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

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

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

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

Solve for $x$ .

Factor the left side of the equation.

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

Factor $x^{2}$ out of $x^{5}$ .

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Multiply $- 3$ by $1$ .

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

Subtract $3$ from $1$ .

$- 2 + 3 \cdot 1 - 1$

Multiply $3$ by $1$ .

$- 2 + 3 - 1$

Add $- 2$ and $3$ .

$1 - 1$

Subtract $1$ from $1$ .

$0$

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.

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

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

$x$

$1$

$x^{3}$

$3 x^{2}$

$3 x$

$1$

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

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

Multiply the new quotient term by the divisor.

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

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

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

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

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

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

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

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

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

Multiply the new quotient term by the divisor.

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

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

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

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

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

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

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

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

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

Multiply the new quotient term by the divisor.

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

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

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

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

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

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

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

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

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

Factor using the perfect square rule.

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

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

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$2 x = 2 \cdot x \cdot 1$

Rewrite the polynomial.

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

Factor using the perfect square trinomial rule $a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ , where $a = x$ and $b = 1$ .

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

Combine like factors.

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

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

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

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

Add $1$ and $2$ .

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

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

$x^{2} = 0$

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

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

Set $x^{2}$ equal to $0$ .

$x^{2} = 0$

Solve $x^{2} = 0$ for $x$ .

Take the square root of both sides of the equation to eliminate the exponent on the left side.

$x = \pm \sqrt{0}$

Simplify $\pm \sqrt{0}$ .

Rewrite $0$ as $0^{2}$ .

$x = \pm \sqrt{0^{2}}$

Pull terms out from under the radical, assuming positive real numbers.

$x = \pm 0$

Plus or minus $0$ is $0$ .

$x = 0$

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

Set ${(x - 1)}^{3}$ equal to $0$ .

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

Solve ${(x - 1)}^{3} = 0$ for $x$ .

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

$x - 1 = 0$

Add $1$ to both sides of the equation.

$x = 1$

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

$x = 0, 1$

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

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