Find the Roots (Zeros) 6x^3-7x^2-6x-1=0

$6 x^{3} - 7 x^{2} - 6 x - 1 = 0$

Step 1

Factor $6 x^{3} - 7 x^{2} - 6 x - 1$ 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.

$p = \pm 1 q = \pm 1, \pm 6, \pm 2, \pm 3$

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

$\pm 1, \pm 0.1\overline{6}, \pm 0.5, \pm 0.\overline{3}$

Substitute $- 0.\overline{3}$ and simplify the expression. In this case, the expression is equal to $0$ so $- 0.\overline{3}$ is a root of the polynomial.

Substitute $- 0.\overline{3}$ into the polynomial.

$6 {(- 0.\overline{3})}^{3} - 7 {(- 0.\overline{3})}^{2} - 6 \cdot - 0.\overline{3} - 1$

Raise $- 0.\overline{3}$ to the power of $3$ .

$6 \cdot - 0.\overline{037} - 7 {(- 0.\overline{3})}^{2} - 6 \cdot - 0.\overline{3} - 1$

Multiply $6$ by $- 0.\overline{037}$ .

$- 0.\overline{2} - 7 {(- 0.\overline{3})}^{2} - 6 \cdot - 0.\overline{3} - 1$

Raise $- 0.\overline{3}$ to the power of $2$ .

$- 0.\overline{2} - 7 \cdot 0.\overline{1} - 6 \cdot - 0.\overline{3} - 1$

Multiply $- 7$ by $0.\overline{1}$ .

$- 0.\overline{2} - 0.\overline{7} - 6 \cdot - 0.\overline{3} - 1$

Subtract $0.\overline{7}$ from $- 0.\overline{2}$ .

$- 1 - 6 \cdot - 0.\overline{3} - 1$

Multiply $- 6$ by $- 0.\overline{3}$ .

$- 1 + 2 - 1$

Add $- 1$ and $2$ .

$1 - 1$

Subtract $1$ from $1$ .

$0$

Since $- 0.\overline{3}$ is a known root, divide the polynomial by $3 x + 1$ to find the quotient polynomial. This polynomial can then be used to find the remaining roots.

$\frac{6 x^{3} - 7 x^{2} - 6 x - 1}{3 x + 1}$

Divide $6 x^{3} - 7 x^{2} - 6 x - 1$ by $3 x + 1$ .

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

$3 x$

$1$

$6 x^{3}$

$7 x^{2}$

$6 x$

$1$

Divide the highest order term in the dividend $6 x^{3}$ by the highest order term in divisor $3 x$ .

					$2 x^{2}$
	$3 x$	+	$1$		$6 x^{3}$	-	$7 x^{2}$	-	$6 x$	-	$1$

Multiply the new quotient term by the divisor.

				$2 x^{2}$
$3 x$	+	$1$		$6 x^{3}$	-	$7 x^{2}$	-	$6 x$	-	$1$
			+	$6 x^{3}$	+	$2 x^{2}$

The expression needs to be subtracted from the dividend, so change all the signs in $6 x^{3} + 2 x^{2}$

				$2 x^{2}$
$3 x$	+	$1$		$6 x^{3}$	-	$7 x^{2}$	-	$6 x$	-	$1$
			-	$6 x^{3}$	-	$2 x^{2}$

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

				$2 x^{2}$
$3 x$	+	$1$		$6 x^{3}$	-	$7 x^{2}$	-	$6 x$	-	$1$
			-	$6 x^{3}$	-	$2 x^{2}$
					-	$9 x^{2}$

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

				$2 x^{2}$
$3 x$	+	$1$		$6 x^{3}$	-	$7 x^{2}$	-	$6 x$	-	$1$
			-	$6 x^{3}$	-	$2 x^{2}$
					-	$9 x^{2}$	-	$6 x$

Divide the highest order term in the dividend $- 9 x^{2}$ by the highest order term in divisor $3 x$ .

				$2 x^{2}$	-	$3 x$
$3 x$	+	$1$		$6 x^{3}$	-	$7 x^{2}$	-	$6 x$	-	$1$
			-	$6 x^{3}$	-	$2 x^{2}$
					-	$9 x^{2}$	-	$6 x$

Multiply the new quotient term by the divisor.

				$2 x^{2}$	-	$3 x$
$3 x$	+	$1$		$6 x^{3}$	-	$7 x^{2}$	-	$6 x$	-	$1$
			-	$6 x^{3}$	-	$2 x^{2}$
					-	$9 x^{2}$	-	$6 x$
					-	$9 x^{2}$	-	$3 x$

The expression needs to be subtracted from the dividend, so change all the signs in $- 9 x^{2} - 3 x$

				$2 x^{2}$	-	$3 x$
$3 x$	+	$1$		$6 x^{3}$	-	$7 x^{2}$	-	$6 x$	-	$1$
			-	$6 x^{3}$	-	$2 x^{2}$
					-	$9 x^{2}$	-	$6 x$
					+	$9 x^{2}$	+	$3 x$

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

				$2 x^{2}$	-	$3 x$
$3 x$	+	$1$		$6 x^{3}$	-	$7 x^{2}$	-	$6 x$	-	$1$
			-	$6 x^{3}$	-	$2 x^{2}$
					-	$9 x^{2}$	-	$6 x$
					+	$9 x^{2}$	+	$3 x$
							-	$3 x$

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

				$2 x^{2}$	-	$3 x$
$3 x$	+	$1$		$6 x^{3}$	-	$7 x^{2}$	-	$6 x$	-	$1$
			-	$6 x^{3}$	-	$2 x^{2}$
					-	$9 x^{2}$	-	$6 x$
					+	$9 x^{2}$	+	$3 x$
							-	$3 x$	-	$1$

Divide the highest order term in the dividend $- 3 x$ by the highest order term in divisor $3 x$ .

				$2 x^{2}$	-	$3 x$	-	$1$
$3 x$	+	$1$		$6 x^{3}$	-	$7 x^{2}$	-	$6 x$	-	$1$
			-	$6 x^{3}$	-	$2 x^{2}$
					-	$9 x^{2}$	-	$6 x$
					+	$9 x^{2}$	+	$3 x$
							-	$3 x$	-	$1$

Multiply the new quotient term by the divisor.

				$2 x^{2}$	-	$3 x$	-	$1$
$3 x$	+	$1$		$6 x^{3}$	-	$7 x^{2}$	-	$6 x$	-	$1$
			-	$6 x^{3}$	-	$2 x^{2}$
					-	$9 x^{2}$	-	$6 x$
					+	$9 x^{2}$	+	$3 x$
							-	$3 x$	-	$1$
							-	$3 x$	-	$1$

The expression needs to be subtracted from the dividend, so change all the signs in $- 3 x - 1$

				$2 x^{2}$	-	$3 x$	-	$1$
$3 x$	+	$1$		$6 x^{3}$	-	$7 x^{2}$	-	$6 x$	-	$1$
			-	$6 x^{3}$	-	$2 x^{2}$
					-	$9 x^{2}$	-	$6 x$
					+	$9 x^{2}$	+	$3 x$
							-	$3 x$	-	$1$
							+	$3 x$	+	$1$

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

				$2 x^{2}$	-	$3 x$	-	$1$
$3 x$	+	$1$		$6 x^{3}$	-	$7 x^{2}$	-	$6 x$	-	$1$
			-	$6 x^{3}$	-	$2 x^{2}$
					-	$9 x^{2}$	-	$6 x$
					+	$9 x^{2}$	+	$3 x$
							-	$3 x$	-	$1$
							+	$3 x$	+	$1$
										$0$

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

$2 x^{2} - 3 x - 1$

Write $6 x^{3} - 7 x^{2} - 6 x - 1$ as a set of factors.

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

Step 2

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

$3 x + 1 = 0$

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

Step 3

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 $x$ .

Subtract $1$ from 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 $1$ .

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

Simplify the right side.

Move the negative in front of the fraction.

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

Step 4

Set $2 x^{2} - 3 x - 1$ equal to $0$ and solve for $x$ .

Set $2 x^{2} - 3 x - 1$ equal to $0$ .

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

Solve $2 x^{2} - 3 x - 1 = 0$ for $x$ .

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Substitute the values $a = 2$ , $b = - 3$ , and $c = - 1$ into the quadratic formula and solve for $x$ .

$\frac{3 \pm \sqrt{{(- 3)}^{2} - 4 \cdot (2 \cdot - 1)}}{2 \cdot 2}$

Simplify.

Simplify the numerator.

Raise $- 3$ to the power of $2$ .

$x = \frac{3 \pm \sqrt{9 - 4 \cdot 2 \cdot - 1}}{2 \cdot 2}$

Multiply $- 4 \cdot 2 \cdot - 1$ .

Multiply $- 4$ by $2$ .

$x = \frac{3 \pm \sqrt{9 - 8 \cdot - 1}}{2 \cdot 2}$

Multiply $- 8$ by $- 1$ .

$x = \frac{3 \pm \sqrt{9 + 8}}{2 \cdot 2}$

Add $9$ and $8$ .

$x = \frac{3 \pm \sqrt{17}}{2 \cdot 2}$

Multiply $2$ by $2$ .

$x = \frac{3 \pm \sqrt{17}}{4}$

The final answer is the combination of both solutions.

$x = \frac{3 + \sqrt{17}}{4}, \frac{3 - \sqrt{17}}{4}$

Step 5

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

$x = - \frac{1}{3}, \frac{3 + \sqrt{17}}{4}, \frac{3 - \sqrt{17}}{4}$

Step 6

The result can be shown in multiple forms.

Exact Form:

$x = - \frac{1}{3}, \frac{3 + \sqrt{17}}{4}, \frac{3 - \sqrt{17}}{4}$

Decimal Form:

$x = - 0.\overline{3}, 1.78077640 \dots, - 0.28077640 \dots$

Step 7

Find the Roots (Zeros) 6x^3-7x^2-6x-1=0

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