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

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

Step 1

Factor the left side of the equation.

Regroup terms.

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

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

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

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

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

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

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

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

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

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

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

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

Factor using the perfect square rule.

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

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

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

$4 x = 2 \cdot x \cdot 2$

Rewrite the polynomial.

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

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

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

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 $a \cdot c = 2 \cdot 2 = 4$ and whose sum is $b = - 5$ .

Factor $- 5$ out of $- 5 x$ .

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

Rewrite $- 5$ as $- 1$ plus $- 4$

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

Apply the distributive property.

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

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

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

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

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

Factor the polynomial by factoring out the greatest common factor, $2 x - 1$ .

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

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

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

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

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

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

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

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

Apply the distributive property.

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

Multiply $x^{3}$ by $x$ by adding the exponents.

Multiply $x^{3}$ by $x$ .

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

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

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

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

Add $3$ and $1$ .

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

Move $- 2$ to the left of $x^{3}$ .

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

Factor.

Rewrite $x^{4} - 2 x^{3} + 2 x - 1$ in a factored form.

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

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

Raise $- 1$ to the power of $4$ .

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

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

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

Multiply $- 2$ by $- 1$ .

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

Add $1$ and $2$ .

$3 + 2 \cdot - 1 - 1$

Multiply $2$ by $- 1$ .

$3 - 2 - 1$

Subtract $2$ from $3$ .

$1 - 1$

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.

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

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

$x$

$1$

$x^{4}$

$2 x^{3}$

$0 x^{2}$

$2 x$

$1$

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

$x^{3}$

$x$

$1$

$x^{4}$

$2 x^{3}$

$0 x^{2}$

$2 x$

$1$

Multiply the new quotient term by the divisor.

$x^{3}$

$x$

$1$

$x^{4}$

$2 x^{3}$

$0 x^{2}$

$2 x$

$1$

$x^{4}$

$x^{3}$

The expression needs to be subtracted from the dividend, so change all the signs in $x^{4} + x^{3}$

$x^{3}$

$x$

$1$

$x^{4}$

$2 x^{3}$

$0 x^{2}$

$2 x$

$1$

$x^{4}$

$x^{3}$

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

$x^{3}$

$x$

$1$

$x^{4}$

$2 x^{3}$

$0 x^{2}$

$2 x$

$1$

$x^{4}$

$x^{3}$

$3 x^{3}$

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

$x^{3}$

$x$

$1$

$x^{4}$

$2 x^{3}$

$0 x^{2}$

$2 x$

$1$

$x^{4}$

$x^{3}$

$3 x^{3}$

$0 x^{2}$

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

$x^{3}$

$3 x^{2}$

$x$

$1$

$x^{4}$

$2 x^{3}$

$0 x^{2}$

$2 x$

$1$

$x^{4}$

$x^{3}$

$3 x^{3}$

$0 x^{2}$

Multiply the new quotient term by the divisor.

$x^{3}$

$3 x^{2}$

$x$

$1$

$x^{4}$

$2 x^{3}$

$0 x^{2}$

$2 x$

$1$

$x^{4}$

$x^{3}$

$3 x^{3}$

$0 x^{2}$

$3 x^{3}$

$3 x^{2}$

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

$x^{3}$

$3 x^{2}$

$x$

$1$

$x^{4}$

$2 x^{3}$

$0 x^{2}$

$2 x$

$1$

$x^{4}$

$x^{3}$

$3 x^{3}$

$0 x^{2}$

$3 x^{3}$

$3 x^{2}$

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

$x^{3}$

$3 x^{2}$

$x$

$1$

$x^{4}$

$2 x^{3}$

$0 x^{2}$

$2 x$

$1$

$x^{4}$

$x^{3}$

$3 x^{3}$

$0 x^{2}$

$3 x^{3}$

$3 x^{2}$

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

$x^{3}$

$3 x^{2}$

$x$

$1$

$x^{4}$

$2 x^{3}$

$0 x^{2}$

$2 x$

$1$

$x^{4}$

$x^{3}$

$3 x^{3}$

$0 x^{2}$

$3 x^{3}$

$3 x^{2}$

$2 x$

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

$x^{3}$

$3 x^{2}$

$3 x$

$x$

$1$

$x^{4}$

$2 x^{3}$

$0 x^{2}$

$2 x$

$1$

$x^{4}$

$x^{3}$

$3 x^{3}$

$0 x^{2}$

$3 x^{3}$

$3 x^{2}$

$2 x$

Multiply the new quotient term by the divisor.

$x^{3}$

$3 x^{2}$

$3 x$

$x$

$1$

$x^{4}$

$2 x^{3}$

$0 x^{2}$

$2 x$

$1$

$x^{4}$

$x^{3}$

$3 x^{3}$

$0 x^{2}$

$3 x^{3}$

$3 x^{2}$

$2 x$

$3 x^{2}$

$3 x$

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

$x^{3}$

$3 x^{2}$

$3 x$

$x$

$1$

$x^{4}$

$2 x^{3}$

$0 x^{2}$

$2 x$

$1$

$x^{4}$

$x^{3}$

$3 x^{3}$

$0 x^{2}$

$3 x^{3}$

$3 x^{2}$

$2 x$

$3 x^{2}$

$3 x$

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

$x^{3}$

$3 x^{2}$

$3 x$

$x$

$1$

$x^{4}$

$2 x^{3}$

$0 x^{2}$

$2 x$

$1$

$x^{4}$

$x^{3}$

$3 x^{3}$

$0 x^{2}$

$3 x^{3}$

$3 x^{2}$

$2 x$

$3 x^{2}$

$3 x$

$x$

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

$x^{3}$

$3 x^{2}$

$3 x$

$x$

$1$

$x^{4}$

$2 x^{3}$

$0 x^{2}$

$2 x$

$1$

$x^{4}$

$x^{3}$

$3 x^{3}$

$0 x^{2}$

$3 x^{3}$

$3 x^{2}$

$2 x$

$3 x^{2}$

$3 x$

$x$

$1$

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

$x^{3}$

$3 x^{2}$

$3 x$

$1$

$x$

$1$

$x^{4}$

$2 x^{3}$

$0 x^{2}$

$2 x$

$1$

$x^{4}$

$x^{3}$

$3 x^{3}$

$0 x^{2}$

$3 x^{3}$

$3 x^{2}$

$2 x$

$3 x^{2}$

$3 x$

$x$

$1$

Multiply the new quotient term by the divisor.

$x^{3}$

$3 x^{2}$

$3 x$

$1$

$x$

$1$

$x^{4}$

$2 x^{3}$

$0 x^{2}$

$2 x$

$1$

$x^{4}$

$x^{3}$

$3 x^{3}$

$0 x^{2}$

$3 x^{3}$

$3 x^{2}$

$2 x$

$3 x^{2}$

$3 x$

$x$

$1$

$x$

$1$

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

$x^{3}$

$3 x^{2}$

$3 x$

$1$

$x$

$1$

$x^{4}$

$2 x^{3}$

$0 x^{2}$

$2 x$

$1$

$x^{4}$

$x^{3}$

$3 x^{3}$

$0 x^{2}$

$3 x^{3}$

$3 x^{2}$

$2 x$

$3 x^{2}$

$3 x$

$x$

$1$

$x$

$1$

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

$x^{3}$

$3 x^{2}$

$3 x$

$1$

$x$

$1$

$x^{4}$

$2 x^{3}$

$0 x^{2}$

$2 x$

$1$

$x^{4}$

$x^{3}$

$3 x^{3}$

$0 x^{2}$

$3 x^{3}$

$3 x^{2}$

$2 x$

$3 x^{2}$

$3 x$

$x$

$1$

$x$

$1$

$0$

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

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

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

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

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.

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

Factor using the perfect square rule.

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

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

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.

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

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

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

Combine like factors.

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

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

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

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

Add $1$ and $2$ .

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

Remove unnecessary parentheses.

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

Step 2

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

$x - 2 = 0$

$x + 1 = 0$

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

Step 3

Set $x - 2$ equal to $0$ and solve for $x$ .

Set $x - 2$ equal to $0$ .

$x - 2 = 0$

Add $2$ to both sides of the equation.

$x = 2$

Step 4

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

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

$x + 1 = 0$

Subtract $1$ from both sides of the equation.

$x = - 1$

Step 5

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.

$x = 1$

Step 6

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

$x = 2, - 1, 1$

Step 7

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

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