Find the Roots (Zeros) 3x^3+x^2-38x+24=0

$3 x^{3} + x^{2} - 38 x + 24 = 0$

Step 1

Factor the left side of the equation.

Factor $3 x^{3} + x^{2} - 38 x + 24$ 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.

$p = \pm 1, \pm 24, \pm 2, \pm 12, \pm 3, \pm 8, \pm 4, \pm 6 q = \pm 1, \pm 3$

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

$\pm 1, \pm 0.\overline{3}, \pm 24, \pm 8, \pm 2, \pm 0.\overline{6}, \pm 12, \pm 4, \pm 3, \pm 2.\overline{6}, \pm 1.\overline{3}, \pm 6$

Substitute $0.\overline{6}$ and simplify the expression. In this case, the expression is equal to $0$ so $0.\overline{6}$ is a root of the polynomial.

Substitute $0.\overline{6}$ into the polynomial.

$3 \cdot {0.\overline{6}}^{3} + {0.\overline{6}}^{2} - 38 \cdot 0.\overline{6} + 24$

Raise $0.\overline{6}$ to the power of $3$ .

$3 \cdot 0.\overline{296} + {0.\overline{6}}^{2} - 38 \cdot 0.\overline{6} + 24$

Multiply $3$ by $0.\overline{296}$ .

$0.\overline{8} + {0.\overline{6}}^{2} - 38 \cdot 0.\overline{6} + 24$

Raise $0.\overline{6}$ to the power of $2$ .

$0.\overline{8} + 0.\overline{4} - 38 \cdot 0.\overline{6} + 24$

Add $0.\overline{8}$ and $0.\overline{4}$ .

$1.\overline{3} - 38 \cdot 0.\overline{6} + 24$

Multiply $- 38$ by $0.\overline{6}$ .

$1.\overline{3} - 25.\overline{3} + 24$

Subtract $25.\overline{3}$ from $1.\overline{3}$ .

$- 24 + 24$

Add $- 24$ and $24$ .

$0$

Since $0.\overline{6}$ is a known root, divide the polynomial by $3 x - 2$ to find the quotient polynomial. This polynomial can then be used to find the remaining roots.

$\frac{3 x^{3} + x^{2} - 38 x + 24}{3 x - 2}$

Divide $3 x^{3} + x^{2} - 38 x + 24$ by $3 x - 2$ .

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

$3 x$

$2$

$3 x^{3}$

$x^{2}$

$38 x$

$24$

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

					$x^{2}$
	$3 x$	-	$2$		$3 x^{3}$	+	$x^{2}$	-	$38 x$	+	$24$

Multiply the new quotient term by the divisor.

				$x^{2}$
$3 x$	-	$2$		$3 x^{3}$	+	$x^{2}$	-	$38 x$	+	$24$
			+	$3 x^{3}$	-	$2 x^{2}$

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

				$x^{2}$
$3 x$	-	$2$		$3 x^{3}$	+	$x^{2}$	-	$38 x$	+	$24$
			-	$3 x^{3}$	+	$2 x^{2}$

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

				$x^{2}$
$3 x$	-	$2$		$3 x^{3}$	+	$x^{2}$	-	$38 x$	+	$24$
			-	$3 x^{3}$	+	$2 x^{2}$
					+	$3 x^{2}$

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

				$x^{2}$
$3 x$	-	$2$		$3 x^{3}$	+	$x^{2}$	-	$38 x$	+	$24$
			-	$3 x^{3}$	+	$2 x^{2}$
					+	$3 x^{2}$	-	$38 x$

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

				$x^{2}$	+	$x$
$3 x$	-	$2$		$3 x^{3}$	+	$x^{2}$	-	$38 x$	+	$24$
			-	$3 x^{3}$	+	$2 x^{2}$
					+	$3 x^{2}$	-	$38 x$

Multiply the new quotient term by the divisor.

				$x^{2}$	+	$x$
$3 x$	-	$2$		$3 x^{3}$	+	$x^{2}$	-	$38 x$	+	$24$
			-	$3 x^{3}$	+	$2 x^{2}$
					+	$3 x^{2}$	-	$38 x$
					+	$3 x^{2}$	-	$2 x$

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

				$x^{2}$	+	$x$
$3 x$	-	$2$		$3 x^{3}$	+	$x^{2}$	-	$38 x$	+	$24$
			-	$3 x^{3}$	+	$2 x^{2}$
					+	$3 x^{2}$	-	$38 x$
					-	$3 x^{2}$	+	$2 x$

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

				$x^{2}$	+	$x$
$3 x$	-	$2$		$3 x^{3}$	+	$x^{2}$	-	$38 x$	+	$24$
			-	$3 x^{3}$	+	$2 x^{2}$
					+	$3 x^{2}$	-	$38 x$
					-	$3 x^{2}$	+	$2 x$
							-	$36 x$

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

				$x^{2}$	+	$x$
$3 x$	-	$2$		$3 x^{3}$	+	$x^{2}$	-	$38 x$	+	$24$
			-	$3 x^{3}$	+	$2 x^{2}$
					+	$3 x^{2}$	-	$38 x$
					-	$3 x^{2}$	+	$2 x$
							-	$36 x$	+	$24$

Divide the highest order term in the dividend $- 36 x$ by the highest order term in divisor $3 x$ .

				$x^{2}$	+	$x$	-	$12$
$3 x$	-	$2$		$3 x^{3}$	+	$x^{2}$	-	$38 x$	+	$24$
			-	$3 x^{3}$	+	$2 x^{2}$
					+	$3 x^{2}$	-	$38 x$
					-	$3 x^{2}$	+	$2 x$
							-	$36 x$	+	$24$

Multiply the new quotient term by the divisor.

				$x^{2}$	+	$x$	-	$12$
$3 x$	-	$2$		$3 x^{3}$	+	$x^{2}$	-	$38 x$	+	$24$
			-	$3 x^{3}$	+	$2 x^{2}$
					+	$3 x^{2}$	-	$38 x$
					-	$3 x^{2}$	+	$2 x$
							-	$36 x$	+	$24$
							-	$36 x$	+	$24$

The expression needs to be subtracted from the dividend, so change all the signs in $- 36 x + 24$

				$x^{2}$	+	$x$	-	$12$
$3 x$	-	$2$		$3 x^{3}$	+	$x^{2}$	-	$38 x$	+	$24$
			-	$3 x^{3}$	+	$2 x^{2}$
					+	$3 x^{2}$	-	$38 x$
					-	$3 x^{2}$	+	$2 x$
							-	$36 x$	+	$24$
							+	$36 x$	-	$24$

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

				$x^{2}$	+	$x$	-	$12$
$3 x$	-	$2$		$3 x^{3}$	+	$x^{2}$	-	$38 x$	+	$24$
			-	$3 x^{3}$	+	$2 x^{2}$
					+	$3 x^{2}$	-	$38 x$
					-	$3 x^{2}$	+	$2 x$
							-	$36 x$	+	$24$
							+	$36 x$	-	$24$
										$0$

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

$x^{2} + x - 12$

Write $3 x^{3} + x^{2} - 38 x + 24$ as a set of factors.

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

Factor $x^{2} + x - 12$ using the AC method.

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 12$ and whose sum is $1$ .

$- 3, 4$

Write the factored form using these integers.

$(3 x - 2) ((x - 3) (x + 4)) = 0$

Remove unnecessary parentheses.

$(3 x - 2) (x - 3) (x + 4) = 0$

Step 2

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

$3 x - 2 = 0$

$x - 3 = 0$

$x + 4 = 0$

Step 3

Set $3 x - 2$ equal to $0$ and solve for $x$ .

Set $3 x - 2$ equal to $0$ .

$3 x - 2 = 0$

Solve $3 x - 2 = 0$ for $x$ .

Add $2$ to both sides of the equation.

$3 x = 2$

Divide each term in $3 x = 2$ by $3$ and simplify.

Divide each term in $3 x = 2$ by $3$ .

$\frac{3 x}{3} = \frac{2}{3}$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 x}{3} = \frac{2}{3}$

Divide $x$ by $1$ .

$x = \frac{2}{3}$

Step 4

Set $x - 3$ equal to $0$ and solve for $x$ .

Set $x - 3$ equal to $0$ .

$x - 3 = 0$

Add $3$ to both sides of the equation.

$x = 3$

Step 5

Set $x + 4$ equal to $0$ and solve for $x$ .

Set $x + 4$ equal to $0$ .

$x + 4 = 0$

Subtract $4$ from both sides of the equation.

$x = - 4$

Step 6

The final solution is all the values that make $(3 x - 2) (x - 3) (x + 4) = 0$ true.

$x = \frac{2}{3}, 3, - 4$

Step 7

Find the Roots (Zeros) 3x^3+x^2-38x+24=0

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