Find the Roots/Zeros Using the Rational Roots Test 3x^3+2x^2-19x+6

$3 x^{3} + 2 x^{2} - 19 x + 6$

Step 1

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.

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

$q = \pm 1, \pm 3$

Step 2

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

$\pm 1, \pm \frac{1}{3}, \pm 2, \pm \frac{2}{3}, \pm 3, \pm 6$

Step 3

Substitute the possible roots one by one into the polynomial to find the actual roots. Simplify to check if the value is $0$ , which means it is a root.

$3 {(2)}^{3} + 2 {(2)}^{2} - 19 \cdot 2 + 6$

Step 4

Simplify the expression. In this case, the expression is equal to $0$ so $x = 2$ is a root of the polynomial.

Simplify each term.

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

$3 \cdot 8 + 2 {(2)}^{2} - 19 \cdot 2 + 6$

Multiply $3$ by $8$ .

$24 + 2 {(2)}^{2} - 19 \cdot 2 + 6$

Multiply $2$ by ${(2)}^{2}$ by adding the exponents.

Multiply $2$ by ${(2)}^{2}$ .

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

$24 + 2^{1} {(2)}^{2} - 19 \cdot 2 + 6$

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

$24 + 2^{1 + 2} - 19 \cdot 2 + 6$

Add $1$ and $2$ .

$24 + 2^{3} - 19 \cdot 2 + 6$

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

$24 + 8 - 19 \cdot 2 + 6$

Multiply $- 19$ by $2$ .

$24 + 8 - 38 + 6$

Simplify by adding and subtracting.

Add $24$ and $8$ .

$32 - 38 + 6$

Subtract $38$ from $32$ .

$- 6 + 6$

Add $- 6$ and $6$ .

$0$

Step 5

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

$\frac{3 x^{3} + 2 x^{2} - 19 x + 6}{x - 2}$

Step 6

Next, find the roots of the remaining polynomial. The order of the polynomial has been reduced by $1$ .

Place the numbers representing the divisor and the dividend into a division-like configuration.

$2$	$3$	$2$	$- 19$	$6$
	\xa0			\xa0

The first number in the dividend $(3)$ is put into the first position of the result area (below the horizontal line).

$2$	$3$	$2$	$- 19$	$6$
	\xa0			\xa0
	$3$

Multiply the newest entry in the result $(3)$ by the divisor $(2)$ and place the result of $(6)$ under the next term in the dividend $(2)$ .

$2$	$3$	$2$	$- 19$	$6$
	\xa0	$6$		\xa0
	$3$

Add the product of the multiplication and the number from the dividend and put the result in the next position on the result line.

$2$	$3$	$2$	$- 19$	$6$
	\xa0	$6$		\xa0
	$3$	$8$

Multiply the newest entry in the result $(8)$ by the divisor $(2)$ and place the result of $(16)$ under the next term in the dividend $(- 19)$ .

$2$	$3$	$2$	$- 19$	$6$
	\xa0	$6$	$16$	\xa0
	$3$	$8$

Add the product of the multiplication and the number from the dividend and put the result in the next position on the result line.

$2$	$3$	$2$	$- 19$	$6$
	\xa0	$6$	$16$	\xa0
	$3$	$8$	$- 3$

Multiply the newest entry in the result $(- 3)$ by the divisor $(2)$ and place the result of $(- 6)$ under the next term in the dividend $(6)$ .

$2$	$3$	$2$	$- 19$	$6$
	\xa0	$6$	$16$	$- 6$
	$3$	$8$	$- 3$

Add the product of the multiplication and the number from the dividend and put the result in the next position on the result line.

$2$	$3$	$2$	$- 19$	$6$
	\xa0	$6$	$16$	$- 6$
	$3$	$8$	$- 3$	$0$

All numbers except the last become the coefficients of the quotient polynomial. The last value in the result line is the remainder.

$3 x^{2} + 8 x - 3$

Step 7

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 $a \cdot c = 3 \cdot - 3 = - 9$ and whose sum is $b = 8$ .

Factor $8$ out of $8 x$ .

$(x - 2) + 3 x^{2} + 8 (x) - 3$

Rewrite $8$ as $- 1$ plus $9$

$(x - 2) + 3 x^{2} + (- 1 + 9) x - 3$

Apply the distributive property.

$(x - 2) + 3 x^{2} - 1 x + 9 x - 3$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(x - 2) (3 x^{2} - 1 x) + 9 x - 3$

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

$(x - 2) + x (3 x - 1) + 3 (3 x - 1)$

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

$(x - 2) + (3 x - 1) (x + 3)$

$(x - 2) (3 x - 1) (x + 3)$

Step 8

Factor the left side of the equation.

Factor $3 x^{3} + 2 x^{2} - 19 x + 6$ 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.

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

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

$\pm 1, \pm 0.\overline{3}, \pm 6, \pm 2, \pm 0.\overline{6}, \pm 3$

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.

$3 \cdot {0.\overline{3}}^{3} + 2 \cdot {0.\overline{3}}^{2} - 19 \cdot 0.\overline{3} + 6$

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

$3 \cdot 0.\overline{037} + 2 \cdot {0.\overline{3}}^{2} - 19 \cdot 0.\overline{3} + 6$

Multiply $3$ by $0.\overline{037}$ .

$0.\overline{1} + 2 \cdot {0.\overline{3}}^{2} - 19 \cdot 0.\overline{3} + 6$

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

$0.\overline{1} + 2 \cdot 0.\overline{1} - 19 \cdot 0.\overline{3} + 6$

Multiply $2$ by $0.\overline{1}$ .

$0.\overline{1} + 0.\overline{2} - 19 \cdot 0.\overline{3} + 6$

Add $0.\overline{1}$ and $0.\overline{2}$ .

$0.\overline{3} - 19 \cdot 0.\overline{3} + 6$

Multiply $- 19$ by $0.\overline{3}$ .

$0.\overline{3} - 6.33333333 + 6$

Subtract $6.33333333$ from $0.\overline{3}$ .

$- 6 + 6$

Add $- 6$ and $6$ .

$0$

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.

$\frac{3 x^{3} + 2 x^{2} - 19 x + 6}{3 x - 1}$

Divide $3 x^{3} + 2 x^{2} - 19 x + 6$ 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 $0$ .

$3 x$

$1$

$3 x^{3}$

$2 x^{2}$

$19 x$

$6$

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

					$x^{2}$
	$3 x$	-	$1$		$3 x^{3}$	+	$2 x^{2}$	-	$19 x$	+	$6$

Multiply the new quotient term by the divisor.

				$x^{2}$
$3 x$	-	$1$		$3 x^{3}$	+	$2 x^{2}$	-	$19 x$	+	$6$
			+	$3 x^{3}$	-	$x^{2}$

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

				$x^{2}$
$3 x$	-	$1$		$3 x^{3}$	+	$2 x^{2}$	-	$19 x$	+	$6$
			-	$3 x^{3}$	+	$x^{2}$

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

				$x^{2}$
$3 x$	-	$1$		$3 x^{3}$	+	$2 x^{2}$	-	$19 x$	+	$6$
			-	$3 x^{3}$	+	$x^{2}$
					+	$3 x^{2}$

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

				$x^{2}$
$3 x$	-	$1$		$3 x^{3}$	+	$2 x^{2}$	-	$19 x$	+	$6$
			-	$3 x^{3}$	+	$x^{2}$
					+	$3 x^{2}$	-	$19 x$

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

				$x^{2}$	+	$x$
$3 x$	-	$1$		$3 x^{3}$	+	$2 x^{2}$	-	$19 x$	+	$6$
			-	$3 x^{3}$	+	$x^{2}$
					+	$3 x^{2}$	-	$19 x$

Multiply the new quotient term by the divisor.

				$x^{2}$	+	$x$
$3 x$	-	$1$		$3 x^{3}$	+	$2 x^{2}$	-	$19 x$	+	$6$
			-	$3 x^{3}$	+	$x^{2}$
					+	$3 x^{2}$	-	$19 x$
					+	$3 x^{2}$	-	$x$

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

				$x^{2}$	+	$x$
$3 x$	-	$1$		$3 x^{3}$	+	$2 x^{2}$	-	$19 x$	+	$6$
			-	$3 x^{3}$	+	$x^{2}$
					+	$3 x^{2}$	-	$19 x$
					-	$3 x^{2}$	+	$x$

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

				$x^{2}$	+	$x$
$3 x$	-	$1$		$3 x^{3}$	+	$2 x^{2}$	-	$19 x$	+	$6$
			-	$3 x^{3}$	+	$x^{2}$
					+	$3 x^{2}$	-	$19 x$
					-	$3 x^{2}$	+	$x$
							-	$18 x$

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

				$x^{2}$	+	$x$
$3 x$	-	$1$		$3 x^{3}$	+	$2 x^{2}$	-	$19 x$	+	$6$
			-	$3 x^{3}$	+	$x^{2}$
					+	$3 x^{2}$	-	$19 x$
					-	$3 x^{2}$	+	$x$
							-	$18 x$	+	$6$

Divide the highest order term in the dividend $- 18 x$ by the highest order term in divisor $3 x$ .

				$x^{2}$	+	$x$	-	$6$
$3 x$	-	$1$		$3 x^{3}$	+	$2 x^{2}$	-	$19 x$	+	$6$
			-	$3 x^{3}$	+	$x^{2}$
					+	$3 x^{2}$	-	$19 x$
					-	$3 x^{2}$	+	$x$
							-	$18 x$	+	$6$

Multiply the new quotient term by the divisor.

				$x^{2}$	+	$x$	-	$6$
$3 x$	-	$1$		$3 x^{3}$	+	$2 x^{2}$	-	$19 x$	+	$6$
			-	$3 x^{3}$	+	$x^{2}$
					+	$3 x^{2}$	-	$19 x$
					-	$3 x^{2}$	+	$x$
							-	$18 x$	+	$6$
							-	$18 x$	+	$6$

The expression needs to be subtracted from the dividend, so change all the signs in $- 18 x + 6$

				$x^{2}$	+	$x$	-	$6$
$3 x$	-	$1$		$3 x^{3}$	+	$2 x^{2}$	-	$19 x$	+	$6$
			-	$3 x^{3}$	+	$x^{2}$
					+	$3 x^{2}$	-	$19 x$
					-	$3 x^{2}$	+	$x$
							-	$18 x$	+	$6$
							+	$18 x$	-	$6$

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

				$x^{2}$	+	$x$	-	$6$
$3 x$	-	$1$		$3 x^{3}$	+	$2 x^{2}$	-	$19 x$	+	$6$
			-	$3 x^{3}$	+	$x^{2}$
					+	$3 x^{2}$	-	$19 x$
					-	$3 x^{2}$	+	$x$
							-	$18 x$	+	$6$
							+	$18 x$	-	$6$
										$0$

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

$x^{2} + x - 6$

Write $3 x^{3} + 2 x^{2} - 19 x + 6$ as a set of factors.

$(3 x - 1) (x^{2} + x - 6) = 0$

Factor $x^{2} + x - 6$ 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 $- 6$ and whose sum is $1$ .

$- 2, 3$

Write the factored form using these integers.

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

Remove unnecessary parentheses.

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

Step 9

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

$3 x - 1 = 0$

$x - 2 = 0$

$x + 3 = 0$

Step 10

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 $1$ .

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

Step 11

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

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

$x - 2 = 0$

Add $2$ to both sides of the equation.

$x = 2$

Step 12

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

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

$x + 3 = 0$

Subtract $3$ from both sides of the equation.

$x = - 3$

Step 13

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

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

Step 14

Find the Roots/Zeros Using the Rational Roots Test 3x^3+2x^2-19x+6

Related Questions

Support LCMGCF.com by disabling your adblocker.