Find the Roots (Zeros) f(x)=6x^3-29x^2-6x+5

$f (x) = 6 x^{3} - 29 x^{2} - 6 x + 5$

Step 1

Set $6 x^{3} - 29 x^{2} - 6 x + 5$ equal to $0$ .

$6 x^{3} - 29 x^{2} - 6 x + 5 = 0$

Step 2

Solve for $x$ .

Factor the left side of the equation.

Factor $6 x^{3} - 29 x^{2} - 6 x + 5$ 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 $- 0.5$ 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.

$6 {(- 0.5)}^{3} - 29 {(- 0.5)}^{2} - 6 \cdot - 0.5 + 5$

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

$6 \cdot - 0.125 - 29 {(- 0.5)}^{2} - 6 \cdot - 0.5 + 5$

Multiply $6$ by $- 0.125$ .

$- 0.75 - 29 {(- 0.5)}^{2} - 6 \cdot - 0.5 + 5$

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

$- 0.75 - 29 \cdot 0.25 - 6 \cdot - 0.5 + 5$

Multiply $- 29$ by $0.25$ .

$- 0.75 - 7.25 - 6 \cdot - 0.5 + 5$

Subtract $7.25$ from $- 0.75$ .

$- 8 - 6 \cdot - 0.5 + 5$

Multiply $- 6$ by $- 0.5$ .

$- 8 + 3 + 5$

Add $- 8$ and $3$ .

$- 5 + 5$

Add $- 5$ and $5$ .

$0$

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.

$\frac{6 x^{3} - 29 x^{2} - 6 x + 5}{2 x + 1}$

Divide $6 x^{3} - 29 x^{2} - 6 x + 5$ 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 $0$ .

$2 x$

$1$

$6 x^{3}$

$29 x^{2}$

$6 x$

$5$

Divide the highest order term in the dividend $6 x^{3}$ by the highest order term in divisor $2 x$ .

					$3 x^{2}$
	$2 x$	+	$1$		$6 x^{3}$	-	$29 x^{2}$	-	$6 x$	+	$5$

Multiply the new quotient term by the divisor.

				$3 x^{2}$
$2 x$	+	$1$		$6 x^{3}$	-	$29 x^{2}$	-	$6 x$	+	$5$
			+	$6 x^{3}$	+	$3 x^{2}$

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

				$3 x^{2}$
$2 x$	+	$1$		$6 x^{3}$	-	$29 x^{2}$	-	$6 x$	+	$5$
			-	$6 x^{3}$	-	$3 x^{2}$

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

				$3 x^{2}$
$2 x$	+	$1$		$6 x^{3}$	-	$29 x^{2}$	-	$6 x$	+	$5$
			-	$6 x^{3}$	-	$3 x^{2}$
					-	$32 x^{2}$

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

				$3 x^{2}$
$2 x$	+	$1$		$6 x^{3}$	-	$29 x^{2}$	-	$6 x$	+	$5$
			-	$6 x^{3}$	-	$3 x^{2}$
					-	$32 x^{2}$	-	$6 x$

Divide the highest order term in the dividend $- 32 x^{2}$ by the highest order term in divisor $2 x$ .

				$3 x^{2}$	-	$16 x$
$2 x$	+	$1$		$6 x^{3}$	-	$29 x^{2}$	-	$6 x$	+	$5$
			-	$6 x^{3}$	-	$3 x^{2}$
					-	$32 x^{2}$	-	$6 x$

Multiply the new quotient term by the divisor.

				$3 x^{2}$	-	$16 x$
$2 x$	+	$1$		$6 x^{3}$	-	$29 x^{2}$	-	$6 x$	+	$5$
			-	$6 x^{3}$	-	$3 x^{2}$
					-	$32 x^{2}$	-	$6 x$
					-	$32 x^{2}$	-	$16 x$

The expression needs to be subtracted from the dividend, so change all the signs in $- 32 x^{2} - 16 x$

				$3 x^{2}$	-	$16 x$
$2 x$	+	$1$		$6 x^{3}$	-	$29 x^{2}$	-	$6 x$	+	$5$
			-	$6 x^{3}$	-	$3 x^{2}$
					-	$32 x^{2}$	-	$6 x$
					+	$32 x^{2}$	+	$16 x$

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

				$3 x^{2}$	-	$16 x$
$2 x$	+	$1$		$6 x^{3}$	-	$29 x^{2}$	-	$6 x$	+	$5$
			-	$6 x^{3}$	-	$3 x^{2}$
					-	$32 x^{2}$	-	$6 x$
					+	$32 x^{2}$	+	$16 x$
							+	$10 x$

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

				$3 x^{2}$	-	$16 x$
$2 x$	+	$1$		$6 x^{3}$	-	$29 x^{2}$	-	$6 x$	+	$5$
			-	$6 x^{3}$	-	$3 x^{2}$
					-	$32 x^{2}$	-	$6 x$
					+	$32 x^{2}$	+	$16 x$
							+	$10 x$	+	$5$

Divide the highest order term in the dividend $10 x$ by the highest order term in divisor $2 x$ .

				$3 x^{2}$	-	$16 x$	+	$5$
$2 x$	+	$1$		$6 x^{3}$	-	$29 x^{2}$	-	$6 x$	+	$5$
			-	$6 x^{3}$	-	$3 x^{2}$
					-	$32 x^{2}$	-	$6 x$
					+	$32 x^{2}$	+	$16 x$
							+	$10 x$	+	$5$

Multiply the new quotient term by the divisor.

				$3 x^{2}$	-	$16 x$	+	$5$
$2 x$	+	$1$		$6 x^{3}$	-	$29 x^{2}$	-	$6 x$	+	$5$
			-	$6 x^{3}$	-	$3 x^{2}$
					-	$32 x^{2}$	-	$6 x$
					+	$32 x^{2}$	+	$16 x$
							+	$10 x$	+	$5$
							+	$10 x$	+	$5$

The expression needs to be subtracted from the dividend, so change all the signs in $10 x + 5$

				$3 x^{2}$	-	$16 x$	+	$5$
$2 x$	+	$1$		$6 x^{3}$	-	$29 x^{2}$	-	$6 x$	+	$5$
			-	$6 x^{3}$	-	$3 x^{2}$
					-	$32 x^{2}$	-	$6 x$
					+	$32 x^{2}$	+	$16 x$
							+	$10 x$	+	$5$
							-	$10 x$	-	$5$

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

				$3 x^{2}$	-	$16 x$	+	$5$
$2 x$	+	$1$		$6 x^{3}$	-	$29 x^{2}$	-	$6 x$	+	$5$
			-	$6 x^{3}$	-	$3 x^{2}$
					-	$32 x^{2}$	-	$6 x$
					+	$32 x^{2}$	+	$16 x$
							+	$10 x$	+	$5$
							-	$10 x$	-	$5$
										$0$

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

$3 x^{2} - 16 x + 5$

Write $6 x^{3} - 29 x^{2} - 6 x + 5$ as a set of factors.

$(2 x + 1) (3 x^{2} - 16 x + 5) = 0$

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 $a \cdot c = 3 \cdot 5 = 15$ and whose sum is $b = - 16$ .

Factor $- 16$ out of $- 16 x$ .

$(2 x + 1) (3 x^{2} - 16 x + 5) = 0$

Rewrite $- 16$ as $- 1$ plus $- 15$

$(2 x + 1) (3 x^{2} + (- 1 - 15) x + 5) = 0$

Apply the distributive property.

$(2 x + 1) (3 x^{2} - 1 x - 15 x + 5) = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(2 x + 1) ((3 x^{2} - 1 x) - 15 x + 5) = 0$

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

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

Factor the polynomial by factoring out the greatest common factor, $3 x - 1$ .

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

Remove unnecessary parentheses.

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

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

$2 x + 1 = 0$

$3 x - 1 = 0$

$x - 5 = 0$

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.

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

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 $x$ .

Add $1$ to 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 $1$ .

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

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

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

Step 3

Find the Roots (Zeros) f(x)=6x^3-29x^2-6x+5

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