Find the Roots/Zeros Using the Rational Roots Test x^3-3x^2+81x-243

$x^{3} - 3 x^{2} + 81 x - 243$

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.

$p = \pm 1, \pm 3, \pm 9, \pm 27, \pm 81, \pm 243$

$q = \pm 1$

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

$\pm 1, \pm 3, \pm 9, \pm 27, \pm 81, \pm 243$

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.

${(3)}^{3} - 3 {(3)}^{2} + 81 (3) - 243$

Simplify the expression. In this case, the expression is equal to $0$ so $x = 3$ is a root of the polynomial.

Simplify each term.

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

$27 - 3 {(3)}^{2} + 81 (3) - 243$

Raise $3$ to the power of $2$ .

$27 - 3 \cdot 9 + 81 (3) - 243$

Multiply $- 3$ by $9$ .

$27 - 27 + 81 (3) - 243$

Multiply $81$ by $3$ .

$27 - 27 + 243 - 243$

Simplify by adding and subtracting.

Subtract $27$ from $27$ .

$0 + 243 - 243$

Add $0$ and $243$ .

$243 - 243$

Subtract $243$ from $243$ .

$0$

Since $3$ is a known root, divide the polynomial by $x - 3$ to find the quotient polynomial. This polynomial can then be used to find the remaining roots.

$\frac{x^{3} - 3 x^{2} + 81 x - 243}{x - 3}$

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.

$3$	$1$	$- 3$	$81$	$- 243$
	\xa0			\xa0

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

$3$	$1$	$- 3$	$81$	$- 243$
	\xa0			\xa0
	$1$

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

$3$	$1$	$- 3$	$81$	$- 243$
	\xa0	$3$		\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.

$3$	$1$	$- 3$	$81$	$- 243$
	\xa0	$3$		\xa0
	$1$	$0$

Multiply the newest entry in the result $(0)$ by the divisor $(3)$ and place the result of $(0)$ under the next term in the dividend $(81)$ .

$3$	$1$	$- 3$	$81$	$- 243$
	\xa0	$3$	$0$	\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.

$3$	$1$	$- 3$	$81$	$- 243$
	\xa0	$3$	$0$	\xa0
	$1$	$0$	$81$

Multiply the newest entry in the result $(81)$ by the divisor $(3)$ and place the result of $(243)$ under the next term in the dividend $(- 243)$ .

$3$	$1$	$- 3$	$81$	$- 243$
	\xa0	$3$	$0$	$243$
	$1$	$0$	$81$

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

$3$	$1$	$- 3$	$81$	$- 243$
	\xa0	$3$	$0$	$243$
	$1$	$0$	$81$	$0$

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

$1 x^{2} + 0 x + 81$

Simplify the quotient polynomial.

$x^{2} + 81$

Solve the equation to find any remaining roots.

Subtract $81$ from both sides of the equation.

$x^{2} = - 81$

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

$x = \pm \sqrt{- 81}$

Simplify $\pm \sqrt{- 81}$ .

Rewrite $- 81$ as $- 1 (81)$ .

$x = \pm \sqrt{- 1 (81)}$

Rewrite $\sqrt{- 1 (81)}$ as $\sqrt{- 1} \cdot \sqrt{81}$ .

$x = \pm \sqrt{- 1} \cdot \sqrt{81}$

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

$x = \pm i \cdot \sqrt{81}$

Rewrite $81$ as $9^{2}$ .

$x = \pm i \cdot \sqrt{9^{2}}$

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

$x = \pm i \cdot 9$

Move $9$ to the left of $i$ .

$x = \pm 9 i$

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.

$x = 9 i$

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

$x = - 9 i$

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

$x = 9 i, - 9 i$

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

$(x - 3) (x - 9 i) (x + 9 i)$

These are the roots (zeros) of the polynomial $x^{3} - 3 x^{2} + 81 x - 243$ .

$x = 3, 9 i, - 9 i$

Find the Roots/Zeros Using the Rational Roots Test x^3-3x^2+81x-243

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