Find the Roots (Zeros) x^4-5x^3+3x^2+x

$x^{4} - 5 x^{3} + 3 x^{2} + x$

Set $x^{4} - 5 x^{3} + 3 x^{2} + x$ equal to $0$ .

$x^{4} - 5 x^{3} + 3 x^{2} + x = 0$

Solve for $x$ .

Factor the left side of the equation.

Factor $x$ out of $x^{4} - 5 x^{3} + 3 x^{2} + x$ .

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

$x \cdot x^{3} - 5 x^{3} + 3 x^{2} + x = 0$

Factor $x$ out of $- 5 x^{3}$ .

$x \cdot x^{3} + x (- 5 x^{2}) + 3 x^{2} + x = 0$

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

$x \cdot x^{3} + x (- 5 x^{2}) + x (3 x) + x = 0$

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

$x \cdot x^{3} + x (- 5 x^{2}) + x (3 x) + x = 0$

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

$x \cdot x^{3} + x (- 5 x^{2}) + x (3 x) + x \cdot 1 = 0$

Factor $x$ out of $x \cdot x^{3} + x (- 5 x^{2})$ .

$x (x^{3} - 5 x^{2}) + x (3 x) + x \cdot 1 = 0$

Factor $x$ out of $x (x^{3} - 5 x^{2}) + x (3 x)$ .

$x (x^{3} - 5 x^{2} + 3 x) + x \cdot 1 = 0$

Factor $x$ out of $x (x^{3} - 5 x^{2} + 3 x) + x \cdot 1$ .

$x (x^{3} - 5 x^{2} + 3 x + 1) = 0$

Factor.

Factor $x^{3} - 5 x^{2} + 3 x + 1$ 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.

$1^{3} - 5 \cdot 1^{2} + 3 \cdot 1 + 1$

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

$1 - 5 \cdot 1^{2} + 3 \cdot 1 + 1$

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

$1 - 5 \cdot 1 + 3 \cdot 1 + 1$

Multiply $- 5$ by $1$ .

$1 - 5 + 3 \cdot 1 + 1$

Subtract $5$ from $1$ .

$- 4 + 3 \cdot 1 + 1$

Multiply $3$ by $1$ .

$- 4 + 3 + 1$

Add $- 4$ and $3$ .

$- 1 + 1$

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.

$\frac{x^{3} - 5 x^{2} + 3 x + 1}{x - 1}$

Divide $x^{3} - 5 x^{2} + 3 x + 1$ 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 $0$ .

$x$

$1$

$x^{3}$

$5 x^{2}$

$3 x$

$1$

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

					$x^{2}$
	$x$	-	$1$		$x^{3}$	-	$5 x^{2}$	+	$3 x$	+	$1$

Multiply the new quotient term by the divisor.

				$x^{2}$
$x$	-	$1$		$x^{3}$	-	$5 x^{2}$	+	$3 x$	+	$1$
			+	$x^{3}$	-	$x^{2}$

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

				$x^{2}$
$x$	-	$1$		$x^{3}$	-	$5 x^{2}$	+	$3 x$	+	$1$
			-	$x^{3}$	+	$x^{2}$

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

				$x^{2}$
$x$	-	$1$		$x^{3}$	-	$5 x^{2}$	+	$3 x$	+	$1$
			-	$x^{3}$	+	$x^{2}$
					-	$4 x^{2}$

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

				$x^{2}$
$x$	-	$1$		$x^{3}$	-	$5 x^{2}$	+	$3 x$	+	$1$
			-	$x^{3}$	+	$x^{2}$
					-	$4 x^{2}$	+	$3 x$

Divide the highest order term in the dividend $- 4 x^{2}$ by the highest order term in divisor $x$ .

				$x^{2}$	-	$4 x$
$x$	-	$1$		$x^{3}$	-	$5 x^{2}$	+	$3 x$	+	$1$
			-	$x^{3}$	+	$x^{2}$
					-	$4 x^{2}$	+	$3 x$

Multiply the new quotient term by the divisor.

				$x^{2}$	-	$4 x$
$x$	-	$1$		$x^{3}$	-	$5 x^{2}$	+	$3 x$	+	$1$
			-	$x^{3}$	+	$x^{2}$
					-	$4 x^{2}$	+	$3 x$
					-	$4 x^{2}$	+	$4 x$

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

				$x^{2}$	-	$4 x$
$x$	-	$1$		$x^{3}$	-	$5 x^{2}$	+	$3 x$	+	$1$
			-	$x^{3}$	+	$x^{2}$
					-	$4 x^{2}$	+	$3 x$
					+	$4 x^{2}$	-	$4 x$

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

				$x^{2}$	-	$4 x$
$x$	-	$1$		$x^{3}$	-	$5 x^{2}$	+	$3 x$	+	$1$
			-	$x^{3}$	+	$x^{2}$
					-	$4 x^{2}$	+	$3 x$
					+	$4 x^{2}$	-	$4 x$
							-	$x$

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

				$x^{2}$	-	$4 x$
$x$	-	$1$		$x^{3}$	-	$5 x^{2}$	+	$3 x$	+	$1$
			-	$x^{3}$	+	$x^{2}$
					-	$4 x^{2}$	+	$3 x$
					+	$4 x^{2}$	-	$4 x$
							-	$x$	+	$1$

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

				$x^{2}$	-	$4 x$	-	$1$
$x$	-	$1$		$x^{3}$	-	$5 x^{2}$	+	$3 x$	+	$1$
			-	$x^{3}$	+	$x^{2}$
					-	$4 x^{2}$	+	$3 x$
					+	$4 x^{2}$	-	$4 x$
							-	$x$	+	$1$

Multiply the new quotient term by the divisor.

				$x^{2}$	-	$4 x$	-	$1$
$x$	-	$1$		$x^{3}$	-	$5 x^{2}$	+	$3 x$	+	$1$
			-	$x^{3}$	+	$x^{2}$
					-	$4 x^{2}$	+	$3 x$
					+	$4 x^{2}$	-	$4 x$
							-	$x$	+	$1$
							-	$x$	+	$1$

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

				$x^{2}$	-	$4 x$	-	$1$
$x$	-	$1$		$x^{3}$	-	$5 x^{2}$	+	$3 x$	+	$1$
			-	$x^{3}$	+	$x^{2}$
					-	$4 x^{2}$	+	$3 x$
					+	$4 x^{2}$	-	$4 x$
							-	$x$	+	$1$
							+	$x$	-	$1$

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

				$x^{2}$	-	$4 x$	-	$1$
$x$	-	$1$		$x^{3}$	-	$5 x^{2}$	+	$3 x$	+	$1$
			-	$x^{3}$	+	$x^{2}$
					-	$4 x^{2}$	+	$3 x$
					+	$4 x^{2}$	-	$4 x$
							-	$x$	+	$1$
							+	$x$	-	$1$
										$0$

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

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

Write $x^{3} - 5 x^{2} + 3 x + 1$ as a set of factors.

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

Remove unnecessary parentheses.

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

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

$x = 0$

$x - 1 = 0$

$x^{2} - 4 x - 1 = 0$

Set $x$ equal to $0$ .

$x = 0$

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.

$x = 1$

Set $x^{2} - 4 x - 1$ equal to $0$ and solve for $x$ .

Set $x^{2} - 4 x - 1$ equal to $0$ .

$x^{2} - 4 x - 1 = 0$

Solve $x^{2} - 4 x - 1 = 0$ for $x$ .

Use the quadratic formula to find the solutions.

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

Substitute the values $a = 1$ , $b = - 4$ , and $c = - 1$ into the quadratic formula and solve for $x$ .

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

Simplify.

Simplify the numerator.

Raise $- 4$ to the power of $2$ .

$x = \frac{4 \pm \sqrt{16 - 4 \cdot 1 \cdot - 1}}{2 \cdot 1}$

Multiply $- 4$ by $1$ .

$x = \frac{4 \pm \sqrt{16 - 4 \cdot - 1}}{2 \cdot 1}$

Multiply $- 4$ by $- 1$ .

$x = \frac{4 \pm \sqrt{16 + 4}}{2 \cdot 1}$

Add $16$ and $4$ .

$x = \frac{4 \pm \sqrt{20}}{2 \cdot 1}$

Rewrite $20$ as $2^{2} \cdot 5$ .

Factor $4$ out of $20$ .

$x = \frac{4 \pm \sqrt{4 (5)}}{2 \cdot 1}$

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

$x = \frac{4 \pm \sqrt{2^{2} \cdot 5}}{2 \cdot 1}$

Pull terms out from under the radical.

$x = \frac{4 \pm 2 \sqrt{5}}{2 \cdot 1}$

Multiply $2$ by $1$ .

$x = \frac{4 \pm 2 \sqrt{5}}{2}$

Simplify $\frac{4 \pm 2 \sqrt{5}}{2}$ .

$x = 2 \pm \sqrt{5}$

The final answer is the combination of both solutions.

$x = 2 + \sqrt{5}, 2 - \sqrt{5}$

The final solution is all the values that make $x (x - 1) (x^{2} - 4 x - 1) = 0$ true.

$x = 0, 1, 2 + \sqrt{5}, 2 - \sqrt{5}$

The result can be shown in multiple forms.

Exact Form:

$x = 0, 1, 2 + \sqrt{5}, 2 - \sqrt{5}$

Decimal Form:

$x = 0, 1, 4.23606797 \dots, - 0.23606797 \dots$

Find the Roots (Zeros) x^4-5x^3+3x^2+x

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