Find the Roots/Zeros Using the Rational Roots Test f(x)=2x^3+7x^2-28x+12

$f (x) = 2 x^{3} + 7 x^{2} - 28 x + 12$

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.

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

$q = \pm 1, \pm 2$

Step 2

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 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.

$2 {(\frac{1}{2})}^{3} + 7 {(\frac{1}{2})}^{2} - 28 (\frac{1}{2}) + 12$

Step 4

Simplify the expression. In this case, the expression is equal to $0$ so $x = \frac{1}{2}$ is a root of the polynomial.

Simplify each term.

Apply the product rule to $\frac{1}{2}$ .

$2 \frac{1^{3}}{2^{3}} + 7 {(\frac{1}{2})}^{2} - 28 (\frac{1}{2}) + 12$

One to any power is one.

$2 \frac{1}{2^{3}} + 7 {(\frac{1}{2})}^{2} - 28 (\frac{1}{2}) + 12$

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

$2 (\frac{1}{8}) + 7 {(\frac{1}{2})}^{2} - 28 (\frac{1}{2}) + 12$

Cancel the common factor of $2$ .

Factor $2$ out of $8$ .

$2 \frac{1}{2 (4)} + 7 {(\frac{1}{2})}^{2} - 28 (\frac{1}{2}) + 12$

Cancel the common factor.

$2 \frac{1}{2 \cdot 4} + 7 {(\frac{1}{2})}^{2} - 28 (\frac{1}{2}) + 12$

Rewrite the expression.

$\frac{1}{4} + 7 {(\frac{1}{2})}^{2} - 28 (\frac{1}{2}) + 12$

Apply the product rule to $\frac{1}{2}$ .

$\frac{1}{4} + 7 \frac{1^{2}}{2^{2}} - 28 (\frac{1}{2}) + 12$

One to any power is one.

$\frac{1}{4} + 7 \frac{1}{2^{2}} - 28 (\frac{1}{2}) + 12$

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

$\frac{1}{4} + 7 (\frac{1}{4}) - 28 (\frac{1}{2}) + 12$

Combine $7$ and $\frac{1}{4}$ .

$\frac{1}{4} + \frac{7}{4} - 28 (\frac{1}{2}) + 12$

Cancel the common factor of $2$ .

Factor $2$ out of $- 28$ .

$\frac{1}{4} + \frac{7}{4} + 2 (- 14) \frac{1}{2} + 12$

Cancel the common factor.

$\frac{1}{4} + \frac{7}{4} + 2 \cdot - 14 \frac{1}{2} + 12$

Rewrite the expression.

$\frac{1}{4} + \frac{7}{4} - 14 + 12$

Combine fractions.

Combine the numerators over the common denominator.

$- 14 + 12 + \frac{1 + 7}{4}$

Add $1$ and $7$ .

$- 14 + 12 + \frac{8}{4}$

Find the common denominator.

Write $- 14$ as a fraction with denominator $1$ .

$\frac{- 14}{1} + 12 + \frac{8}{4}$

Multiply $\frac{- 14}{1}$ by $\frac{4}{4}$ .

$\frac{- 14}{1} \cdot \frac{4}{4} + 12 + \frac{8}{4}$

Multiply $\frac{- 14}{1}$ by $\frac{4}{4}$ .

$\frac{- 14 \cdot 4}{4} + 12 + \frac{8}{4}$

Write $12$ as a fraction with denominator $1$ .

$\frac{- 14 \cdot 4}{4} + \frac{12}{1} + \frac{8}{4}$

Multiply $\frac{12}{1}$ by $\frac{4}{4}$ .

$\frac{- 14 \cdot 4}{4} + \frac{12}{1} \cdot \frac{4}{4} + \frac{8}{4}$

Multiply $\frac{12}{1}$ by $\frac{4}{4}$ .

$\frac{- 14 \cdot 4}{4} + \frac{12 \cdot 4}{4} + \frac{8}{4}$

Combine the numerators over the common denominator.

$\frac{- 14 \cdot 4 + 12 \cdot 4 + 8}{4}$

Simplify each term.

Multiply $- 14$ by $4$ .

$\frac{- 56 + 12 \cdot 4 + 8}{4}$

Multiply $12$ by $4$ .

$\frac{- 56 + 48 + 8}{4}$

Simplify the expression.

Add $- 56$ and $48$ .

$\frac{- 8 + 8}{4}$

Add $- 8$ and $8$ .

$\frac{0}{4}$

Divide $0$ by $4$ .

$0$

Step 5

Since $\frac{1}{2}$ is a known root, divide the polynomial by $x - \frac{1}{2}$ to find the quotient polynomial. This polynomial can then be used to find the remaining roots.

$\frac{2 x^{3} + 7 x^{2} - 28 x + 12}{x - \frac{1}{2}}$

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.

$\frac{1}{2}$	$2$	$7$	$- 28$	$12$
	\xa0			\xa0

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

$\frac{1}{2}$	$2$	$7$	$- 28$	$12$
	\xa0			\xa0
	$2$

Multiply the newest entry in the result $(2)$ by the divisor $(\frac{1}{2})$ and place the result of $(1)$ under the next term in the dividend $(7)$ .

$\frac{1}{2}$	$2$	$7$	$- 28$	$12$
	\xa0	$1$		\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.

$\frac{1}{2}$	$2$	$7$	$- 28$	$12$
	\xa0	$1$		\xa0
	$2$	$8$

Multiply the newest entry in the result $(8)$ by the divisor $(\frac{1}{2})$ and place the result of $(4)$ under the next term in the dividend $(- 28)$ .

$\frac{1}{2}$	$2$	$7$	$- 28$	$12$
	\xa0	$1$	$4$	\xa0
	$2$	$8$

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

$\frac{1}{2}$	$2$	$7$	$- 28$	$12$
	\xa0	$1$	$4$	\xa0
	$2$	$8$	$- 24$

Multiply the newest entry in the result $(- 24)$ by the divisor $(\frac{1}{2})$ and place the result of $(- 12)$ under the next term in the dividend $(12)$ .

$\frac{1}{2}$	$2$	$7$	$- 28$	$12$
	\xa0	$1$	$4$	$- 12$
	$2$	$8$	$- 24$

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

$\frac{1}{2}$	$2$	$7$	$- 28$	$12$
	\xa0	$1$	$4$	$- 12$
	$2$	$8$	$- 24$	$0$

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

$2 x^{2} + 8 x - 24$

Step 7

Factor $2$ out of $2 x^{2} + 8 x - 24$ .

Factor $2$ out of $2 x^{2}$ .

$(x - \frac{1}{2}) (2 (x^{2}) + 8 x - 24)$

Factor $2$ out of $8 x$ .

$(x - \frac{1}{2}) (2 (x^{2}) + 2 (4 x) - 24)$

Factor $2$ out of $- 24$ .

$(x - \frac{1}{2}) (2 x^{2} + 2 (4 x) + 2 \cdot - 12)$

Factor $2$ out of $2 x^{2} + 2 (4 x)$ .

$(x - \frac{1}{2}) (2 (x^{2} + 4 x) + 2 \cdot - 12)$

Factor $2$ out of $2 (x^{2} + 4 x) + 2 \cdot - 12$ .

$(x - \frac{1}{2}) (2 (x^{2} + 4 x - 12))$

Step 8

Factor.

Factor $x^{2} + 4 x - 12$ 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 $4$ .

$- 2, 6$

Write the factored form using these integers.

$(x - \frac{1}{2}) (2 ((x - 2) (x + 6)))$

Remove unnecessary parentheses.

$(x - \frac{1}{2}) (2 (x - 2) (x + 6))$

Step 9

Factor the left side of the equation.

Factor $2 x^{3} + 7 x^{2} - 28 x + 12$ 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.

$2 \cdot {0.5}^{3} + 7 \cdot {0.5}^{2} - 28 \cdot 0.5 + 12$

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

$2 \cdot 0.125 + 7 \cdot {0.5}^{2} - 28 \cdot 0.5 + 12$

Multiply $2$ by $0.125$ .

$0.25 + 7 \cdot {0.5}^{2} - 28 \cdot 0.5 + 12$

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

$0.25 + 7 \cdot 0.25 - 28 \cdot 0.5 + 12$

Multiply $7$ by $0.25$ .

$0.25 + 1.75 - 28 \cdot 0.5 + 12$

Add $0.25$ and $1.75$ .

$2 - 28 \cdot 0.5 + 12$

Multiply $- 28$ by $0.5$ .

$2 - 14 + 12$

Subtract $14$ from $2$ .

$- 12 + 12$

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.

$\frac{2 x^{3} + 7 x^{2} - 28 x + 12}{2 x - 1}$

Divide $2 x^{3} + 7 x^{2} - 28 x + 12$ 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 $0$ .

$2 x$

$1$

$2 x^{3}$

$7 x^{2}$

$28 x$

$12$

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

					$x^{2}$
	$2 x$	-	$1$		$2 x^{3}$	+	$7 x^{2}$	-	$28 x$	+	$12$

Multiply the new quotient term by the divisor.

				$x^{2}$
$2 x$	-	$1$		$2 x^{3}$	+	$7 x^{2}$	-	$28 x$	+	$12$
			+	$2 x^{3}$	-	$x^{2}$

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

				$x^{2}$
$2 x$	-	$1$		$2 x^{3}$	+	$7 x^{2}$	-	$28 x$	+	$12$
			-	$2 x^{3}$	+	$x^{2}$

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

				$x^{2}$
$2 x$	-	$1$		$2 x^{3}$	+	$7 x^{2}$	-	$28 x$	+	$12$
			-	$2 x^{3}$	+	$x^{2}$
					+	$8 x^{2}$

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

				$x^{2}$
$2 x$	-	$1$		$2 x^{3}$	+	$7 x^{2}$	-	$28 x$	+	$12$
			-	$2 x^{3}$	+	$x^{2}$
					+	$8 x^{2}$	-	$28 x$

Divide the highest order term in the dividend $8 x^{2}$ by the highest order term in divisor $2 x$ .

				$x^{2}$	+	$4 x$
$2 x$	-	$1$		$2 x^{3}$	+	$7 x^{2}$	-	$28 x$	+	$12$
			-	$2 x^{3}$	+	$x^{2}$
					+	$8 x^{2}$	-	$28 x$

Multiply the new quotient term by the divisor.

				$x^{2}$	+	$4 x$
$2 x$	-	$1$		$2 x^{3}$	+	$7 x^{2}$	-	$28 x$	+	$12$
			-	$2 x^{3}$	+	$x^{2}$
					+	$8 x^{2}$	-	$28 x$
					+	$8 x^{2}$	-	$4 x$

The expression needs to be subtracted from the dividend, so change all the signs in $8 x^{2} - 4 x$

				$x^{2}$	+	$4 x$
$2 x$	-	$1$		$2 x^{3}$	+	$7 x^{2}$	-	$28 x$	+	$12$
			-	$2 x^{3}$	+	$x^{2}$
					+	$8 x^{2}$	-	$28 x$
					-	$8 x^{2}$	+	$4 x$

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

				$x^{2}$	+	$4 x$
$2 x$	-	$1$		$2 x^{3}$	+	$7 x^{2}$	-	$28 x$	+	$12$
			-	$2 x^{3}$	+	$x^{2}$
					+	$8 x^{2}$	-	$28 x$
					-	$8 x^{2}$	+	$4 x$
							-	$24 x$

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

				$x^{2}$	+	$4 x$
$2 x$	-	$1$		$2 x^{3}$	+	$7 x^{2}$	-	$28 x$	+	$12$
			-	$2 x^{3}$	+	$x^{2}$
					+	$8 x^{2}$	-	$28 x$
					-	$8 x^{2}$	+	$4 x$
							-	$24 x$	+	$12$

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

				$x^{2}$	+	$4 x$	-	$12$
$2 x$	-	$1$		$2 x^{3}$	+	$7 x^{2}$	-	$28 x$	+	$12$
			-	$2 x^{3}$	+	$x^{2}$
					+	$8 x^{2}$	-	$28 x$
					-	$8 x^{2}$	+	$4 x$
							-	$24 x$	+	$12$

Multiply the new quotient term by the divisor.

				$x^{2}$	+	$4 x$	-	$12$
$2 x$	-	$1$		$2 x^{3}$	+	$7 x^{2}$	-	$28 x$	+	$12$
			-	$2 x^{3}$	+	$x^{2}$
					+	$8 x^{2}$	-	$28 x$
					-	$8 x^{2}$	+	$4 x$
							-	$24 x$	+	$12$
							-	$24 x$	+	$12$

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

				$x^{2}$	+	$4 x$	-	$12$
$2 x$	-	$1$		$2 x^{3}$	+	$7 x^{2}$	-	$28 x$	+	$12$
			-	$2 x^{3}$	+	$x^{2}$
					+	$8 x^{2}$	-	$28 x$
					-	$8 x^{2}$	+	$4 x$
							-	$24 x$	+	$12$
							+	$24 x$	-	$12$

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

				$x^{2}$	+	$4 x$	-	$12$
$2 x$	-	$1$		$2 x^{3}$	+	$7 x^{2}$	-	$28 x$	+	$12$
			-	$2 x^{3}$	+	$x^{2}$
					+	$8 x^{2}$	-	$28 x$
					-	$8 x^{2}$	+	$4 x$
							-	$24 x$	+	$12$
							+	$24 x$	-	$12$
										$0$

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

$x^{2} + 4 x - 12$

Write $2 x^{3} + 7 x^{2} - 28 x + 12$ as a set of factors.

$(2 x - 1) (x^{2} + 4 x - 12) = 0$

Factor $x^{2} + 4 x - 12$ 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 $4$ .

$- 2, 6$

Write the factored form using these integers.

$(2 x - 1) ((x - 2) (x + 6)) = 0$

Remove unnecessary parentheses.

$(2 x - 1) (x - 2) (x + 6) = 0$

Step 10

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

$2 x - 1 = 0$

$x - 2 = 0$

$x + 6 = 0$

Step 11

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 $1$ .

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

Step 12

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 13

Set $x + 6$ equal to $0$ and solve for $x$ .

Set $x + 6$ equal to $0$ .

$x + 6 = 0$

Subtract $6$ from both sides of the equation.

$x = - 6$

Step 14

The final solution is all the values that make $(2 x - 1) (x - 2) (x + 6) = 0$ true.

$x = \frac{1}{2}, 2, - 6$

Step 15

Find the Roots/Zeros Using the Rational Roots Test f(x)=2x^3+7x^2-28x+12

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