Find the Roots/Zeros Using the Rational Roots Test x^3-x^2+x-1

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

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.

$p = \pm 1$

$q = \pm 1$

Step 2

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

$\pm 1$

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.

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

Step 4

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

Remove parentheses.

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

Simplify each term.

One to any power is one.

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

One to any power is one.

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

Multiply $- 1$ by $1$ .

$1 - 1 + 1 - 1$

Simplify by adding and subtracting.

Subtract $1$ from $1$ .

$0 + 1 - 1$

Add $0$ and $1$ .

$1 - 1$

Subtract $1$ from $1$ .

$0$

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.

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

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.

$1$	$1$	$- 1$	$1$	$- 1$
	\xa0			\xa0

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

$1$	$1$	$- 1$	$1$	$- 1$
	\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 $(- 1)$ .

$1$	$1$	$- 1$	$1$	$- 1$
	\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.

$1$	$1$	$- 1$	$1$	$- 1$
	\xa0	$1$		\xa0
	$1$	$0$

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

$1$	$1$	$- 1$	$1$	$- 1$
	\xa0	$1$	$0$	\xa0
	$1$	$0$

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

$1$	$1$	$- 1$	$1$	$- 1$
	\xa0	$1$	$0$	\xa0
	$1$	$0$	$1$

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 $(- 1)$ .

$1$	$1$	$- 1$	$1$	$- 1$
	\xa0	$1$	$0$	$1$
	$1$	$0$	$1$

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

$1$	$1$	$- 1$	$1$	$- 1$
	\xa0	$1$	$0$	$1$
	$1$	$0$	$1$	$0$

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

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

Simplify the quotient polynomial.

$x^{2} + 1$

Step 7

Solve the equation to find any remaining roots.

Subtract $1$ from both sides of the equation.

$x^{2} = - 1$

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

$x = \pm \sqrt{- 1}$

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

$x = \pm i$

The complete solution is the result of both the positive and negative portions of the solution.

First, use the positive value of the $\pm$ to find the first solution.

$x = i$

Next, use the negative value of the $\pm$ to find the second solution.

$x = - i$

The complete solution is the result of both the positive and negative portions of the solution.

$x = i, - i$

Step 8

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

$(x - 1) (x - i) (x + i)$

Step 9

These are the roots (zeros) of the polynomial $x^{3} - x^{2} + x - 1$ .

$x = 1, i, - i$

Step 10

Find the Roots/Zeros Using the Rational Roots Test x^3-x^2+x-1

Related Questions

Support LCMGCF.com by disabling your adblocker.