Find the Roots/Zeros Using the Rational Roots Test x^3-6x^2+13x-20

$x^{3} - 6 x^{2} + 13 x - 20$

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.

$p = \pm 1, \pm 2, \pm 4, \pm 5, \pm 10, \pm 20$

$q = \pm 1$

Step 2

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

$\pm 1, \pm 2, \pm 4, \pm 5, \pm 10, \pm 20$

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.

${(4)}^{3} - 6 {(4)}^{2} + 13 (4) - 20$

Step 4

Simplify the expression. In this case, the expression is equal to $0$ so $x = 4$ is a root of the polynomial.

Simplify each term.

Raise $4$ to the power of $3$ .

$64 - 6 {(4)}^{2} + 13 (4) - 20$

Raise $4$ to the power of $2$ .

$64 - 6 \cdot 16 + 13 (4) - 20$

Multiply $- 6$ by $16$ .

$64 - 96 + 13 (4) - 20$

Multiply $13$ by $4$ .

$64 - 96 + 52 - 20$

Simplify by adding and subtracting.

Subtract $96$ from $64$ .

$- 32 + 52 - 20$

Add $- 32$ and $52$ .

$20 - 20$

Subtract $20$ from $20$ .

$0$

Step 5

Since $4$ is a known root, divide the polynomial by $x - 4$ to find the quotient polynomial. This polynomial can then be used to find the remaining roots.

$\frac{x^{3} - 6 x^{2} + 13 x - 20}{x - 4}$

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.

$4$	$1$	$- 6$	$13$	$- 20$
	\xa0			\xa0

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

$4$	$1$	$- 6$	$13$	$- 20$
	\xa0			\xa0
	$1$

Multiply the newest entry in the result $(1)$ by the divisor $(4)$ and place the result of $(4)$ under the next term in the dividend $(- 6)$ .

$4$	$1$	$- 6$	$13$	$- 20$
	\xa0	$4$		\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.

$4$	$1$	$- 6$	$13$	$- 20$
	\xa0	$4$		\xa0
	$1$	$- 2$

Multiply the newest entry in the result $(- 2)$ by the divisor $(4)$ and place the result of $(- 8)$ under the next term in the dividend $(13)$ .

$4$	$1$	$- 6$	$13$	$- 20$
	\xa0	$4$	$- 8$	\xa0
	$1$	$- 2$

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

$4$	$1$	$- 6$	$13$	$- 20$
	\xa0	$4$	$- 8$	\xa0
	$1$	$- 2$	$5$

Multiply the newest entry in the result $(5)$ by the divisor $(4)$ and place the result of $(20)$ under the next term in the dividend $(- 20)$ .

$4$	$1$	$- 6$	$13$	$- 20$
	\xa0	$4$	$- 8$	$20$
	$1$	$- 2$	$5$

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

$4$	$1$	$- 6$	$13$	$- 20$
	\xa0	$4$	$- 8$	$20$
	$1$	$- 2$	$5$	$0$

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

$1 x^{2} + - 2 x + 5$

Simplify the quotient polynomial.

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

Step 7

Solve the equation to find any remaining roots.

Use the quadratic formula to find the solutions.

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

Substitute the values $a = 1$ , $b = - 2$ , and $c = 5$ into the quadratic formula and solve for $x$ .

$\frac{2 \pm \sqrt{{(- 2)}^{2} - 4 \cdot (1 \cdot 5)}}{2 \cdot 1}$

Simplify.

Simplify the numerator.

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

$x = \frac{2 \pm \sqrt{4 - 4 \cdot 1 \cdot 5}}{2 \cdot 1}$

Multiply $- 4 \cdot 1 \cdot 5$ .

Multiply $- 4$ by $1$ .

$x = \frac{2 \pm \sqrt{4 - 4 \cdot 5}}{2 \cdot 1}$

Multiply $- 4$ by $5$ .

$x = \frac{2 \pm \sqrt{4 - 20}}{2 \cdot 1}$

Subtract $20$ from $4$ .

$x = \frac{2 \pm \sqrt{- 16}}{2 \cdot 1}$

Rewrite $- 16$ as $- 1 (16)$ .

$x = \frac{2 \pm \sqrt{- 1 \cdot 16}}{2 \cdot 1}$

Rewrite $\sqrt{- 1 (16)}$ as $\sqrt{- 1} \cdot \sqrt{16}$ .

$x = \frac{2 \pm \sqrt{- 1} \cdot \sqrt{16}}{2 \cdot 1}$

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

$x = \frac{2 \pm i \cdot \sqrt{16}}{2 \cdot 1}$

Rewrite $16$ as $4^{2}$ .

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

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

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

Move $4$ to the left of $i$ .

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

Multiply $2$ by $1$ .

$x = \frac{2 \pm 4 i}{2}$

Simplify $\frac{2 \pm 4 i}{2}$ .

$x = 1 \pm 2 i$

The final answer is the combination of both solutions.

$x = 1 + 2 i, 1 - 2 i$

Step 8

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

$(x - 4) (x - 1 - 2 i) (x - 1 + 2 i)$

Step 9

These are the roots (zeros) of the polynomial $x^{3} - 6 x^{2} + 13 x - 20$ .

$x = 4, 1 + 2 i, 1 - 2 i$

Step 10

Find the Roots/Zeros Using the Rational Roots Test x^3-6x^2+13x-20

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