Find the Roots/Zeros Using the Rational Roots Test x^5-2x^4+2x^3-3x+2

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

Step 1

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.

$p = \pm 1, \pm 2$

$q = \pm 1$

Step 2

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

$\pm 1, \pm 2$

Step 3

Substitute the possible roots one by one into the polynomial to find the actual roots. Simplify to check if the value is $0$ , which means it is a root.

${(1)}^{5} - 2 {(1)}^{4} + 2 {(1)}^{3} - 3 \cdot 1 + 2$

Step 4

Simplify the expression. In this case, the expression is equal to $0$ so $x = 1$ is a root of the polynomial.

Simplify each term.

One to any power is one.

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

One to any power is one.

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

Multiply $- 2$ by $1$ .

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

One to any power is one.

$1 - 2 + 2 \cdot 1 - 3 \cdot 1 + 2$

Multiply $2$ by $1$ .

$1 - 2 + 2 - 3 \cdot 1 + 2$

Multiply $- 3$ by $1$ .

$1 - 2 + 2 - 3 + 2$

$1 - 2 + 2 - 3 + 2$

Simplify by adding and subtracting.

Subtract $2$ from $1$ .

$- 1 + 2 - 3 + 2$

Add $- 1$ and $2$ .

$1 - 3 + 2$

Subtract $3$ from $1$ .

$- 2 + 2$

Add $- 2$ and $2$ .

$0$

$0$

$0$

Step 5

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.

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

Step 6

Next, find the roots of the remaining polynomial. The order of the polynomial has been reduced by $1$ .

Place the numbers representing the divisor and the dividend into a division-like configuration.

$1$	$1$	$- 2$	$2$	$0$	$- 3$	$2$
	\xa0					\xa0

The first number in the dividend $(1)$ is put into the first position of the result area (below the horizontal line).

$1$	$1$	$- 2$	$2$	$0$	$- 3$	$2$
	\xa0					\xa0
	$1$

Multiply the newest entry in the result $(1)$ by the divisor $(1)$ and place the result of $(1)$ under the next term in the dividend $(- 2)$ .

$1$	$1$	$- 2$	$2$	$0$	$- 3$	$2$
	\xa0	$1$				\xa0
	$1$

Add the product of the multiplication and the number from the dividend and put the result in the next position on the result line.

$1$	$1$	$- 2$	$2$	$0$	$- 3$	$2$
	\xa0	$1$				\xa0
	$1$	$- 1$

Multiply the newest entry in the result $(- 1)$ by the divisor $(1)$ and place the result of $(- 1)$ under the next term in the dividend $(2)$ .

$1$	$1$	$- 2$	$2$	$0$	$- 3$	$2$
	\xa0	$1$	$- 1$			\xa0
	$1$	$- 1$

Add the product of the multiplication and the number from the dividend and put the result in the next position on the result line.

$1$	$1$	$- 2$	$2$	$0$	$- 3$	$2$
	\xa0	$1$	$- 1$			\xa0
	$1$	$- 1$	$1$

Multiply the newest entry in the result $(1)$ by the divisor $(1)$ and place the result of $(1)$ under the next term in the dividend $(0)$ .

$1$	$1$	$- 2$	$2$	$0$	$- 3$	$2$
	\xa0	$1$	$- 1$	$1$		\xa0
	$1$	$- 1$	$1$

Add the product of the multiplication and the number from the dividend and put the result in the next position on the result line.

$1$	$1$	$- 2$	$2$	$0$	$- 3$	$2$
	\xa0	$1$	$- 1$	$1$		\xa0
	$1$	$- 1$	$1$	$1$

Multiply the newest entry in the result $(1)$ by the divisor $(1)$ and place the result of $(1)$ under the next term in the dividend $(- 3)$ .

$1$	$1$	$- 2$	$2$	$0$	$- 3$	$2$
	\xa0	$1$	$- 1$	$1$	$1$	\xa0
	$1$	$- 1$	$1$	$1$

Add the product of the multiplication and the number from the dividend and put the result in the next position on the result line.

$1$	$1$	$- 2$	$2$	$0$	$- 3$	$2$
	\xa0	$1$	$- 1$	$1$	$1$	\xa0
	$1$	$- 1$	$1$	$1$	$- 2$

Multiply the newest entry in the result $(- 2)$ by the divisor $(1)$ and place the result of $(- 2)$ under the next term in the dividend $(2)$ .

$1$	$1$	$- 2$	$2$	$0$	$- 3$	$2$
	\xa0	$1$	$- 1$	$1$	$1$	$- 2$
	$1$	$- 1$	$1$	$1$	$- 2$

Add the product of the multiplication and the number from the dividend and put the result in the next position on the result line.

$1$	$1$	$- 2$	$2$	$0$	$- 3$	$2$
	\xa0	$1$	$- 1$	$1$	$1$	$- 2$
	$1$	$- 1$	$1$	$1$	$- 2$	$0$

All numbers except the last become the coefficients of the quotient polynomial. The last value in the result line is the remainder.

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

Simplify the quotient polynomial.

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

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

Step 7

Solve the equation to find any remaining roots.

Factor the left side of the equation.

Regroup terms.

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

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

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

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

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

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

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

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

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

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.

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

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

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

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

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

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

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

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

Factor.

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

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

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

$- 1 - 1 + 2$

Subtract $1$ from $- 1$ .

$- 2 + 2$

Add $- 2$ and $2$ .

$0$

$0$

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.

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

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

$x$

+

$1$

$x^{3}$

+

$0 x^{2}$

+

$x$

+

$2$

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

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

Multiply the new quotient term by the divisor.

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

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

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

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

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

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

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

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

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

Multiply the new quotient term by the divisor.

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

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

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

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

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

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

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

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

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

Multiply the new quotient term by the divisor.

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

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

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

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

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

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

$x^{2} - x + 2$

$x^{2} - x + 2$

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

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

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

Remove unnecessary parentheses.

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

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

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

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

$x - 1 = 0$

$x + 1 = 0$

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

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.

$x = 1$

$x = 1$

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.

$x = - 1$

$x = - 1$

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

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

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

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

Use the quadratic formula to find the solutions.

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

Substitute the values $a = 1$ , $b = - 1$ , and $c = 2$ into the quadratic formula and solve for $x$ .

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

Simplify.

Simplify the numerator.

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

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

Multiply $- 4 \cdot 1 \cdot 2$ .

Multiply $- 4$ by $1$ .

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

Multiply $- 4$ by $2$ .

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

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

Subtract $8$ from $1$ .

$x = \frac{1 \pm \sqrt{- 7}}{2 \cdot 1}$

Rewrite $- 7$ as $- 1 (7)$ .

$x = \frac{1 \pm \sqrt{- 1 \cdot 7}}{2 \cdot 1}$

Rewrite $\sqrt{- 1 (7)}$ as $\sqrt{- 1} \cdot \sqrt{7}$ .

$x = \frac{1 \pm \sqrt{- 1} \cdot \sqrt{7}}{2 \cdot 1}$

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

$x = \frac{1 \pm i \sqrt{7}}{2 \cdot 1}$

$x = \frac{1 \pm i \sqrt{7}}{2 \cdot 1}$

Multiply $2$ by $1$ .

$x = \frac{1 \pm i \sqrt{7}}{2}$

$x = \frac{1 \pm i \sqrt{7}}{2}$

The final answer is the combination of both solutions.

$x = \frac{1 + i \sqrt{7}}{2}, \frac{1 - i \sqrt{7}}{2}$

$x = \frac{1 + i \sqrt{7}}{2}, \frac{1 - i \sqrt{7}}{2}$

$x = \frac{1 + i \sqrt{7}}{2}, \frac{1 - i \sqrt{7}}{2}$

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

$x = 1, - 1, \frac{1 + i \sqrt{7}}{2}, \frac{1 - i \sqrt{7}}{2}$

$x = 1, - 1, \frac{1 + i \sqrt{7}}{2}, \frac{1 - i \sqrt{7}}{2}$

Step 8

The polynomial can be written as a set of linear factors.

$(x - 1) (x + 1) (x - \frac{1 + i \sqrt{7}}{2}) (x - \frac{1 - i \sqrt{7}}{2})$

Step 9

These are the roots (zeros) of the polynomial $x^{5} - 2 x^{4} + 2 x^{3} - 3 x + 2$ .

$x = 1, - 1, \frac{1 + i \sqrt{7}}{2}, \frac{1 - i \sqrt{7}}{2}$

Step 10

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