Find the Roots/Zeros Using the Rational Roots Test 3x^4-10x^3-45x^2-20x+12

$3 x^{4} - 10 x^{3} - 45 x^{2} - 20 x + 12$

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.

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

$q = \pm 1, \pm 3$

Step 2

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

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

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.

$3 {(- 1)}^{4} - 10 {(- 1)}^{3} - 45 {(- 1)}^{2} - 20 \cdot - 1 + 12$

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 $4$ .

$3 \cdot 1 - 10 {(- 1)}^{3} - 45 {(- 1)}^{2} - 20 \cdot - 1 + 12$

Multiply $3$ by $1$ .

$3 - 10 {(- 1)}^{3} - 45 {(- 1)}^{2} - 20 \cdot - 1 + 12$

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

$3 - 10 \cdot - 1 - 45 {(- 1)}^{2} - 20 \cdot - 1 + 12$

Multiply $- 10$ by $- 1$ .

$3 + 10 - 45 {(- 1)}^{2} - 20 \cdot - 1 + 12$

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

$3 + 10 - 45 \cdot 1 - 20 \cdot - 1 + 12$

Multiply $- 45$ by $1$ .

$3 + 10 - 45 - 20 \cdot - 1 + 12$

Multiply $- 20$ by $- 1$ .

$3 + 10 - 45 + 20 + 12$

Simplify by adding and subtracting.

Add $3$ and $10$ .

$13 - 45 + 20 + 12$

Subtract $45$ from $13$ .

$- 32 + 20 + 12$

Add $- 32$ and $20$ .

$- 12 + 12$

Add $- 12$ and $12$ .

$0$

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.

$\frac{3 x^{4} - 10 x^{3} - 45 x^{2} - 20 x + 12}{x + 1}$

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.

$- 1$	$3$	$- 10$	$- 45$	$- 20$	$12$
	\xa0				\xa0

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

$- 1$	$3$	$- 10$	$- 45$	$- 20$	$12$
	\xa0				\xa0
	$3$

Multiply the newest entry in the result $(3)$ by the divisor $(- 1)$ and place the result of $(- 3)$ under the next term in the dividend $(- 10)$ .

$- 1$	$3$	$- 10$	$- 45$	$- 20$	$12$
	\xa0	$- 3$			\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.

$- 1$	$3$	$- 10$	$- 45$	$- 20$	$12$
	\xa0	$- 3$			\xa0
	$3$	$- 13$

Multiply the newest entry in the result $(- 13)$ by the divisor $(- 1)$ and place the result of $(13)$ under the next term in the dividend $(- 45)$ .

$- 1$	$3$	$- 10$	$- 45$	$- 20$	$12$
	\xa0	$- 3$	$13$		\xa0
	$3$	$- 13$

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

$- 1$	$3$	$- 10$	$- 45$	$- 20$	$12$
	\xa0	$- 3$	$13$		\xa0
	$3$	$- 13$	$- 32$

Multiply the newest entry in the result $(- 32)$ by the divisor $(- 1)$ and place the result of $(32)$ under the next term in the dividend $(- 20)$ .

$- 1$	$3$	$- 10$	$- 45$	$- 20$	$12$
	\xa0	$- 3$	$13$	$32$	\xa0
	$3$	$- 13$	$- 32$

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

$- 1$	$3$	$- 10$	$- 45$	$- 20$	$12$
	\xa0	$- 3$	$13$	$32$	\xa0
	$3$	$- 13$	$- 32$	$12$

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 $(12)$ .

$- 1$	$3$	$- 10$	$- 45$	$- 20$	$12$
	\xa0	$- 3$	$13$	$32$	$- 12$
	$3$	$- 13$	$- 32$	$12$

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

$- 1$	$3$	$- 10$	$- 45$	$- 20$	$12$
	\xa0	$- 3$	$13$	$32$	$- 12$
	$3$	$- 13$	$- 32$	$12$	$0$

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

$3 x^{3} + - 13 x^{2} + (- 32) x + 12$

Simplify the quotient polynomial.

$3 x^{3} - 13 x^{2} - 32 x + 12$

Step 7

Solve the equation to find any remaining roots.

Factor the left side of the equation.

Factor $3 x^{3} - 13 x^{2} - 32 x + 12$ 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.

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

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

$\pm 1, \pm 0.\overline{3}, \pm 12, \pm 4, \pm 2, \pm 0.\overline{6}, \pm 6, \pm 3, \pm 1.\overline{3}$

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.

$3 \cdot {0.\overline{3}}^{3} - 13 \cdot {0.\overline{3}}^{2} - 32 \cdot 0.\overline{3} + 12$

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

$3 \cdot 0.\overline{037} - 13 \cdot {0.\overline{3}}^{2} - 32 \cdot 0.\overline{3} + 12$

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

$0.\overline{1} - 13 \cdot {0.\overline{3}}^{2} - 32 \cdot 0.\overline{3} + 12$

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

$0.\overline{1} - 13 \cdot 0.\overline{1} - 32 \cdot 0.\overline{3} + 12$

Multiply $- 13$ by $0.\overline{1}$ .

$0.\overline{1} - 1.\overline{4} - 32 \cdot 0.\overline{3} + 12$

Subtract $1.\overline{4}$ from $0.\overline{1}$ .

$- 1.\overline{3} - 32 \cdot 0.\overline{3} + 12$

Multiply $- 32$ by $0.\overline{3}$ .

$- 1.\overline{3} - 10.\overline{6} + 12$

Subtract $10.\overline{6}$ from $- 1.\overline{3}$ .

$- 12 + 12$

Add $- 12$ and $12$ .

$0$

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.

$\frac{3 x^{3} - 13 x^{2} - 32 x + 12}{3 x - 1}$

Divide $3 x^{3} - 13 x^{2} - 32 x + 12$ 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 $0$ .

$3 x$

$1$

$3 x^{3}$

$13 x^{2}$

$32 x$

$12$

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

					$x^{2}$
	$3 x$	-	$1$		$3 x^{3}$	-	$13 x^{2}$	-	$32 x$	+	$12$

Multiply the new quotient term by the divisor.

				$x^{2}$
$3 x$	-	$1$		$3 x^{3}$	-	$13 x^{2}$	-	$32 x$	+	$12$
			+	$3 x^{3}$	-	$x^{2}$

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

				$x^{2}$
$3 x$	-	$1$		$3 x^{3}$	-	$13 x^{2}$	-	$32 x$	+	$12$
			-	$3 x^{3}$	+	$x^{2}$

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

				$x^{2}$
$3 x$	-	$1$		$3 x^{3}$	-	$13 x^{2}$	-	$32 x$	+	$12$
			-	$3 x^{3}$	+	$x^{2}$
					-	$12 x^{2}$

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

				$x^{2}$
$3 x$	-	$1$		$3 x^{3}$	-	$13 x^{2}$	-	$32 x$	+	$12$
			-	$3 x^{3}$	+	$x^{2}$
					-	$12 x^{2}$	-	$32 x$

Divide the highest order term in the dividend $- 12 x^{2}$ by the highest order term in divisor $3 x$ .

				$x^{2}$	-	$4 x$
$3 x$	-	$1$		$3 x^{3}$	-	$13 x^{2}$	-	$32 x$	+	$12$
			-	$3 x^{3}$	+	$x^{2}$
					-	$12 x^{2}$	-	$32 x$

Multiply the new quotient term by the divisor.

				$x^{2}$	-	$4 x$
$3 x$	-	$1$		$3 x^{3}$	-	$13 x^{2}$	-	$32 x$	+	$12$
			-	$3 x^{3}$	+	$x^{2}$
					-	$12 x^{2}$	-	$32 x$
					-	$12 x^{2}$	+	$4 x$

The expression needs to be subtracted from the dividend, so change all the signs in $- 12 x^{2} + 4 x$

				$x^{2}$	-	$4 x$
$3 x$	-	$1$		$3 x^{3}$	-	$13 x^{2}$	-	$32 x$	+	$12$
			-	$3 x^{3}$	+	$x^{2}$
					-	$12 x^{2}$	-	$32 x$
					+	$12 x^{2}$	-	$4 x$

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

				$x^{2}$	-	$4 x$
$3 x$	-	$1$		$3 x^{3}$	-	$13 x^{2}$	-	$32 x$	+	$12$
			-	$3 x^{3}$	+	$x^{2}$
					-	$12 x^{2}$	-	$32 x$
					+	$12 x^{2}$	-	$4 x$
							-	$36 x$

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

				$x^{2}$	-	$4 x$
$3 x$	-	$1$		$3 x^{3}$	-	$13 x^{2}$	-	$32 x$	+	$12$
			-	$3 x^{3}$	+	$x^{2}$
					-	$12 x^{2}$	-	$32 x$
					+	$12 x^{2}$	-	$4 x$
							-	$36 x$	+	$12$

Divide the highest order term in the dividend $- 36 x$ by the highest order term in divisor $3 x$ .

				$x^{2}$	-	$4 x$	-	$12$
$3 x$	-	$1$		$3 x^{3}$	-	$13 x^{2}$	-	$32 x$	+	$12$
			-	$3 x^{3}$	+	$x^{2}$
					-	$12 x^{2}$	-	$32 x$
					+	$12 x^{2}$	-	$4 x$
							-	$36 x$	+	$12$

Multiply the new quotient term by the divisor.

				$x^{2}$	-	$4 x$	-	$12$
$3 x$	-	$1$		$3 x^{3}$	-	$13 x^{2}$	-	$32 x$	+	$12$
			-	$3 x^{3}$	+	$x^{2}$
					-	$12 x^{2}$	-	$32 x$
					+	$12 x^{2}$	-	$4 x$
							-	$36 x$	+	$12$
							-	$36 x$	+	$12$

The expression needs to be subtracted from the dividend, so change all the signs in $- 36 x + 12$

				$x^{2}$	-	$4 x$	-	$12$
$3 x$	-	$1$		$3 x^{3}$	-	$13 x^{2}$	-	$32 x$	+	$12$
			-	$3 x^{3}$	+	$x^{2}$
					-	$12 x^{2}$	-	$32 x$
					+	$12 x^{2}$	-	$4 x$
							-	$36 x$	+	$12$
							+	$36 x$	-	$12$

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

				$x^{2}$	-	$4 x$	-	$12$
$3 x$	-	$1$		$3 x^{3}$	-	$13 x^{2}$	-	$32 x$	+	$12$
			-	$3 x^{3}$	+	$x^{2}$
					-	$12 x^{2}$	-	$32 x$
					+	$12 x^{2}$	-	$4 x$
							-	$36 x$	+	$12$
							+	$36 x$	-	$12$
										$0$

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

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

Write $3 x^{3} - 13 x^{2} - 32 x + 12$ as a set of factors.

$(3 x - 1) (x^{2} - 4 x - 12) = 0$

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 $- 4$ .

$- 6, 2$

Write the factored form using these integers.

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

Remove unnecessary parentheses.

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

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

$3 x - 1 = 0$

$x - 6 = 0$

$x + 2 = 0$

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 $1$ .

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

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

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

$x - 6 = 0$

Add $6$ to both sides of the equation.

$x = 6$

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.

$x = - 2$

The final solution is all the values that make $(3 x - 1) (x - 6) (x + 2) = 0$ true.

$x = \frac{1}{3}, 6, - 2$

Step 8

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

$(x + 1) (x - \frac{1}{3}) (x - 6) (x + 2)$

Step 9

These are the roots (zeros) of the polynomial $3 x^{4} - 10 x^{3} - 45 x^{2} - 20 x + 12$ .

$x = - 1, \frac{1}{3}, 6, - 2$

Step 10

Find the Roots/Zeros Using the Rational Roots Test 3x^4-10x^3-45x^2-20x+12

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