Find the Roots/Zeros Using the Rational Roots Test 2x^4-1x^3-26x^2+37x-12

$2 x^{4} - 1 x^{3} - 26 x^{2} + 37 x - 12$

Step 1

Rewrite $- 1 x^{3}$ as $- x^{3}$ .

$y = 2 x^{4} - x^{3} - 26 x^{2} + 37 x - 12$

Step 2

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.

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

$q = \pm 1, \pm 2$

Step 3

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

$\pm 1, \pm \frac{1}{2}, \pm 2, \pm 3, \pm \frac{3}{2}, \pm 4, \pm 6, \pm 12$

Step 4

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.

$2 {(1)}^{4} - {(1)}^{3} - 26 {(1)}^{2} + 37 (1) - 12$

Step 5

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.

$2 \cdot 1 - {(1)}^{3} - 26 {(1)}^{2} + 37 (1) - 12$

Multiply $2$ by $1$ .

$2 - {(1)}^{3} - 26 {(1)}^{2} + 37 (1) - 12$

One to any power is one.

$2 - 1 \cdot 1 - 26 {(1)}^{2} + 37 (1) - 12$

Multiply $- 1$ by $1$ .

$2 - 1 - 26 {(1)}^{2} + 37 (1) - 12$

One to any power is one.

$2 - 1 - 26 \cdot 1 + 37 (1) - 12$

Multiply $- 26$ by $1$ .

$2 - 1 - 26 + 37 (1) - 12$

Multiply $37$ by $1$ .

$2 - 1 - 26 + 37 - 12$

Simplify by adding and subtracting.

Subtract $1$ from $2$ .

$1 - 26 + 37 - 12$

Subtract $26$ from $1$ .

$- 25 + 37 - 12$

Add $- 25$ and $37$ .

$12 - 12$

Subtract $12$ from $12$ .

$0$

Step 6

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.

$\frac{2 x^{4} - x^{3} - 26 x^{2} + 37 x - 12}{x - 1}$

Step 7

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.

$1$	$2$	$- 1$	$- 26$	$37$	$- 12$
	\xa0				\xa0

The first number in the dividend $(2)$ is put into the first position of the result area (below the horizontal line).

$1$	$2$	$- 1$	$- 26$	$37$	$- 12$
	\xa0				\xa0
	$2$

Multiply the newest entry in the result $(2)$ by the divisor $(1)$ and place the result of $(2)$ under the next term in the dividend $(- 1)$ .

$1$	$2$	$- 1$	$- 26$	$37$	$- 12$
	\xa0	$2$			\xa0
	$2$

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

$1$	$2$	$- 1$	$- 26$	$37$	$- 12$
	\xa0	$2$			\xa0
	$2$	$1$

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 $(- 26)$ .

$1$	$2$	$- 1$	$- 26$	$37$	$- 12$
	\xa0	$2$	$1$		\xa0
	$2$	$1$

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

$1$	$2$	$- 1$	$- 26$	$37$	$- 12$
	\xa0	$2$	$1$		\xa0
	$2$	$1$	$- 25$

Multiply the newest entry in the result $(- 25)$ by the divisor $(1)$ and place the result of $(- 25)$ under the next term in the dividend $(37)$ .

$1$	$2$	$- 1$	$- 26$	$37$	$- 12$
	\xa0	$2$	$1$	$- 25$	\xa0
	$2$	$1$	$- 25$

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

$1$	$2$	$- 1$	$- 26$	$37$	$- 12$
	\xa0	$2$	$1$	$- 25$	\xa0
	$2$	$1$	$- 25$	$12$

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 $(- 12)$ .

$1$	$2$	$- 1$	$- 26$	$37$	$- 12$
	\xa0	$2$	$1$	$- 25$	$12$
	$2$	$1$	$- 25$	$12$

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

$1$	$2$	$- 1$	$- 26$	$37$	$- 12$
	\xa0	$2$	$1$	$- 25$	$12$
	$2$	$1$	$- 25$	$12$	$0$

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

$2 x^{3} + 1 x^{2} + (- 25) x + 12$

Simplify the quotient polynomial.

$2 x^{3} + x^{2} - 25 x + 12$

Step 8

Solve the equation to find any remaining roots.

Factor the left side of the equation.

Factor $2 x^{3} + x^{2} - 25 x + 12$ 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.

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

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

$\pm 1, \pm 0.5, \pm 12, \pm 6, \pm 2, \pm 3, \pm 1.5, \pm 4$

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.

$2 \cdot {0.5}^{3} + {0.5}^{2} - 25 \cdot 0.5 + 12$

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

$2 \cdot 0.125 + {0.5}^{2} - 25 \cdot 0.5 + 12$

Multiply $2$ by $0.125$ .

$0.25 + {0.5}^{2} - 25 \cdot 0.5 + 12$

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

$0.25 + 0.25 - 25 \cdot 0.5 + 12$

Add $0.25$ and $0.25$ .

$0.5 - 25 \cdot 0.5 + 12$

Multiply $- 25$ by $0.5$ .

$0.5 - 12.5 + 12$

Subtract $12.5$ from $0.5$ .

$- 12 + 12$

Add $- 12$ and $12$ .

$0$

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.

$\frac{2 x^{3} + x^{2} - 25 x + 12}{2 x - 1}$

Divide $2 x^{3} + x^{2} - 25 x + 12$ 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 $0$ .

$2 x$

$1$

$2 x^{3}$

$x^{2}$

$25 x$

$12$

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

					$x^{2}$
	$2 x$	-	$1$		$2 x^{3}$	+	$x^{2}$	-	$25 x$	+	$12$

Multiply the new quotient term by the divisor.

				$x^{2}$
$2 x$	-	$1$		$2 x^{3}$	+	$x^{2}$	-	$25 x$	+	$12$
			+	$2 x^{3}$	-	$x^{2}$

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

				$x^{2}$
$2 x$	-	$1$		$2 x^{3}$	+	$x^{2}$	-	$25 x$	+	$12$
			-	$2 x^{3}$	+	$x^{2}$

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

				$x^{2}$
$2 x$	-	$1$		$2 x^{3}$	+	$x^{2}$	-	$25 x$	+	$12$
			-	$2 x^{3}$	+	$x^{2}$
					+	$2 x^{2}$

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

				$x^{2}$
$2 x$	-	$1$		$2 x^{3}$	+	$x^{2}$	-	$25 x$	+	$12$
			-	$2 x^{3}$	+	$x^{2}$
					+	$2 x^{2}$	-	$25 x$

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

				$x^{2}$	+	$x$
$2 x$	-	$1$		$2 x^{3}$	+	$x^{2}$	-	$25 x$	+	$12$
			-	$2 x^{3}$	+	$x^{2}$
					+	$2 x^{2}$	-	$25 x$

Multiply the new quotient term by the divisor.

				$x^{2}$	+	$x$
$2 x$	-	$1$		$2 x^{3}$	+	$x^{2}$	-	$25 x$	+	$12$
			-	$2 x^{3}$	+	$x^{2}$
					+	$2 x^{2}$	-	$25 x$
					+	$2 x^{2}$	-	$x$

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

				$x^{2}$	+	$x$
$2 x$	-	$1$		$2 x^{3}$	+	$x^{2}$	-	$25 x$	+	$12$
			-	$2 x^{3}$	+	$x^{2}$
					+	$2 x^{2}$	-	$25 x$
					-	$2 x^{2}$	+	$x$

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

				$x^{2}$	+	$x$
$2 x$	-	$1$		$2 x^{3}$	+	$x^{2}$	-	$25 x$	+	$12$
			-	$2 x^{3}$	+	$x^{2}$
					+	$2 x^{2}$	-	$25 x$
					-	$2 x^{2}$	+	$x$
							-	$24 x$

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

				$x^{2}$	+	$x$
$2 x$	-	$1$		$2 x^{3}$	+	$x^{2}$	-	$25 x$	+	$12$
			-	$2 x^{3}$	+	$x^{2}$
					+	$2 x^{2}$	-	$25 x$
					-	$2 x^{2}$	+	$x$
							-	$24 x$	+	$12$

Divide the highest order term in the dividend $- 24 x$ by the highest order term in divisor $2 x$ .

				$x^{2}$	+	$x$	-	$12$
$2 x$	-	$1$		$2 x^{3}$	+	$x^{2}$	-	$25 x$	+	$12$
			-	$2 x^{3}$	+	$x^{2}$
					+	$2 x^{2}$	-	$25 x$
					-	$2 x^{2}$	+	$x$
							-	$24 x$	+	$12$

Multiply the new quotient term by the divisor.

				$x^{2}$	+	$x$	-	$12$
$2 x$	-	$1$		$2 x^{3}$	+	$x^{2}$	-	$25 x$	+	$12$
			-	$2 x^{3}$	+	$x^{2}$
					+	$2 x^{2}$	-	$25 x$
					-	$2 x^{2}$	+	$x$
							-	$24 x$	+	$12$
							-	$24 x$	+	$12$

The expression needs to be subtracted from the dividend, so change all the signs in $- 24 x + 12$

				$x^{2}$	+	$x$	-	$12$
$2 x$	-	$1$		$2 x^{3}$	+	$x^{2}$	-	$25 x$	+	$12$
			-	$2 x^{3}$	+	$x^{2}$
					+	$2 x^{2}$	-	$25 x$
					-	$2 x^{2}$	+	$x$
							-	$24 x$	+	$12$
							+	$24 x$	-	$12$

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

				$x^{2}$	+	$x$	-	$12$
$2 x$	-	$1$		$2 x^{3}$	+	$x^{2}$	-	$25 x$	+	$12$
			-	$2 x^{3}$	+	$x^{2}$
					+	$2 x^{2}$	-	$25 x$
					-	$2 x^{2}$	+	$x$
							-	$24 x$	+	$12$
							+	$24 x$	-	$12$
										$0$

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

$x^{2} + x - 12$

Write $2 x^{3} + x^{2} - 25 x + 12$ as a set of factors.

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

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.

$(2 x - 1) ((x - 3) (x + 4)) = 0$

Remove unnecessary parentheses.

$(2 x - 1) (x - 3) (x + 4) = 0$

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

$2 x - 1 = 0$

$x - 3 = 0$

$x + 4 = 0$

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 $x$ .

Add $1$ to 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 $1$ .

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

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.

$x = 3$

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.

$x = - 4$

The final solution is all the values that make $(2 x - 1) (x - 3) (x + 4) = 0$ true.

$x = \frac{1}{2}, 3, - 4$

Step 9

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

$(x - 1) (x - \frac{1}{2}) (x - 3) (x + 4)$

Step 10

These are the roots (zeros) of the polynomial $2 x^{4} - x^{3} - 26 x^{2} + 37 x - 12$ .

$x = 1, \frac{1}{2}, 3, - 4$

Step 11

Find the Roots/Zeros Using the Rational Roots Test 2x^4-1x^3-26x^2+37x-12

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