Find the Roots/Zeros Using the Rational Roots Test 3x^4-4x^3+x^2+6x-2

$3 x^{4} - 4 x^{3} + x^{2} + 6 x - 2$

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.

$p = \pm 1, \pm 2$

$q = \pm 1, \pm 3$

Step 2

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

$\pm 1, \pm \frac{1}{3}, \pm 2, \pm \frac{2}{3}$

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.

$3 {(- 1)}^{4} - 4 {(- 1)}^{3} + {(- 1)}^{2} + 6 (- 1) - 2$

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 $4$ .

$3 \cdot 1 - 4 {(- 1)}^{3} + {(- 1)}^{2} + 6 (- 1) - 2$

Multiply $3$ by $1$ .

$3 - 4 {(- 1)}^{3} + {(- 1)}^{2} + 6 (- 1) - 2$

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

$3 - 4 \cdot - 1 + {(- 1)}^{2} + 6 (- 1) - 2$

Multiply $- 4$ by $- 1$ .

$3 + 4 + {(- 1)}^{2} + 6 (- 1) - 2$

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

$3 + 4 + 1 + 6 (- 1) - 2$

Multiply $6$ by $- 1$ .

$3 + 4 + 1 - 6 - 2$

Simplify by adding and subtracting.

Add $3$ and $4$ .

$7 + 1 - 6 - 2$

Add $7$ and $1$ .

$8 - 6 - 2$

Subtract $6$ from $8$ .

$2 - 2$

Subtract $2$ from $2$ .

$0$

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.

$\frac{3 x^{4} - 4 x^{3} + x^{2} + 6 x - 2}{x + 1}$

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.

$- 1$	$3$	$- 4$	$1$	$6$	$- 2$
	\xa0				\xa0

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

$- 1$	$3$	$- 4$	$1$	$6$	$- 2$
	\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 $(- 4)$ .

$- 1$	$3$	$- 4$	$1$	$6$	$- 2$
	\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.

$- 1$	$3$	$- 4$	$1$	$6$	$- 2$
	\xa0	$- 3$			\xa0
	$3$	$- 7$

Multiply the newest entry in the result $(- 7)$ by the divisor $(- 1)$ and place the result of $(7)$ under the next term in the dividend $(1)$ .

$- 1$	$3$	$- 4$	$1$	$6$	$- 2$
	\xa0	$- 3$	$7$		\xa0
	$3$	$- 7$

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

$- 1$	$3$	$- 4$	$1$	$6$	$- 2$
	\xa0	$- 3$	$7$		\xa0
	$3$	$- 7$	$8$

Multiply the newest entry in the result $(8)$ by the divisor $(- 1)$ and place the result of $(- 8)$ under the next term in the dividend $(6)$ .

$- 1$	$3$	$- 4$	$1$	$6$	$- 2$
	\xa0	$- 3$	$7$	$- 8$	\xa0
	$3$	$- 7$	$8$

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

$- 1$	$3$	$- 4$	$1$	$6$	$- 2$
	\xa0	$- 3$	$7$	$- 8$	\xa0
	$3$	$- 7$	$8$	$- 2$

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 $(- 2)$ .

$- 1$	$3$	$- 4$	$1$	$6$	$- 2$
	\xa0	$- 3$	$7$	$- 8$	$2$
	$3$	$- 7$	$8$	$- 2$

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

$- 1$	$3$	$- 4$	$1$	$6$	$- 2$
	\xa0	$- 3$	$7$	$- 8$	$2$
	$3$	$- 7$	$8$	$- 2$	$0$

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

$3 x^{3} + - 7 x^{2} + (8) x - 2$

Simplify the quotient polynomial.

$3 x^{3} - 7 x^{2} + 8 x - 2$

Step 7

Solve the equation to find any remaining roots.

Factor $3 x^{3} - 7 x^{2} + 8 x - 2$ 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.

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

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

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

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.

$3 \cdot {0.\overline{3}}^{3} - 7 \cdot {0.\overline{3}}^{2} + 8 \cdot 0.\overline{3} - 2$

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

$3 \cdot 0.\overline{037} - 7 \cdot {0.\overline{3}}^{2} + 8 \cdot 0.\overline{3} - 2$

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

$0.\overline{1} - 7 \cdot {0.\overline{3}}^{2} + 8 \cdot 0.\overline{3} - 2$

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

$0.\overline{1} - 7 \cdot 0.\overline{1} + 8 \cdot 0.\overline{3} - 2$

Multiply $- 7$ by $0.\overline{1}$ .

$0.\overline{1} - 0.\overline{7} + 8 \cdot 0.\overline{3} - 2$

Subtract $0.\overline{7}$ from $0.\overline{1}$ .

$- 0.\overline{6} + 8 \cdot 0.\overline{3} - 2$

Multiply $8$ by $0.\overline{3}$ .

$- 0.\overline{6} + 2.\overline{6} - 2$

Add $- 0.\overline{6}$ and $2.\overline{6}$ .

$2 - 2$

Subtract $2$ from $2$ .

$0$

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.

$\frac{3 x^{3} - 7 x^{2} + 8 x - 2}{3 x - 1}$

Divide $3 x^{3} - 7 x^{2} + 8 x - 2$ 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 $0$ .

$3 x$

$1$

$3 x^{3}$

$7 x^{2}$

$8 x$

$2$

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

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

Multiply the new quotient term by the divisor.

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

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

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

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

				$x^{2}$
$3 x$	-	$1$		$3 x^{3}$	-	$7 x^{2}$	+	$8 x$	-	$2$
			-	$3 x^{3}$	+	$x^{2}$
					-	$6 x^{2}$

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

				$x^{2}$
$3 x$	-	$1$		$3 x^{3}$	-	$7 x^{2}$	+	$8 x$	-	$2$
			-	$3 x^{3}$	+	$x^{2}$
					-	$6 x^{2}$	+	$8 x$

Divide the highest order term in the dividend $- 6 x^{2}$ by the highest order term in divisor $3 x$ .

				$x^{2}$	-	$2 x$
$3 x$	-	$1$		$3 x^{3}$	-	$7 x^{2}$	+	$8 x$	-	$2$
			-	$3 x^{3}$	+	$x^{2}$
					-	$6 x^{2}$	+	$8 x$

Multiply the new quotient term by the divisor.

				$x^{2}$	-	$2 x$
$3 x$	-	$1$		$3 x^{3}$	-	$7 x^{2}$	+	$8 x$	-	$2$
			-	$3 x^{3}$	+	$x^{2}$
					-	$6 x^{2}$	+	$8 x$
					-	$6 x^{2}$	+	$2 x$

The expression needs to be subtracted from the dividend, so change all the signs in $- 6 x^{2} + 2 x$

				$x^{2}$	-	$2 x$
$3 x$	-	$1$		$3 x^{3}$	-	$7 x^{2}$	+	$8 x$	-	$2$
			-	$3 x^{3}$	+	$x^{2}$
					-	$6 x^{2}$	+	$8 x$
					+	$6 x^{2}$	-	$2 x$

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

				$x^{2}$	-	$2 x$
$3 x$	-	$1$		$3 x^{3}$	-	$7 x^{2}$	+	$8 x$	-	$2$
			-	$3 x^{3}$	+	$x^{2}$
					-	$6 x^{2}$	+	$8 x$
					+	$6 x^{2}$	-	$2 x$
							+	$6 x$

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

				$x^{2}$	-	$2 x$
$3 x$	-	$1$		$3 x^{3}$	-	$7 x^{2}$	+	$8 x$	-	$2$
			-	$3 x^{3}$	+	$x^{2}$
					-	$6 x^{2}$	+	$8 x$
					+	$6 x^{2}$	-	$2 x$
							+	$6 x$	-	$2$

Divide the highest order term in the dividend $6 x$ by the highest order term in divisor $3 x$ .

				$x^{2}$	-	$2 x$	+	$2$
$3 x$	-	$1$		$3 x^{3}$	-	$7 x^{2}$	+	$8 x$	-	$2$
			-	$3 x^{3}$	+	$x^{2}$
					-	$6 x^{2}$	+	$8 x$
					+	$6 x^{2}$	-	$2 x$
							+	$6 x$	-	$2$

Multiply the new quotient term by the divisor.

				$x^{2}$	-	$2 x$	+	$2$
$3 x$	-	$1$		$3 x^{3}$	-	$7 x^{2}$	+	$8 x$	-	$2$
			-	$3 x^{3}$	+	$x^{2}$
					-	$6 x^{2}$	+	$8 x$
					+	$6 x^{2}$	-	$2 x$
							+	$6 x$	-	$2$
							+	$6 x$	-	$2$

The expression needs to be subtracted from the dividend, so change all the signs in $6 x - 2$

				$x^{2}$	-	$2 x$	+	$2$
$3 x$	-	$1$		$3 x^{3}$	-	$7 x^{2}$	+	$8 x$	-	$2$
			-	$3 x^{3}$	+	$x^{2}$
					-	$6 x^{2}$	+	$8 x$
					+	$6 x^{2}$	-	$2 x$
							+	$6 x$	-	$2$
							-	$6 x$	+	$2$

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

				$x^{2}$	-	$2 x$	+	$2$
$3 x$	-	$1$		$3 x^{3}$	-	$7 x^{2}$	+	$8 x$	-	$2$
			-	$3 x^{3}$	+	$x^{2}$
					-	$6 x^{2}$	+	$8 x$
					+	$6 x^{2}$	-	$2 x$
							+	$6 x$	-	$2$
							-	$6 x$	+	$2$
										$0$

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

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

Write $3 x^{3} - 7 x^{2} + 8 x - 2$ as a set of factors.

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

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

$3 x - 1 = 0$

$x^{2} - 2 x + 2 = 0$

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 $x^{2} - 2 x + 2$ equal to $0$ and solve for $x$ .

Set $x^{2} - 2 x + 2$ equal to $0$ .

$x^{2} - 2 x + 2 = 0$

Solve $x^{2} - 2 x + 2 = 0$ for $x$ .

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Substitute the values $a = 1$ , $b = - 2$ , and $c = 2$ into the quadratic formula and solve for $x$ .

$\frac{2 \pm \sqrt{{(- 2)}^{2} - 4 \cdot (1 \cdot 2)}}{2 \cdot 1}$

Simplify.

Simplify the numerator.

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

$x = \frac{2 \pm \sqrt{4 - 4 \cdot 1 \cdot 2}}{2 \cdot 1}$

Multiply $- 4 \cdot 1 \cdot 2$ .

Multiply $- 4$ by $1$ .

$x = \frac{2 \pm \sqrt{4 - 4 \cdot 2}}{2 \cdot 1}$

Multiply $- 4$ by $2$ .

$x = \frac{2 \pm \sqrt{4 - 8}}{2 \cdot 1}$

Subtract $8$ from $4$ .

$x = \frac{2 \pm \sqrt{- 4}}{2 \cdot 1}$

Rewrite $- 4$ as $- 1 (4)$ .

$x = \frac{2 \pm \sqrt{- 1 \cdot 4}}{2 \cdot 1}$

Rewrite $\sqrt{- 1 (4)}$ as $\sqrt{- 1} \cdot \sqrt{4}$ .

$x = \frac{2 \pm \sqrt{- 1} \cdot \sqrt{4}}{2 \cdot 1}$

Rewrite $\sqrt{- 1}$ as $i$ .

$x = \frac{2 \pm i \cdot \sqrt{4}}{2 \cdot 1}$

Rewrite $4$ as $2^{2}$ .

$x = \frac{2 \pm i \cdot \sqrt{2^{2}}}{2 \cdot 1}$

Pull terms out from under the radical, assuming positive real numbers.

$x = \frac{2 \pm i \cdot 2}{2 \cdot 1}$

Move $2$ to the left of $i$ .

$x = \frac{2 \pm 2 i}{2 \cdot 1}$

Multiply $2$ by $1$ .

$x = \frac{2 \pm 2 i}{2}$

Simplify $\frac{2 \pm 2 i}{2}$ .

$x = 1 \pm i$

The final answer is the combination of both solutions.

$x = 1 + i, 1 - i$

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

$x = \frac{1}{3}, 1 + i, 1 - i$

Step 8

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

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

Step 9

These are the roots (zeros) of the polynomial $3 x^{4} - 4 x^{3} + x^{2} + 6 x - 2$ .

$x = - 1, \frac{1}{3}, 1 + i, 1 - i$

Step 10

Find the Roots/Zeros Using the Rational Roots Test 3x^4-4x^3+x^2+6x-2

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