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

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

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

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

Solve for $x$ .

Factor the left side of the equation.

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

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

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

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

$x \cdot x^{4} + x (- x^{3}) - 3 x^{3} + 5 x^{2} - 2 x = 0$

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

$x \cdot x^{4} + x (- x^{3}) + x (- 3 x^{2}) + 5 x^{2} - 2 x = 0$

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

$x \cdot x^{4} + x (- x^{3}) + x (- 3 x^{2}) + x (5 x) - 2 x = 0$

Factor $x$ out of $- 2 x$ .

$x \cdot x^{4} + x (- x^{3}) + x (- 3 x^{2}) + x (5 x) + x \cdot - 2 = 0$

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

$x (x^{4} - x^{3}) + x (- 3 x^{2}) + x (5 x) + x \cdot - 2 = 0$

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

$x (x^{4} - x^{3} - 3 x^{2}) + x (5 x) + x \cdot - 2 = 0$

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

$x (x^{4} - x^{3} - 3 x^{2} + 5 x) + x \cdot - 2 = 0$

Factor $x$ out of $x (x^{4} - x^{3} - 3 x^{2} + 5 x) + x \cdot - 2$ .

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

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

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

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

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

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

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

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

Factor by grouping.

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = - 3 \cdot - 2 = 6$ and whose sum is $b = 5$ .

Factor $5$ out of $5 x$ .

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

Rewrite $5$ as $2$ plus $3$

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

Apply the distributive property.

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

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

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

Factor out the greatest common factor (GCF) from each group.

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

Factor the polynomial by factoring out the greatest common factor, $- 3 x + 2$ .

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

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

Factor $x - 1$ out of $x^{3} (x - 1)$ .

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

Factor $x - 1$ out of $(- 3 x + 2) (x - 1)$ .

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

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

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

Factor.

Rewrite $x^{3} - 3 x + 2$ in a factored form.

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

$p = \pm 1, \pm 2 q = \pm 1$

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

$\pm 1, \pm 2$

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.

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

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

$1 - 3 \cdot 1 + 2$

Multiply $- 3$ by $1$ .

$1 - 3 + 2$

Subtract $3$ from $1$ .

$- 2 + 2$

Add $- 2$ and $2$ .

$0$

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.

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

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

$x$

$1$

$x^{3}$

$0 x^{2}$

$3 x$

$2$

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

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

Multiply the new quotient term by the divisor.

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

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

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

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

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

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

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

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

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

Multiply the new quotient term by the divisor.

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

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

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

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

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

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

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

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

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

Multiply the new quotient term by the divisor.

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

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

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

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

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

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

$x^{2} + x - 2$

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

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

Factor $x^{2} + x - 2$ using the AC method.

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 2$ and whose sum is $1$ .

$- 1, 2$

Write the factored form using these integers.

$x ((x - 1) ((x - 1) ((x - 1) (x + 2)))) = 0$

Remove unnecessary parentheses.

$x ((x - 1) ((x - 1) (x - 1) (x + 2))) = 0$

Combine like factors.

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

$x ((x - 1) ((x - 1) (x - 1) (x + 2))) = 0$

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

$x ((x - 1) ((x - 1) (x - 1) (x + 2))) = 0$

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

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

Add $1$ and $1$ .

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

Remove unnecessary parentheses.

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

Factor.

Multiply $x - 1$ by ${(x - 1)}^{2}$ by adding the exponents.

Multiply $x - 1$ by ${(x - 1)}^{2}$ .

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

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

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

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

Add $1$ and $2$ .

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

Remove unnecessary parentheses.

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

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

$x = 0$

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

$x + 2 = 0$

Set $x$ equal to $0$ .

$x = 0$

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.

$x = 1$

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.

$x = - 2$

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

$x = 0, 1, - 2$

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

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