Find the Roots (Zeros) x^4-4x^3=6x^2-12x

$x^{4} - 4 x^{3} = 6 x^{2} - 12 x$

Step 1

Move all terms containing $x$ to the left side of the equation.

Subtract $6 x^{2}$ from both sides of the equation.

$x^{4} - 4 x^{3} - 6 x^{2} = - 12 x$

Add $12 x$ to both sides of the equation.

$x^{4} - 4 x^{3} - 6 x^{2} + 12 x = 0$

Step 2

Factor the left side of the equation.

Factor $x$ out of $x^{4} - 4 x^{3} - 6 x^{2} + 12 x$ .

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

$x \cdot x^{3} - 4 x^{3} - 6 x^{2} + 12 x = 0$

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

$x \cdot x^{3} + x (- 4 x^{2}) - 6 x^{2} + 12 x = 0$

Factor $x$ out of $- 6 x^{2}$ .

$x \cdot x^{3} + x (- 4 x^{2}) + x (- 6 x) + 12 x = 0$

Factor $x$ out of $12 x$ .

$x \cdot x^{3} + x (- 4 x^{2}) + x (- 6 x) + x \cdot 12 = 0$

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

$x (x^{3} - 4 x^{2}) + x (- 6 x) + x \cdot 12 = 0$

Factor $x$ out of $x (x^{3} - 4 x^{2}) + x (- 6 x)$ .

$x (x^{3} - 4 x^{2} - 6 x) + x \cdot 12 = 0$

Factor $x$ out of $x (x^{3} - 4 x^{2} - 6 x) + x \cdot 12$ .

$x (x^{3} - 4 x^{2} - 6 x + 12) = 0$

Factor.

Factor $x^{3} - 4 x^{2} - 6 x + 12$ 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.

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

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

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

Substitute $- 2$ and simplify the expression. In this case, the expression is equal to $0$ so $- 2$ is a root of the polynomial.

Substitute $- 2$ into the polynomial.

${(- 2)}^{3} - 4 {(- 2)}^{2} - 6 \cdot - 2 + 12$

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

$- 8 - 4 {(- 2)}^{2} - 6 \cdot - 2 + 12$

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

$- 8 - 4 \cdot 4 - 6 \cdot - 2 + 12$

Multiply $- 4$ by $4$ .

$- 8 - 16 - 6 \cdot - 2 + 12$

Subtract $16$ from $- 8$ .

$- 24 - 6 \cdot - 2 + 12$

Multiply $- 6$ by $- 2$ .

$- 24 + 12 + 12$

Add $- 24$ and $12$ .

$- 12 + 12$

Add $- 12$ and $12$ .

$0$

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

$\frac{x^{3} - 4 x^{2} - 6 x + 12}{x + 2}$

Divide $x^{3} - 4 x^{2} - 6 x + 12$ by $x + 2$ .

Set up the polynomials to be divided. If there is not a term for every exponent, insert one with a value of $0$ .

$x$

$2$

$x^{3}$

$4 x^{2}$

$6 x$

$12$

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

					$x^{2}$
	$x$	+	$2$		$x^{3}$	-	$4 x^{2}$	-	$6 x$	+	$12$

Multiply the new quotient term by the divisor.

				$x^{2}$
$x$	+	$2$		$x^{3}$	-	$4 x^{2}$	-	$6 x$	+	$12$
			+	$x^{3}$	+	$2 x^{2}$

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

				$x^{2}$
$x$	+	$2$		$x^{3}$	-	$4 x^{2}$	-	$6 x$	+	$12$
			-	$x^{3}$	-	$2 x^{2}$

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

				$x^{2}$
$x$	+	$2$		$x^{3}$	-	$4 x^{2}$	-	$6 x$	+	$12$
			-	$x^{3}$	-	$2 x^{2}$
					-	$6 x^{2}$

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

				$x^{2}$
$x$	+	$2$		$x^{3}$	-	$4 x^{2}$	-	$6 x$	+	$12$
			-	$x^{3}$	-	$2 x^{2}$
					-	$6 x^{2}$	-	$6 x$

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

				$x^{2}$	-	$6 x$
$x$	+	$2$		$x^{3}$	-	$4 x^{2}$	-	$6 x$	+	$12$
			-	$x^{3}$	-	$2 x^{2}$
					-	$6 x^{2}$	-	$6 x$

Multiply the new quotient term by the divisor.

				$x^{2}$	-	$6 x$
$x$	+	$2$		$x^{3}$	-	$4 x^{2}$	-	$6 x$	+	$12$
			-	$x^{3}$	-	$2 x^{2}$
					-	$6 x^{2}$	-	$6 x$
					-	$6 x^{2}$	-	$12 x$

The expression needs to be subtracted from the dividend, so change all the signs in $- 6 x^{2} - 12 x$

				$x^{2}$	-	$6 x$
$x$	+	$2$		$x^{3}$	-	$4 x^{2}$	-	$6 x$	+	$12$
			-	$x^{3}$	-	$2 x^{2}$
					-	$6 x^{2}$	-	$6 x$
					+	$6 x^{2}$	+	$12 x$

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

				$x^{2}$	-	$6 x$
$x$	+	$2$		$x^{3}$	-	$4 x^{2}$	-	$6 x$	+	$12$
			-	$x^{3}$	-	$2 x^{2}$
					-	$6 x^{2}$	-	$6 x$
					+	$6 x^{2}$	+	$12 x$
							+	$6 x$

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

				$x^{2}$	-	$6 x$
$x$	+	$2$		$x^{3}$	-	$4 x^{2}$	-	$6 x$	+	$12$
			-	$x^{3}$	-	$2 x^{2}$
					-	$6 x^{2}$	-	$6 x$
					+	$6 x^{2}$	+	$12 x$
							+	$6 x$	+	$12$

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

				$x^{2}$	-	$6 x$	+	$6$
$x$	+	$2$		$x^{3}$	-	$4 x^{2}$	-	$6 x$	+	$12$
			-	$x^{3}$	-	$2 x^{2}$
					-	$6 x^{2}$	-	$6 x$
					+	$6 x^{2}$	+	$12 x$
							+	$6 x$	+	$12$

Multiply the new quotient term by the divisor.

				$x^{2}$	-	$6 x$	+	$6$
$x$	+	$2$		$x^{3}$	-	$4 x^{2}$	-	$6 x$	+	$12$
			-	$x^{3}$	-	$2 x^{2}$
					-	$6 x^{2}$	-	$6 x$
					+	$6 x^{2}$	+	$12 x$
							+	$6 x$	+	$12$
							+	$6 x$	+	$12$

The expression needs to be subtracted from the dividend, so change all the signs in $6 x + 12$

				$x^{2}$	-	$6 x$	+	$6$
$x$	+	$2$		$x^{3}$	-	$4 x^{2}$	-	$6 x$	+	$12$
			-	$x^{3}$	-	$2 x^{2}$
					-	$6 x^{2}$	-	$6 x$
					+	$6 x^{2}$	+	$12 x$
							+	$6 x$	+	$12$
							-	$6 x$	-	$12$

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

				$x^{2}$	-	$6 x$	+	$6$
$x$	+	$2$		$x^{3}$	-	$4 x^{2}$	-	$6 x$	+	$12$
			-	$x^{3}$	-	$2 x^{2}$
					-	$6 x^{2}$	-	$6 x$
					+	$6 x^{2}$	+	$12 x$
							+	$6 x$	+	$12$
							-	$6 x$	-	$12$
										$0$

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

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

Write $x^{3} - 4 x^{2} - 6 x + 12$ as a set of factors.

$x ((x + 2) (x^{2} - 6 x + 6)) = 0$

Remove unnecessary parentheses.

$x (x + 2) (x^{2} - 6 x + 6) = 0$

Step 3

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

$x = 0$

$x + 2 = 0$

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

Step 4

Set $x$ equal to $0$ .

$x = 0$

Step 5

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.

$x = - 2$

Step 6

Set $x^{2} - 6 x + 6$ equal to $0$ and solve for $x$ .

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

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

Solve $x^{2} - 6 x + 6 = 0$ for $x$ .

Use the quadratic formula to find the solutions.

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

Substitute the values $a = 1$ , $b = - 6$ , and $c = 6$ into the quadratic formula and solve for $x$ .

$\frac{6 \pm \sqrt{{(- 6)}^{2} - 4 \cdot (1 \cdot 6)}}{2 \cdot 1}$

Simplify.

Simplify the numerator.

Raise $- 6$ to the power of $2$ .

$x = \frac{6 \pm \sqrt{36 - 4 \cdot 1 \cdot 6}}{2 \cdot 1}$

Multiply $- 4 \cdot 1 \cdot 6$ .

Multiply $- 4$ by $1$ .

$x = \frac{6 \pm \sqrt{36 - 4 \cdot 6}}{2 \cdot 1}$

Multiply $- 4$ by $6$ .

$x = \frac{6 \pm \sqrt{36 - 24}}{2 \cdot 1}$

Subtract $24$ from $36$ .

$x = \frac{6 \pm \sqrt{12}}{2 \cdot 1}$

Rewrite $12$ as $2^{2} \cdot 3$ .

Factor $4$ out of $12$ .

$x = \frac{6 \pm \sqrt{4 (3)}}{2 \cdot 1}$

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

$x = \frac{6 \pm \sqrt{2^{2} \cdot 3}}{2 \cdot 1}$

Pull terms out from under the radical.

$x = \frac{6 \pm 2 \sqrt{3}}{2 \cdot 1}$

Multiply $2$ by $1$ .

$x = \frac{6 \pm 2 \sqrt{3}}{2}$

Simplify $\frac{6 \pm 2 \sqrt{3}}{2}$ .

$x = 3 \pm \sqrt{3}$

The final answer is the combination of both solutions.

$x = 3 + \sqrt{3}, 3 - \sqrt{3}$

Step 7

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

$x = 0, - 2, 3 + \sqrt{3}, 3 - \sqrt{3}$

Step 8

The result can be shown in multiple forms.

Exact Form:

$x = 0, - 2, 3 + \sqrt{3}, 3 - \sqrt{3}$

Decimal Form:

$x = 0, - 2, 4.73205080 \dots, 1.26794919 \dots$

Step 9

Find the Roots (Zeros) x^4-4x^3=6x^2-12x

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