Find the Roots/Zeros Using the Rational Roots Test x^4-7x^3+11x^2+7x-12

$x^{4} - 7 x^{3} + 11 x^{2} + 7 x - 12$

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.

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

$q = \pm 1$

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

$\pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 12$

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.

${(1)}^{4} - 7 {(1)}^{3} + 11 {(1)}^{2} + 7 (1) - 12$

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

Simplify each term.

One to any power is one.

$1 - 7 {(1)}^{3} + 11 {(1)}^{2} + 7 (1) - 12$

One to any power is one.

$1 - 7 \cdot 1 + 11 {(1)}^{2} + 7 (1) - 12$

Multiply $- 7$ by $1$ .

$1 - 7 + 11 {(1)}^{2} + 7 (1) - 12$

One to any power is one.

$1 - 7 + 11 \cdot 1 + 7 (1) - 12$

Multiply $11$ by $1$ .

$1 - 7 + 11 + 7 (1) - 12$

Multiply $7$ by $1$ .

$1 - 7 + 11 + 7 - 12$

Simplify by adding and subtracting.

Subtract $7$ from $1$ .

$- 6 + 11 + 7 - 12$

Add $- 6$ and $11$ .

$5 + 7 - 12$

Add $5$ and $7$ .

$12 - 12$

Subtract $12$ from $12$ .

$0$

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.

$\frac{x^{4} - 7 x^{3} + 11 x^{2} + 7 x - 12}{x - 1}$

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.

$1$	$1$	$- 7$	$11$	$7$	$- 12$
	\xa0				\xa0

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

$1$	$1$	$- 7$	$11$	$7$	$- 12$
	\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 $(- 7)$ .

$1$	$1$	$- 7$	$11$	$7$	$- 12$
	\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.

$1$	$1$	$- 7$	$11$	$7$	$- 12$
	\xa0	$1$			\xa0
	$1$	$- 6$

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 $(11)$ .

$1$	$1$	$- 7$	$11$	$7$	$- 12$
	\xa0	$1$	$- 6$		\xa0
	$1$	$- 6$

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

$1$	$1$	$- 7$	$11$	$7$	$- 12$
	\xa0	$1$	$- 6$		\xa0
	$1$	$- 6$	$5$

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 $(7)$ .

$1$	$1$	$- 7$	$11$	$7$	$- 12$
	\xa0	$1$	$- 6$	$5$	\xa0
	$1$	$- 6$	$5$

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

$1$	$1$	$- 7$	$11$	$7$	$- 12$
	\xa0	$1$	$- 6$	$5$	\xa0
	$1$	$- 6$	$5$	$12$

Multiply the newest entry in the result $(12)$ by the divisor $(1)$ and place the result of $(12)$ under the next term in the dividend $(- 12)$ .

$1$	$1$	$- 7$	$11$	$7$	$- 12$
	\xa0	$1$	$- 6$	$5$	$12$
	$1$	$- 6$	$5$	$12$

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

$1$	$1$	$- 7$	$11$	$7$	$- 12$
	\xa0	$1$	$- 6$	$5$	$12$
	$1$	$- 6$	$5$	$12$	$0$

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

$1 x^{3} + - 6 x^{2} + (5) x + 12$

Simplify the quotient polynomial.

$x^{3} - 6 x^{2} + 5 x + 12$

Solve the equation to find any remaining roots.

Factor the left side of the equation.

Factor $x^{3} - 6 x^{2} + 5 x + 12$ 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.

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

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

$\pm 1, \pm 12, \pm 2, \pm 6, \pm 3, \pm 4$

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.

${(- 1)}^{3} - 6 {(- 1)}^{2} + 5 \cdot - 1 + 12$

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

$- 1 - 6 {(- 1)}^{2} + 5 \cdot - 1 + 12$

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

$- 1 - 6 \cdot 1 + 5 \cdot - 1 + 12$

Multiply $- 6$ by $1$ .

$- 1 - 6 + 5 \cdot - 1 + 12$

Subtract $6$ from $- 1$ .

$- 7 + 5 \cdot - 1 + 12$

Multiply $5$ by $- 1$ .

$- 7 - 5 + 12$

Subtract $5$ from $- 7$ .

$- 12 + 12$

Add $- 12$ and $12$ .

$0$

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.

$\frac{x^{3} - 6 x^{2} + 5 x + 12}{x + 1}$

Divide $x^{3} - 6 x^{2} + 5 x + 12$ 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 $0$ .

$x$

$1$

$x^{3}$

$6 x^{2}$

$5 x$

$12$

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

					$x^{2}$
	$x$	+	$1$		$x^{3}$	-	$6 x^{2}$	+	$5 x$	+	$12$

Multiply the new quotient term by the divisor.

				$x^{2}$
$x$	+	$1$		$x^{3}$	-	$6 x^{2}$	+	$5 x$	+	$12$
			+	$x^{3}$	+	$x^{2}$

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

				$x^{2}$
$x$	+	$1$		$x^{3}$	-	$6 x^{2}$	+	$5 x$	+	$12$
			-	$x^{3}$	-	$x^{2}$

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

				$x^{2}$
$x$	+	$1$		$x^{3}$	-	$6 x^{2}$	+	$5 x$	+	$12$
			-	$x^{3}$	-	$x^{2}$
					-	$7 x^{2}$

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

				$x^{2}$
$x$	+	$1$		$x^{3}$	-	$6 x^{2}$	+	$5 x$	+	$12$
			-	$x^{3}$	-	$x^{2}$
					-	$7 x^{2}$	+	$5 x$

Divide the highest order term in the dividend $- 7 x^{2}$ by the highest order term in divisor $x$ .

				$x^{2}$	-	$7 x$
$x$	+	$1$		$x^{3}$	-	$6 x^{2}$	+	$5 x$	+	$12$
			-	$x^{3}$	-	$x^{2}$
					-	$7 x^{2}$	+	$5 x$

Multiply the new quotient term by the divisor.

				$x^{2}$	-	$7 x$
$x$	+	$1$		$x^{3}$	-	$6 x^{2}$	+	$5 x$	+	$12$
			-	$x^{3}$	-	$x^{2}$
					-	$7 x^{2}$	+	$5 x$
					-	$7 x^{2}$	-	$7 x$

The expression needs to be subtracted from the dividend, so change all the signs in $- 7 x^{2} - 7 x$

				$x^{2}$	-	$7 x$
$x$	+	$1$		$x^{3}$	-	$6 x^{2}$	+	$5 x$	+	$12$
			-	$x^{3}$	-	$x^{2}$
					-	$7 x^{2}$	+	$5 x$
					+	$7 x^{2}$	+	$7 x$

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

				$x^{2}$	-	$7 x$
$x$	+	$1$		$x^{3}$	-	$6 x^{2}$	+	$5 x$	+	$12$
			-	$x^{3}$	-	$x^{2}$
					-	$7 x^{2}$	+	$5 x$
					+	$7 x^{2}$	+	$7 x$
							+	$12 x$

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

				$x^{2}$	-	$7 x$
$x$	+	$1$		$x^{3}$	-	$6 x^{2}$	+	$5 x$	+	$12$
			-	$x^{3}$	-	$x^{2}$
					-	$7 x^{2}$	+	$5 x$
					+	$7 x^{2}$	+	$7 x$
							+	$12 x$	+	$12$

Divide the highest order term in the dividend $12 x$ by the highest order term in divisor $x$ .

				$x^{2}$	-	$7 x$	+	$12$
$x$	+	$1$		$x^{3}$	-	$6 x^{2}$	+	$5 x$	+	$12$
			-	$x^{3}$	-	$x^{2}$
					-	$7 x^{2}$	+	$5 x$
					+	$7 x^{2}$	+	$7 x$
							+	$12 x$	+	$12$

Multiply the new quotient term by the divisor.

				$x^{2}$	-	$7 x$	+	$12$
$x$	+	$1$		$x^{3}$	-	$6 x^{2}$	+	$5 x$	+	$12$
			-	$x^{3}$	-	$x^{2}$
					-	$7 x^{2}$	+	$5 x$
					+	$7 x^{2}$	+	$7 x$
							+	$12 x$	+	$12$
							+	$12 x$	+	$12$

The expression needs to be subtracted from the dividend, so change all the signs in $12 x + 12$

				$x^{2}$	-	$7 x$	+	$12$
$x$	+	$1$		$x^{3}$	-	$6 x^{2}$	+	$5 x$	+	$12$
			-	$x^{3}$	-	$x^{2}$
					-	$7 x^{2}$	+	$5 x$
					+	$7 x^{2}$	+	$7 x$
							+	$12 x$	+	$12$
							-	$12 x$	-	$12$

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

				$x^{2}$	-	$7 x$	+	$12$
$x$	+	$1$		$x^{3}$	-	$6 x^{2}$	+	$5 x$	+	$12$
			-	$x^{3}$	-	$x^{2}$
					-	$7 x^{2}$	+	$5 x$
					+	$7 x^{2}$	+	$7 x$
							+	$12 x$	+	$12$
							-	$12 x$	-	$12$
										$0$

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

$x^{2} - 7 x + 12$

Write $x^{3} - 6 x^{2} + 5 x + 12$ as a set of factors.

$(x + 1) (x^{2} - 7 x + 12) = 0$

Factor $x^{2} - 7 x + 12$ 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 $- 7$ .

$- 4, - 3$

Write the factored form using these integers.

$(x + 1) ((x - 4) (x - 3)) = 0$

Remove unnecessary parentheses.

$(x + 1) (x - 4) (x - 3) = 0$

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

$x + 1 = 0$

$x - 4 = 0$

$x - 3 = 0$

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.

$x = - 1$

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

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

$x - 4 = 0$

Add $4$ to both sides of the equation.

$x = 4$

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.

$x = 3$

The final solution is all the values that make $(x + 1) (x - 4) (x - 3) = 0$ true.

$x = - 1, 4, 3$

The polynomial can be written as a set of linear factors.

$(x - 1) (x + 1) (x - 4) (x - 3)$

These are the roots (zeros) of the polynomial $x^{4} - 7 x^{3} + 11 x^{2} + 7 x - 12$ .

$x = 1, - 1, 4, 3$

Find the Roots/Zeros Using the Rational Roots Test x^4-7x^3+11x^2+7x-12

Related Questions

Support LCMGCF.com by disabling your adblocker.