Find the Roots/Zeros Using the Rational Roots Test x^3+6x^2+11x+6

$x^{3} + 6 x^{2} + 11 x + 6$

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.

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

$q = \pm 1$

Step 2

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

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

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.

${(- 1)}^{3} + 6 {(- 1)}^{2} + 11 (- 1) + 6$

Step 4

Simplify the expression. In this case, the expression is equal to $0$ so $x = - 1$ is a root of the polynomial.

Simplify each term.

Raise $- 1$ to the power of $3$ .

$- 1 + 6 {(- 1)}^{2} + 11 (- 1) + 6$

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

$- 1 + 6 \cdot 1 + 11 (- 1) + 6$

Multiply $6$ by $1$ .

$- 1 + 6 + 11 (- 1) + 6$

Multiply $11$ by $- 1$ .

$- 1 + 6 - 11 + 6$

Simplify by adding and subtracting.

Add $- 1$ and $6$ .

$5 - 11 + 6$

Subtract $11$ from $5$ .

$- 6 + 6$

Add $- 6$ and $6$ .

$0$

Step 5

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

$\frac{x^{3} + 6 x^{2} + 11 x + 6}{x + 1}$

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.

$- 1$	$1$	$6$	$11$	$6$
	\xa0			\xa0

The first number in the dividend $(1)$ is put into the first position of the result area (below the horizontal line).

$- 1$	$1$	$6$	$11$	$6$
	\xa0			\xa0
	$1$

Multiply the newest entry in the result $(1)$ by the divisor $(- 1)$ and place the result of $(- 1)$ under the next term in the dividend $(6)$ .

$- 1$	$1$	$6$	$11$	$6$
	\xa0	$- 1$		\xa0
	$1$

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

$- 1$	$1$	$6$	$11$	$6$
	\xa0	$- 1$		\xa0
	$1$	$5$

Multiply the newest entry in the result $(5)$ by the divisor $(- 1)$ and place the result of $(- 5)$ under the next term in the dividend $(11)$ .

$- 1$	$1$	$6$	$11$	$6$
	\xa0	$- 1$	$- 5$	\xa0
	$1$	$5$

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

$- 1$	$1$	$6$	$11$	$6$
	\xa0	$- 1$	$- 5$	\xa0
	$1$	$5$	$6$

Multiply the newest entry in the result $(6)$ by the divisor $(- 1)$ and place the result of $(- 6)$ under the next term in the dividend $(6)$ .

$- 1$	$1$	$6$	$11$	$6$
	\xa0	$- 1$	$- 5$	$- 6$
	$1$	$5$	$6$

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

$- 1$	$1$	$6$	$11$	$6$
	\xa0	$- 1$	$- 5$	$- 6$
	$1$	$5$	$6$	$0$

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

$1 x^{2} + 5 x + 6$

Simplify the quotient polynomial.

$x^{2} + 5 x + 6$

Step 7

Factor $x^{2} + 5 x + 6$ 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 $5$ .

$2, 3$

Write the factored form using these integers.

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

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

Step 8

Factor the left side of the equation.

Factor $x^{3} + 6 x^{2} + 11 x + 6$ 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.

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

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

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

Substitute $- 1$ and simplify the expression. In this case, the expression is equal to $0$ so $- 1$ is a root of the polynomial.

Substitute $- 1$ into the polynomial.

${(- 1)}^{3} + 6 {(- 1)}^{2} + 11 \cdot - 1 + 6$

Raise $- 1$ to the power of $3$ .

$- 1 + 6 {(- 1)}^{2} + 11 \cdot - 1 + 6$

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

$- 1 + 6 \cdot 1 + 11 \cdot - 1 + 6$

Multiply $6$ by $1$ .

$- 1 + 6 + 11 \cdot - 1 + 6$

Add $- 1$ and $6$ .

$5 + 11 \cdot - 1 + 6$

Multiply $11$ by $- 1$ .

$5 - 11 + 6$

Subtract $11$ from $5$ .

$- 6 + 6$

Add $- 6$ and $6$ .

$0$

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

$\frac{x^{3} + 6 x^{2} + 11 x + 6}{x + 1}$

Divide $x^{3} + 6 x^{2} + 11 x + 6$ by $x + 1$ .

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

$x$

$1$

$x^{3}$

$6 x^{2}$

$11 x$

$6$

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

					$x^{2}$
	$x$	+	$1$		$x^{3}$	+	$6 x^{2}$	+	$11 x$	+	$6$

Multiply the new quotient term by the divisor.

				$x^{2}$
$x$	+	$1$		$x^{3}$	+	$6 x^{2}$	+	$11 x$	+	$6$
			+	$x^{3}$	+	$x^{2}$

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

				$x^{2}$
$x$	+	$1$		$x^{3}$	+	$6 x^{2}$	+	$11 x$	+	$6$
			-	$x^{3}$	-	$x^{2}$

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

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

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

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

Divide the highest order term in the dividend $5 x^{2}$ by the highest order term in divisor $x$ .

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

Multiply the new quotient term by the divisor.

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

The expression needs to be subtracted from the dividend, so change all the signs in $5 x^{2} + 5 x$

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

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

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

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

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

Divide the highest order term in the dividend $6 x$ by the highest order term in divisor $x$ .

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

Multiply the new quotient term by the divisor.

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

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

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

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

				$x^{2}$	+	$5 x$	+	$6$
$x$	+	$1$		$x^{3}$	+	$6 x^{2}$	+	$11 x$	+	$6$
			-	$x^{3}$	-	$x^{2}$
					+	$5 x^{2}$	+	$11 x$
					-	$5 x^{2}$	-	$5 x$
							+	$6 x$	+	$6$
							-	$6 x$	-	$6$
										$0$

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

$x^{2} + 5 x + 6$

Write $x^{3} + 6 x^{2} + 11 x + 6$ as a set of factors.

$(x + 1) (x^{2} + 5 x + 6) = 0$

Factor $x^{2} + 5 x + 6$ 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 $5$ .

$2, 3$

Write the factored form using these integers.

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

Remove unnecessary parentheses.

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

Step 9

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

$x + 1 = 0$

$x + 2 = 0$

$x + 3 = 0$

Step 10

Set $x + 1$ equal to $0$ and solve for $x$ .

Set $x + 1$ equal to $0$ .

$x + 1 = 0$

Subtract $1$ from both sides of the equation.

$x = - 1$

Step 11

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

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

$x + 2 = 0$

Subtract $2$ from 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 $(x + 1) (x + 2) (x + 3) = 0$ true.

$x = - 1, - 2, - 3$

Step 14

Find the Roots/Zeros Using the Rational Roots Test x^3+6x^2+11x+6

Related Questions

Support LCMGCF.com by disabling your adblocker.