Find the Roots (Zeros) x^4-5x^3+9x^2-9x=0

$x^{4} - 5 x^{3} + 9 x^{2} - 9 x = 0$

Factor the left side of the equation.

Factor $x$ out of $x^{4} - 5 x^{3} + 9 x^{2} - 9 x$ .

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

$x \cdot x^{3} - 5 x^{3} + 9 x^{2} - 9 x = 0$

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

$x \cdot x^{3} + x (- 5 x^{2}) + 9 x^{2} - 9 x = 0$

Factor $x$ out of $9 x^{2}$ .

$x \cdot x^{3} + x (- 5 x^{2}) + x (9 x) - 9 x = 0$

Factor $x$ out of $- 9 x$ .

$x \cdot x^{3} + x (- 5 x^{2}) + x (9 x) + x \cdot - 9 = 0$

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

$x (x^{3} - 5 x^{2}) + x (9 x) + x \cdot - 9 = 0$

Factor $x$ out of $x (x^{3} - 5 x^{2}) + x (9 x)$ .

$x (x^{3} - 5 x^{2} + 9 x) + x \cdot - 9 = 0$

Factor $x$ out of $x (x^{3} - 5 x^{2} + 9 x) + x \cdot - 9$ .

$x (x^{3} - 5 x^{2} + 9 x - 9) = 0$

Factor.

Factor $x^{3} - 5 x^{2} + 9 x - 9$ 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.

$p = \pm 1, \pm 9, \pm 3 q = \pm 1$

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

$\pm 1, \pm 9, \pm 3$

Substitute $3$ and simplify the expression. In this case, the expression is equal to $0$ so $3$ is a root of the polynomial.

Substitute $3$ into the polynomial.

$3^{3} - 5 \cdot 3^{2} + 9 \cdot 3 - 9$

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

$27 - 5 \cdot 3^{2} + 9 \cdot 3 - 9$

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

$27 - 5 \cdot 9 + 9 \cdot 3 - 9$

Multiply $- 5$ by $9$ .

$27 - 45 + 9 \cdot 3 - 9$

Subtract $45$ from $27$ .

$- 18 + 9 \cdot 3 - 9$

Multiply $9$ by $3$ .

$- 18 + 27 - 9$

Add $- 18$ and $27$ .

$9 - 9$

Subtract $9$ from $9$ .

$0$

Since $3$ is a known root, divide the polynomial by $x - 3$ to find the quotient polynomial. This polynomial can then be used to find the remaining roots.

$\frac{x^{3} - 5 x^{2} + 9 x - 9}{x - 3}$

Divide $x^{3} - 5 x^{2} + 9 x - 9$ by $x - 3$ .

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

$x$

$3$

$x^{3}$

$5 x^{2}$

$9 x$

$9$

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

					$x^{2}$
	$x$	-	$3$		$x^{3}$	-	$5 x^{2}$	+	$9 x$	-	$9$

Multiply the new quotient term by the divisor.

				$x^{2}$
$x$	-	$3$		$x^{3}$	-	$5 x^{2}$	+	$9 x$	-	$9$
			+	$x^{3}$	-	$3 x^{2}$

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

				$x^{2}$
$x$	-	$3$		$x^{3}$	-	$5 x^{2}$	+	$9 x$	-	$9$
			-	$x^{3}$	+	$3 x^{2}$

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

				$x^{2}$
$x$	-	$3$		$x^{3}$	-	$5 x^{2}$	+	$9 x$	-	$9$
			-	$x^{3}$	+	$3 x^{2}$
					-	$2 x^{2}$

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

				$x^{2}$
$x$	-	$3$		$x^{3}$	-	$5 x^{2}$	+	$9 x$	-	$9$
			-	$x^{3}$	+	$3 x^{2}$
					-	$2 x^{2}$	+	$9 x$

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

				$x^{2}$	-	$2 x$
$x$	-	$3$		$x^{3}$	-	$5 x^{2}$	+	$9 x$	-	$9$
			-	$x^{3}$	+	$3 x^{2}$
					-	$2 x^{2}$	+	$9 x$

Multiply the new quotient term by the divisor.

				$x^{2}$	-	$2 x$
$x$	-	$3$		$x^{3}$	-	$5 x^{2}$	+	$9 x$	-	$9$
			-	$x^{3}$	+	$3 x^{2}$
					-	$2 x^{2}$	+	$9 x$
					-	$2 x^{2}$	+	$6 x$

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

				$x^{2}$	-	$2 x$
$x$	-	$3$		$x^{3}$	-	$5 x^{2}$	+	$9 x$	-	$9$
			-	$x^{3}$	+	$3 x^{2}$
					-	$2 x^{2}$	+	$9 x$
					+	$2 x^{2}$	-	$6 x$

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

				$x^{2}$	-	$2 x$
$x$	-	$3$		$x^{3}$	-	$5 x^{2}$	+	$9 x$	-	$9$
			-	$x^{3}$	+	$3 x^{2}$
					-	$2 x^{2}$	+	$9 x$
					+	$2 x^{2}$	-	$6 x$
							+	$3 x$

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

				$x^{2}$	-	$2 x$
$x$	-	$3$		$x^{3}$	-	$5 x^{2}$	+	$9 x$	-	$9$
			-	$x^{3}$	+	$3 x^{2}$
					-	$2 x^{2}$	+	$9 x$
					+	$2 x^{2}$	-	$6 x$
							+	$3 x$	-	$9$

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

				$x^{2}$	-	$2 x$	+	$3$
$x$	-	$3$		$x^{3}$	-	$5 x^{2}$	+	$9 x$	-	$9$
			-	$x^{3}$	+	$3 x^{2}$
					-	$2 x^{2}$	+	$9 x$
					+	$2 x^{2}$	-	$6 x$
							+	$3 x$	-	$9$

Multiply the new quotient term by the divisor.

				$x^{2}$	-	$2 x$	+	$3$
$x$	-	$3$		$x^{3}$	-	$5 x^{2}$	+	$9 x$	-	$9$
			-	$x^{3}$	+	$3 x^{2}$
					-	$2 x^{2}$	+	$9 x$
					+	$2 x^{2}$	-	$6 x$
							+	$3 x$	-	$9$
							+	$3 x$	-	$9$

The expression needs to be subtracted from the dividend, so change all the signs in $3 x - 9$

				$x^{2}$	-	$2 x$	+	$3$
$x$	-	$3$		$x^{3}$	-	$5 x^{2}$	+	$9 x$	-	$9$
			-	$x^{3}$	+	$3 x^{2}$
					-	$2 x^{2}$	+	$9 x$
					+	$2 x^{2}$	-	$6 x$
							+	$3 x$	-	$9$
							-	$3 x$	+	$9$

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

				$x^{2}$	-	$2 x$	+	$3$
$x$	-	$3$		$x^{3}$	-	$5 x^{2}$	+	$9 x$	-	$9$
			-	$x^{3}$	+	$3 x^{2}$
					-	$2 x^{2}$	+	$9 x$
					+	$2 x^{2}$	-	$6 x$
							+	$3 x$	-	$9$
							-	$3 x$	+	$9$
										$0$

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

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

Write $x^{3} - 5 x^{2} + 9 x - 9$ as a set of factors.

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

Remove unnecessary parentheses.

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

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

$x = 0$

$x - 3 = 0$

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

Set $x$ equal to $0$ .

$x = 0$

Set $x - 3$ equal to $0$ and solve for $x$ .

Set $x - 3$ equal to $0$ .

$x - 3 = 0$

Add $3$ to both sides of the equation.

$x = 3$

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

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

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

Solve $x^{2} - 2 x + 3 = 0$ 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 $c = 3$ into the quadratic formula and solve for $x$ .

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

Simplify.

Simplify the numerator.

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

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

Multiply $- 4$ by $1$ .

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

Multiply $- 4$ by $3$ .

$x = \frac{2 \pm \sqrt{4 - 12}}{2 \cdot 1}$

Subtract $12$ from $4$ .

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

Rewrite $- 8$ as $- 1 (8)$ .

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

Rewrite $\sqrt{- 1 (8)}$ as $\sqrt{- 1} \cdot \sqrt{8}$ .

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

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

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

Rewrite $8$ as $2^{2} \cdot 2$ .

Factor $4$ out of $8$ .

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

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

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

Pull terms out from under the radical.

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

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

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

Multiply $2$ by $1$ .

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

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

$x = 1 \pm i \sqrt{2}$

The final answer is the combination of both solutions.

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

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

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

Find the Roots (Zeros) x^4-5x^3+9x^2-9x=0

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