{q}$ where $p$ is a factor of the constant and $q$ is a factor of the leading coefficient.

$p = \pm 1 q = \pm 1, \pm 8, \pm 2, \pm 4$

Find every combination of $\pm \frac{p}{q}$ . These are the possible roots of the polynomial function.

$\pm 1, \pm 0.125, \pm 0.5, \pm 0.25$

Substitute $0.5$ and simplify the expression. In this case, the expression is equal to $0$ so $0.5$ is a root of the polynomial.

Substitute $0.5$ into the polynomial.

$8 \cdot {0.5}^{3} - 12 \cdot {0.5}^{2} + 6 \cdot 0.5 - 1$

Raise $0.5$ to the power of $3$ .

$8 \cdot 0.125 - 12 \cdot {0.5}^{2} + 6 \cdot 0.5 - 1$

Multiply $8$ by $0.125$ .

$1 - 12 \cdot {0.5}^{2} + 6 \cdot 0.5 - 1$

Raise $0.5$ to the power of $2$ .

$1 - 12 \cdot 0.25 + 6 \cdot 0.5 - 1$

Multiply $- 12$ by $0.25$ .

$1 - 3 + 6 \cdot 0.5 - 1$

Subtract $3$ from $1$ .

$- 2 + 6 \cdot 0.5 - 1$

Multiply $6$ by $0.5$ .

$- 2 + 3 - 1$

Add $- 2$ and $3$ .

$1 - 1$

Subtract $1$ from $1$ .

$0$

Since $0.5$ is a known root, divide the polynomial by $2 x - 1$ to find the quotient polynomial. This polynomial can then be used to find the remaining roots.

$\frac{8 x^{3} - 12 x^{2} + 6 x - 1}{2 x - 1}$

Divide $8 x^{3} - 12 x^{2} + 6 x - 1$ by $2 x - 1$ .

Set up the polynomials to be divided. If there is not a term for every exponent, insert one with a value of $0$ .

$2 x$

$1$

$8 x^{3}$

$12 x^{2}$

$6 x$

$1$

Divide the highest order term in the dividend $8 x^{3}$ by the highest order term in divisor $2 x$ .

					$4 x^{2}$
	$2 x$	-	$1$		$8 x^{3}$	-	$12 x^{2}$	+	$6 x$	-	$1$

Multiply the new quotient term by the divisor.

				$4 x^{2}$
$2 x$	-	$1$		$8 x^{3}$	-	$12 x^{2}$	+	$6 x$	-	$1$
			+	$8 x^{3}$	-	$4 x^{2}$

The expression needs to be subtracted from the dividend, so change all the signs in $8 x^{3} - 4 x^{2}$

				$4 x^{2}$
$2 x$	-	$1$		$8 x^{3}$	-	$12 x^{2}$	+	$6 x$	-	$1$
			-	$8 x^{3}$	+	$4 x^{2}$

After changing the signs, add the last dividend from the multiplied polynomial to find the new dividend.

				$4 x^{2}$
$2 x$	-	$1$		$8 x^{3}$	-	$12 x^{2}$	+	$6 x$	-	$1$
			-	$8 x^{3}$	+	$4 x^{2}$
					-	$8 x^{2}$

Pull the next terms from the original dividend down into the current dividend.

				$4 x^{2}$
$2 x$	-	$1$		$8 x^{3}$	-	$12 x^{2}$	+	$6 x$	-	$1$
			-	$8 x^{3}$	+	$4 x^{2}$
					-	$8 x^{2}$	+	$6 x$

Divide the highest order term in the dividend $- 8 x^{2}$ by the highest order term in divisor $2 x$ .

				$4 x^{2}$	-	$4 x$
$2 x$	-	$1$		$8 x^{3}$	-	$12 x^{2}$	+	$6 x$	-	$1$
			-	$8 x^{3}$	+	$4 x^{2}$
					-	$8 x^{2}$	+	$6 x$

Multiply the new quotient term by the divisor.

				$4 x^{2}$	-	$4 x$
$2 x$	-	$1$		$8 x^{3}$	-	$12 x^{2}$	+	$6 x$	-	$1$
			-	$8 x^{3}$	+	$4 x^{2}$
					-	$8 x^{2}$	+	$6 x$
					-	$8 x^{2}$	+	$4 x$

The expression needs to be subtracted from the dividend, so change all the signs in $- 8 x^{2} + 4 x$

				$4 x^{2}$	-	$4 x$
$2 x$	-	$1$		$8 x^{3}$	-	$12 x^{2}$	+	$6 x$	-	$1$
			-	$8 x^{3}$	+	$4 x^{2}$
					-	$8 x^{2}$	+	$6 x$
					+	$8 x^{2}$	-	$4 x$

After changing the signs, add the last dividend from the multiplied polynomial to find the new dividend.

				$4 x^{2}$	-	$4 x$
$2 x$	-	$1$		$8 x^{3}$	-	$12 x^{2}$	+	$6 x$	-	$1$
			-	$8 x^{3}$	+	$4 x^{2}$
					-	$8 x^{2}$	+	$6 x$
					+	$8 x^{2}$	-	$4 x$
							+	$2 x$

Pull the next terms from the original dividend down into the current dividend.

				$4 x^{2}$	-	$4 x$
$2 x$	-	$1$		$8 x^{3}$	-	$12 x^{2}$	+	$6 x$	-	$1$
			-	$8 x^{3}$	+	$4 x^{2}$
					-	$8 x^{2}$	+	$6 x$
					+	$8 x^{2}$	-	$4 x$
							+	$2 x$	-	$1$

Divide the highest order term in the dividend $2 x$ by the highest order term in divisor $2 x$ .

				$4 x^{2}$	-	$4 x$	+	$1$
$2 x$	-	$1$		$8 x^{3}$	-	$12 x^{2}$	+	$6 x$	-	$1$
			-	$8 x^{3}$	+	$4 x^{2}$
					-	$8 x^{2}$	+	$6 x$
					+	$8 x^{2}$	-	$4 x$
							+	$2 x$	-	$1$

Multiply the new quotient term by the divisor.

				$4 x^{2}$	-	$4 x$	+	$1$
$2 x$	-	$1$		$8 x^{3}$	-	$12 x^{2}$	+	$6 x$	-	$1$
			-	$8 x^{3}$	+	$4 x^{2}$
					-	$8 x^{2}$	+	$6 x$
					+	$8 x^{2}$	-	$4 x$
							+	$2 x$	-	$1$
							+	$2 x$	-	$1$

The expression needs to be subtracted from the dividend, so change all the signs in $2 x - 1$

				$4 x^{2}$	-	$4 x$	+	$1$
$2 x$	-	$1$		$8 x^{3}$	-	$12 x^{2}$	+	$6 x$	-	$1$
			-	$8 x^{3}$	+	$4 x^{2}$
					-	$8 x^{2}$	+	$6 x$
					+	$8 x^{2}$	-	$4 x$
							+	$2 x$	-	$1$
							-	$2 x$	+	$1$

After changing the signs, add the last dividend from the multiplied polynomial to find the new dividend.

				$4 x^{2}$	-	$4 x$	+	$1$
$2 x$	-	$1$		$8 x^{3}$	-	$12 x^{2}$	+	$6 x$	-	$1$
			-	$8 x^{3}$	+	$4 x^{2}$
					-	$8 x^{2}$	+	$6 x$
					+	$8 x^{2}$	-	$4 x$
							+	$2 x$	-	$1$
							-	$2 x$	+	$1$
										$0$

Since the remander is $0$ , the final answer is the quotient.

$4 x^{2} - 4 x + 1$

Write $8 x^{3} - 12 x^{2} + 6 x - 1$ as a set of factors.

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

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 $b = 1$ .

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

Combine like factors.

Raise $2 x - 1$ to the power of $1$ .

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

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

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

Add $1$ and $2$ .

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

Set the $2 x - 1$ equal to $0$ .

$2 x - 1 = 0$

Solve 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 $1$ .

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

Step 3

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

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