Solve for x x^4+2x^3-4x^2+x=0

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

Step 1

Factor the left side of the equation.

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

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

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

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

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

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

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

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

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

Factor $x$ out of $x^{1}$ .

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

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

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

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

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

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

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

Factor.

Factor $x^{3} + 2 x^{2} - 4 x + 1$ 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.

$p = \pm 1 q = \pm 1$

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

$\pm 1$

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.

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

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

$1 + 2 \cdot 1^{2} - 4 \cdot 1 + 1$

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

$1 + 2 \cdot 1 - 4 \cdot 1 + 1$

Multiply $2$ by $1$ .

$1 + 2 - 4 \cdot 1 + 1$

Add $1$ and $2$ .

$3 - 4 \cdot 1 + 1$

Multiply $- 4$ by $1$ .

$3 - 4 + 1$

Subtract $4$ from $3$ .

$- 1 + 1$

Add $- 1$ and $1$ .

$0$

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.

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

Divide $x^{3} + 2 x^{2} - 4 x + 1$ 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 $0$ .

$x$

$1$

$x^{3}$

$2 x^{2}$

$4 x$

$1$

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

					$x^{2}$
	$x$	-	$1$		$x^{3}$	+	$2 x^{2}$	-	$4 x$	+	$1$

Multiply the new quotient term by the divisor.

				$x^{2}$
$x$	-	$1$		$x^{3}$	+	$2 x^{2}$	-	$4 x$	+	$1$
			+	$x^{3}$	-	$x^{2}$

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

				$x^{2}$
$x$	-	$1$		$x^{3}$	+	$2 x^{2}$	-	$4 x$	+	$1$
			-	$x^{3}$	+	$x^{2}$

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

				$x^{2}$
$x$	-	$1$		$x^{3}$	+	$2 x^{2}$	-	$4 x$	+	$1$
			-	$x^{3}$	+	$x^{2}$
					+	$3 x^{2}$

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

				$x^{2}$
$x$	-	$1$		$x^{3}$	+	$2 x^{2}$	-	$4 x$	+	$1$
			-	$x^{3}$	+	$x^{2}$
					+	$3 x^{2}$	-	$4 x$

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

				$x^{2}$	+	$3 x$
$x$	-	$1$		$x^{3}$	+	$2 x^{2}$	-	$4 x$	+	$1$
			-	$x^{3}$	+	$x^{2}$
					+	$3 x^{2}$	-	$4 x$

Multiply the new quotient term by the divisor.

				$x^{2}$	+	$3 x$
$x$	-	$1$		$x^{3}$	+	$2 x^{2}$	-	$4 x$	+	$1$
			-	$x^{3}$	+	$x^{2}$
					+	$3 x^{2}$	-	$4 x$
					+	$3 x^{2}$	-	$3 x$

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

				$x^{2}$	+	$3 x$
$x$	-	$1$		$x^{3}$	+	$2 x^{2}$	-	$4 x$	+	$1$
			-	$x^{3}$	+	$x^{2}$
					+	$3 x^{2}$	-	$4 x$
					-	$3 x^{2}$	+	$3 x$

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

				$x^{2}$	+	$3 x$
$x$	-	$1$		$x^{3}$	+	$2 x^{2}$	-	$4 x$	+	$1$
			-	$x^{3}$	+	$x^{2}$
					+	$3 x^{2}$	-	$4 x$
					-	$3 x^{2}$	+	$3 x$
							-	$x$

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

				$x^{2}$	+	$3 x$
$x$	-	$1$		$x^{3}$	+	$2 x^{2}$	-	$4 x$	+	$1$
			-	$x^{3}$	+	$x^{2}$
					+	$3 x^{2}$	-	$4 x$
					-	$3 x^{2}$	+	$3 x$
							-	$x$	+	$1$

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

				$x^{2}$	+	$3 x$	-	$1$
$x$	-	$1$		$x^{3}$	+	$2 x^{2}$	-	$4 x$	+	$1$
			-	$x^{3}$	+	$x^{2}$
					+	$3 x^{2}$	-	$4 x$
					-	$3 x^{2}$	+	$3 x$
							-	$x$	+	$1$

Multiply the new quotient term by the divisor.

				$x^{2}$	+	$3 x$	-	$1$
$x$	-	$1$		$x^{3}$	+	$2 x^{2}$	-	$4 x$	+	$1$
			-	$x^{3}$	+	$x^{2}$
					+	$3 x^{2}$	-	$4 x$
					-	$3 x^{2}$	+	$3 x$
							-	$x$	+	$1$
							-	$x$	+	$1$

The expression needs to be subtracted from the dividend, so change all the signs in $- x + 1$

				$x^{2}$	+	$3 x$	-	$1$
$x$	-	$1$		$x^{3}$	+	$2 x^{2}$	-	$4 x$	+	$1$
			-	$x^{3}$	+	$x^{2}$
					+	$3 x^{2}$	-	$4 x$
					-	$3 x^{2}$	+	$3 x$
							-	$x$	+	$1$
							+	$x$	-	$1$

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

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

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

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

Write $x^{3} + 2 x^{2} - 4 x + 1$ as a set of factors.

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

Remove unnecessary parentheses.

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

Step 2

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

$x = 0$

$x - 1 = 0$

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

Step 3

Set $x$ equal to $0$ .

$x = 0$

Step 4

Set $x - 1$ equal to $0$ and solve for $x$ .

Set $x - 1$ equal to $0$ .

$x - 1 = 0$

Add $1$ to both sides of the equation.

$x = 1$

Step 5

Set $x^{2} + 3 x - 1$ equal to $0$ and solve for $x$ .

Set $x^{2} + 3 x - 1$ equal to $0$ .

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

Solve $x^{2} + 3 x - 1 = 0$ for $x$ .

Use the quadratic formula to find the solutions.

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

Substitute the values $a = 1$ , $b = 3$ , and $c = - 1$ into the quadratic formula and solve for $x$ .

$\frac{- 3 \pm \sqrt{3^{2} - 4 \cdot (1 \cdot - 1)}}{2 \cdot 1}$

Simplify.

Simplify the numerator.

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

$x = \frac{- 3 \pm \sqrt{9 - 4 \cdot 1 \cdot - 1}}{2 \cdot 1}$

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

Multiply $- 4$ by $1$ .

$x = \frac{- 3 \pm \sqrt{9 - 4 \cdot - 1}}{2 \cdot 1}$

Multiply $- 4$ by $- 1$ .

$x = \frac{- 3 \pm \sqrt{9 + 4}}{2 \cdot 1}$

Add $9$ and $4$ .

$x = \frac{- 3 \pm \sqrt{13}}{2 \cdot 1}$

Multiply $2$ by $1$ .

$x = \frac{- 3 \pm \sqrt{13}}{2}$

The final answer is the combination of both solutions.

$x = - \frac{3 - \sqrt{13}}{2}, - \frac{3 + \sqrt{13}}{2}$

Step 6

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

$x = 0, 1, - \frac{3 - \sqrt{13}}{2}, - \frac{3 + \sqrt{13}}{2}$

Step 7

The result can be shown in multiple forms.

Exact Form:

$x = 0, 1, - \frac{3 - \sqrt{13}}{2}, - \frac{3 + \sqrt{13}}{2}$

Decimal Form:

$x = 0, 1, 0.30277563 \dots, - 3.30277563 \dots$

Solve for x x^4+2x^3-4x^2+x=0

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