Find the Roots (Zeros) f(x)=6x^3-31x^2+4x+5

$f (x) = 6 x^{3} - 31 x^{2} + 4 x + 5$

Set $6 x^{3} - 31 x^{2} + 4 x + 5$ equal to $0$ .

$6 x^{3} - 31 x^{2} + 4 x + 5 = 0$

Solve for $x$ .

Factor the left side of the equation.

Factor $6 x^{3} - 31 x^{2} + 4 x + 5$ 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.

$p = \pm 1, \pm 5 q = \pm 1, \pm 6, \pm 2, \pm 3$

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

$\pm 1, \pm 0.1\overline{6}, \pm 0.5, \pm 0.\overline{3}, \pm 5, \pm 0.8\overline{3}, \pm 2.5, \pm 1.\overline{6}$

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.

$6 {(- 0.\overline{3})}^{3} - 31 {(- 0.\overline{3})}^{2} + 4 \cdot - 0.\overline{3} + 5$

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

$6 \cdot - 0.\overline{037} - 31 {(- 0.\overline{3})}^{2} + 4 \cdot - 0.\overline{3} + 5$

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

$- 0.\overline{2} - 31 {(- 0.\overline{3})}^{2} + 4 \cdot - 0.\overline{3} + 5$

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

$- 0.\overline{2} - 31 \cdot 0.\overline{1} + 4 \cdot - 0.\overline{3} + 5$

Multiply $- 31$ by $0.\overline{1}$ .

$- 0.\overline{2} - 3.\overline{4} + 4 \cdot - 0.\overline{3} + 5$

Subtract $3.\overline{4}$ from $- 0.\overline{2}$ .

$- 3.\overline{6} + 4 \cdot - 0.\overline{3} + 5$

Multiply $4$ by $- 0.\overline{3}$ .

$- 3.\overline{6} - 1.\overline{3} + 5$

Subtract $1.\overline{3}$ from $- 3.\overline{6}$ .

$- 5 + 5$

Add $- 5$ and $5$ .

$0$

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.

$\frac{6 x^{3} - 31 x^{2} + 4 x + 5}{3 x + 1}$

Divide $6 x^{3} - 31 x^{2} + 4 x + 5$ 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 $0$ .

$3 x$

$1$

$6 x^{3}$

$31 x^{2}$

$4 x$

$5$

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

					$2 x^{2}$
	$3 x$	+	$1$		$6 x^{3}$	-	$31 x^{2}$	+	$4 x$	+	$5$

Multiply the new quotient term by the divisor.

				$2 x^{2}$
$3 x$	+	$1$		$6 x^{3}$	-	$31 x^{2}$	+	$4 x$	+	$5$
			+	$6 x^{3}$	+	$2 x^{2}$

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

				$2 x^{2}$
$3 x$	+	$1$		$6 x^{3}$	-	$31 x^{2}$	+	$4 x$	+	$5$
			-	$6 x^{3}$	-	$2 x^{2}$

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

				$2 x^{2}$
$3 x$	+	$1$		$6 x^{3}$	-	$31 x^{2}$	+	$4 x$	+	$5$
			-	$6 x^{3}$	-	$2 x^{2}$
					-	$33 x^{2}$

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

				$2 x^{2}$
$3 x$	+	$1$		$6 x^{3}$	-	$31 x^{2}$	+	$4 x$	+	$5$
			-	$6 x^{3}$	-	$2 x^{2}$
					-	$33 x^{2}$	+	$4 x$

Divide the highest order term in the dividend $- 33 x^{2}$ by the highest order term in divisor $3 x$ .

				$2 x^{2}$	-	$11 x$
$3 x$	+	$1$		$6 x^{3}$	-	$31 x^{2}$	+	$4 x$	+	$5$
			-	$6 x^{3}$	-	$2 x^{2}$
					-	$33 x^{2}$	+	$4 x$

Multiply the new quotient term by the divisor.

				$2 x^{2}$	-	$11 x$
$3 x$	+	$1$		$6 x^{3}$	-	$31 x^{2}$	+	$4 x$	+	$5$
			-	$6 x^{3}$	-	$2 x^{2}$
					-	$33 x^{2}$	+	$4 x$
					-	$33 x^{2}$	-	$11 x$

The expression needs to be subtracted from the dividend, so change all the signs in $- 33 x^{2} - 11 x$

				$2 x^{2}$	-	$11 x$
$3 x$	+	$1$		$6 x^{3}$	-	$31 x^{2}$	+	$4 x$	+	$5$
			-	$6 x^{3}$	-	$2 x^{2}$
					-	$33 x^{2}$	+	$4 x$
					+	$33 x^{2}$	+	$11 x$

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

				$2 x^{2}$	-	$11 x$
$3 x$	+	$1$		$6 x^{3}$	-	$31 x^{2}$	+	$4 x$	+	$5$
			-	$6 x^{3}$	-	$2 x^{2}$
					-	$33 x^{2}$	+	$4 x$
					+	$33 x^{2}$	+	$11 x$
							+	$15 x$

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

				$2 x^{2}$	-	$11 x$
$3 x$	+	$1$		$6 x^{3}$	-	$31 x^{2}$	+	$4 x$	+	$5$
			-	$6 x^{3}$	-	$2 x^{2}$
					-	$33 x^{2}$	+	$4 x$
					+	$33 x^{2}$	+	$11 x$
							+	$15 x$	+	$5$

Divide the highest order term in the dividend $15 x$ by the highest order term in divisor $3 x$ .

				$2 x^{2}$	-	$11 x$	+	$5$
$3 x$	+	$1$		$6 x^{3}$	-	$31 x^{2}$	+	$4 x$	+	$5$
			-	$6 x^{3}$	-	$2 x^{2}$
					-	$33 x^{2}$	+	$4 x$
					+	$33 x^{2}$	+	$11 x$
							+	$15 x$	+	$5$

Multiply the new quotient term by the divisor.

				$2 x^{2}$	-	$11 x$	+	$5$
$3 x$	+	$1$		$6 x^{3}$	-	$31 x^{2}$	+	$4 x$	+	$5$
			-	$6 x^{3}$	-	$2 x^{2}$
					-	$33 x^{2}$	+	$4 x$
					+	$33 x^{2}$	+	$11 x$
							+	$15 x$	+	$5$
							+	$15 x$	+	$5$

The expression needs to be subtracted from the dividend, so change all the signs in $15 x + 5$

				$2 x^{2}$	-	$11 x$	+	$5$
$3 x$	+	$1$		$6 x^{3}$	-	$31 x^{2}$	+	$4 x$	+	$5$
			-	$6 x^{3}$	-	$2 x^{2}$
					-	$33 x^{2}$	+	$4 x$
					+	$33 x^{2}$	+	$11 x$
							+	$15 x$	+	$5$
							-	$15 x$	-	$5$

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

				$2 x^{2}$	-	$11 x$	+	$5$
$3 x$	+	$1$		$6 x^{3}$	-	$31 x^{2}$	+	$4 x$	+	$5$
			-	$6 x^{3}$	-	$2 x^{2}$
					-	$33 x^{2}$	+	$4 x$
					+	$33 x^{2}$	+	$11 x$
							+	$15 x$	+	$5$
							-	$15 x$	-	$5$
										$0$

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

$2 x^{2} - 11 x + 5$

Write $6 x^{3} - 31 x^{2} + 4 x + 5$ as a set of factors.

$(3 x + 1) (2 x^{2} - 11 x + 5) = 0$

Factor by grouping.

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = 2 \cdot 5 = 10$ and whose sum is $b = - 11$ .

Factor $- 11$ out of $- 11 x$ .

$(3 x + 1) (2 x^{2} - 11 x + 5) = 0$

Rewrite $- 11$ as $- 1$ plus $- 10$

$(3 x + 1) (2 x^{2} + (- 1 - 10) x + 5) = 0$

Apply the distributive property.

$(3 x + 1) (2 x^{2} - 1 x - 10 x + 5) = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(3 x + 1) ((2 x^{2} - 1 x) - 10 x + 5) = 0$

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

$(3 x + 1) (x (2 x - 1) - 5 (2 x - 1)) = 0$

Factor the polynomial by factoring out the greatest common factor, $2 x - 1$ .

$(3 x + 1) ((2 x - 1) (x - 5)) = 0$

Remove unnecessary parentheses.

$(3 x + 1) (2 x - 1) (x - 5) = 0$

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

$3 x + 1 = 0$

$2 x - 1 = 0$

$x - 5 = 0$

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.

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

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 $1$ .

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

Set $x - 5$ equal to $0$ and solve for $x$ .

Set $x - 5$ equal to $0$ .

$x - 5 = 0$

Add $5$ to both sides of the equation.

$x = 5$

The final solution is all the values that make $(3 x + 1) (2 x - 1) (x - 5) = 0$ true.

$x = - \frac{1}{3}, \frac{1}{2}, 5$

Find the Roots (Zeros) f(x)=6x^3-31x^2+4x+5

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