Find the Roots (Zeros) 3x^4-4x^3+x^2+6x-2

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

Set $3 x^{4} - 4 x^{3} + x^{2} + 6 x - 2$ equal to $0$ .

$3 x^{4} - 4 x^{3} + x^{2} + 6 x - 2 = 0$

Solve for $x$ .

Factor the left side of the equation.

Regroup terms.

$6 x - 2 + 3 x^{4} - 4 x^{3} + x^{2} = 0$

Factor $2$ out of $6 x - 2$ .

Factor $2$ out of $6 x$ .

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

Factor $2$ out of $- 2$ .

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

Factor $2$ out of $2 (3 x) + 2 (- 1)$ .

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

Factor $x^{2}$ out of $3 x^{4} - 4 x^{3} + x^{2}$ .

Factor $x^{2}$ out of $3 x^{4}$ .

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

Factor $x^{2}$ out of $- 4 x^{3}$ .

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

Multiply by $1$ .

$2 (3 x - 1) + x^{2} (3 x^{2}) + x^{2} (- 4 x) + x^{2} \cdot 1 = 0$

Factor $x^{2}$ out of $x^{2} (3 x^{2}) + x^{2} (- 4 x)$ .

$2 (3 x - 1) + x^{2} (3 x^{2} - 4 x) + x^{2} \cdot 1 = 0$

Factor $x^{2}$ out of $x^{2} (3 x^{2} - 4 x) + x^{2} \cdot 1$ .

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

Factor.

Factor by grouping.

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = 3 \cdot 1 = 3$ and whose sum is $b = - 4$ .

Factor $- 4$ out of $- 4 x$ .

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

Rewrite $- 4$ as $- 1$ plus $- 3$

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

Apply the distributive property.

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

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

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

Factor out the greatest common factor (GCF) from each group.

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

Factor the polynomial by factoring out the greatest common factor, $3 x - 1$ .

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

Remove unnecessary parentheses.

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

Factor $3 x - 1$ out of $2 (3 x - 1) + x^{2} (3 x - 1) (x - 1)$ .

Factor $3 x - 1$ out of $2 (3 x - 1)$ .

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

Factor $3 x - 1$ out of $x^{2} (3 x - 1) (x - 1)$ .

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

Factor $3 x - 1$ out of $(3 x - 1) \cdot 2 + (3 x - 1) (x^{2} (x - 1))$ .

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

Apply the distributive property.

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

Multiply $x^{2}$ by $x$ by adding the exponents.

Multiply $x^{2}$ by $x$ .

Raise $x$ to the power of $1$ .

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

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

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

Add $2$ and $1$ .

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

Move $- 1$ to the left of $x^{2}$ .

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

Rewrite $- 1 x^{2}$ as $- x^{2}$ .

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

Reorder terms.

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

Factor.

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

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

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

$\pm 1, \pm 2$

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.

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

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

$- 1 - {(- 1)}^{2} + 2$

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

$- 1 - 1 \cdot 1 + 2$

Multiply $- 1$ by $1$ .

$- 1 - 1 + 2$

Subtract $1$ from $- 1$ .

$- 2 + 2$

Add $- 2$ and $2$ .

$0$

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.

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

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

$x$

$1$

$x^{3}$

$x^{2}$

$0 x$

$2$

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

					$x^{2}$
	$x$	+	$1$		$x^{3}$	-	$x^{2}$	+	$0 x$	+	$2$

Multiply the new quotient term by the divisor.

				$x^{2}$
$x$	+	$1$		$x^{3}$	-	$x^{2}$	+	$0 x$	+	$2$
			+	$x^{3}$	+	$x^{2}$

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

				$x^{2}$
$x$	+	$1$		$x^{3}$	-	$x^{2}$	+	$0 x$	+	$2$
			-	$x^{3}$	-	$x^{2}$

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

				$x^{2}$
$x$	+	$1$		$x^{3}$	-	$x^{2}$	+	$0 x$	+	$2$
			-	$x^{3}$	-	$x^{2}$
					-	$2 x^{2}$

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

				$x^{2}$
$x$	+	$1$		$x^{3}$	-	$x^{2}$	+	$0 x$	+	$2$
			-	$x^{3}$	-	$x^{2}$
					-	$2 x^{2}$	+	$0 x$

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

				$x^{2}$	-	$2 x$
$x$	+	$1$		$x^{3}$	-	$x^{2}$	+	$0 x$	+	$2$
			-	$x^{3}$	-	$x^{2}$
					-	$2 x^{2}$	+	$0 x$

Multiply the new quotient term by the divisor.

				$x^{2}$	-	$2 x$
$x$	+	$1$		$x^{3}$	-	$x^{2}$	+	$0 x$	+	$2$
			-	$x^{3}$	-	$x^{2}$
					-	$2 x^{2}$	+	$0 x$
					-	$2 x^{2}$	-	$2 x$

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

				$x^{2}$	-	$2 x$
$x$	+	$1$		$x^{3}$	-	$x^{2}$	+	$0 x$	+	$2$
			-	$x^{3}$	-	$x^{2}$
					-	$2 x^{2}$	+	$0 x$
					+	$2 x^{2}$	+	$2 x$

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

				$x^{2}$	-	$2 x$
$x$	+	$1$		$x^{3}$	-	$x^{2}$	+	$0 x$	+	$2$
			-	$x^{3}$	-	$x^{2}$
					-	$2 x^{2}$	+	$0 x$
					+	$2 x^{2}$	+	$2 x$
							+	$2 x$

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

				$x^{2}$	-	$2 x$
$x$	+	$1$		$x^{3}$	-	$x^{2}$	+	$0 x$	+	$2$
			-	$x^{3}$	-	$x^{2}$
					-	$2 x^{2}$	+	$0 x$
					+	$2 x^{2}$	+	$2 x$
							+	$2 x$	+	$2$

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

				$x^{2}$	-	$2 x$	+	$2$
$x$	+	$1$		$x^{3}$	-	$x^{2}$	+	$0 x$	+	$2$
			-	$x^{3}$	-	$x^{2}$
					-	$2 x^{2}$	+	$0 x$
					+	$2 x^{2}$	+	$2 x$
							+	$2 x$	+	$2$

Multiply the new quotient term by the divisor.

				$x^{2}$	-	$2 x$	+	$2$
$x$	+	$1$		$x^{3}$	-	$x^{2}$	+	$0 x$	+	$2$
			-	$x^{3}$	-	$x^{2}$
					-	$2 x^{2}$	+	$0 x$
					+	$2 x^{2}$	+	$2 x$
							+	$2 x$	+	$2$
							+	$2 x$	+	$2$

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

				$x^{2}$	-	$2 x$	+	$2$
$x$	+	$1$		$x^{3}$	-	$x^{2}$	+	$0 x$	+	$2$
			-	$x^{3}$	-	$x^{2}$
					-	$2 x^{2}$	+	$0 x$
					+	$2 x^{2}$	+	$2 x$
							+	$2 x$	+	$2$
							-	$2 x$	-	$2$

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

				$x^{2}$	-	$2 x$	+	$2$
$x$	+	$1$		$x^{3}$	-	$x^{2}$	+	$0 x$	+	$2$
			-	$x^{3}$	-	$x^{2}$
					-	$2 x^{2}$	+	$0 x$
					+	$2 x^{2}$	+	$2 x$
							+	$2 x$	+	$2$
							-	$2 x$	-	$2$
										$0$

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

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

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

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

Remove unnecessary parentheses.

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

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

$3 x - 1 = 0$

$x + 1 = 0$

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

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 $x + 1$ 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 $x^{2} - 2 x + 2$ 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 $- 4$ 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 $(3 x - 1) (x + 1) (x^{2} - 2 x + 2) = 0$ true.

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

Find the Roots (Zeros) 3x^4-4x^3+x^2+6x-2

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