Find the Roots/Zeros Using the Rational Roots Test f(x)=x^3-7x^2+13x-6

$f (x) = x^{3} - 7 x^{2} + 13 x - 6$

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.

$p = \pm 1, \pm 2, \pm 3, \pm 6$

$q = \pm 1$

Step 2

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

$\pm 1, \pm 2, \pm 3, \pm 6$

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.

${(2)}^{3} - 7 {(2)}^{2} + 13 (2) - 6$

Step 4

Simplify the expression. In this case, the expression is equal to $0$ so $x = 2$ is a root of the polynomial.

Simplify each term.

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

$8 - 7 {(2)}^{2} + 13 (2) - 6$

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

$8 - 7 \cdot 4 + 13 (2) - 6$

Multiply $- 7$ by $4$ .

$8 - 28 + 13 (2) - 6$

Multiply $13$ by $2$ .

$8 - 28 + 26 - 6$

Simplify by adding and subtracting.

Subtract $28$ from $8$ .

$- 20 + 26 - 6$

Add $- 20$ and $26$ .

$6 - 6$

Subtract $6$ from $6$ .

$0$

Step 5

Since $2$ is a known root, divide the polynomial by $x - 2$ to find the quotient polynomial. This polynomial can then be used to find the remaining roots.

$\frac{x^{3} - 7 x^{2} + 13 x - 6}{x - 2}$

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.

$2$	$1$	$- 7$	$13$	$- 6$
	\xa0			\xa0

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

$2$	$1$	$- 7$	$13$	$- 6$
	\xa0			\xa0
	$1$

Multiply the newest entry in the result $(1)$ by the divisor $(2)$ and place the result of $(2)$ under the next term in the dividend $(- 7)$ .

$2$	$1$	$- 7$	$13$	$- 6$
	\xa0	$2$		\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.

$2$	$1$	$- 7$	$13$	$- 6$
	\xa0	$2$		\xa0
	$1$	$- 5$

Multiply the newest entry in the result $(- 5)$ by the divisor $(2)$ and place the result of $(- 10)$ under the next term in the dividend $(13)$ .

$2$	$1$	$- 7$	$13$	$- 6$
	\xa0	$2$	$- 10$	\xa0
	$1$	$- 5$

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

$2$	$1$	$- 7$	$13$	$- 6$
	\xa0	$2$	$- 10$	\xa0
	$1$	$- 5$	$3$

Multiply the newest entry in the result $(3)$ by the divisor $(2)$ and place the result of $(6)$ under the next term in the dividend $(- 6)$ .

$2$	$1$	$- 7$	$13$	$- 6$
	\xa0	$2$	$- 10$	$6$
	$1$	$- 5$	$3$

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

$2$	$1$	$- 7$	$13$	$- 6$
	\xa0	$2$	$- 10$	$6$
	$1$	$- 5$	$3$	$0$

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

$1 x^{2} + - 5 x + 3$

Simplify the quotient polynomial.

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

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 $a = 1$ , $b = - 5$ , and $c = 3$ into the quadratic formula and solve for $x$ .

$\frac{5 \pm \sqrt{{(- 5)}^{2} - 4 \cdot (1 \cdot 3)}}{2 \cdot 1}$

Simplify.

Simplify the numerator.

Raise $- 5$ to the power of $2$ .

$x = \frac{5 \pm \sqrt{25 - 4 \cdot 1 \cdot 3}}{2 \cdot 1}$

Multiply $- 4 \cdot 1 \cdot 3$ .

Multiply $- 4$ by $1$ .

$x = \frac{5 \pm \sqrt{25 - 4 \cdot 3}}{2 \cdot 1}$

Multiply $- 4$ by $3$ .

$x = \frac{5 \pm \sqrt{25 - 12}}{2 \cdot 1}$

Subtract $12$ from $25$ .

$x = \frac{5 \pm \sqrt{13}}{2 \cdot 1}$

Multiply $2$ by $1$ .

$x = \frac{5 \pm \sqrt{13}}{2}$

Simplify the expression to solve for the $+$ portion of the $\pm$ .

Simplify the numerator.

Raise $- 5$ to the power of $2$ .

$x = \frac{5 \pm \sqrt{25 - 4 \cdot 1 \cdot 3}}{2 \cdot 1}$

Multiply $- 4 \cdot 1 \cdot 3$ .

Multiply $- 4$ by $1$ .

$x = \frac{5 \pm \sqrt{25 - 4 \cdot 3}}{2 \cdot 1}$

Multiply $- 4$ by $3$ .

$x = \frac{5 \pm \sqrt{25 - 12}}{2 \cdot 1}$

Subtract $12$ from $25$ .

$x = \frac{5 \pm \sqrt{13}}{2 \cdot 1}$

Multiply $2$ by $1$ .

$x = \frac{5 \pm \sqrt{13}}{2}$

Change the $\pm$ to $+$ .

$x = \frac{5 + \sqrt{13}}{2}$

Simplify the expression to solve for the $-$ portion of the $\pm$ .

Simplify the numerator.

Raise $- 5$ to the power of $2$ .

$x = \frac{5 \pm \sqrt{25 - 4 \cdot 1 \cdot 3}}{2 \cdot 1}$

Multiply $- 4 \cdot 1 \cdot 3$ .

Multiply $- 4$ by $1$ .

$x = \frac{5 \pm \sqrt{25 - 4 \cdot 3}}{2 \cdot 1}$

Multiply $- 4$ by $3$ .

$x = \frac{5 \pm \sqrt{25 - 12}}{2 \cdot 1}$

Subtract $12$ from $25$ .

$x = \frac{5 \pm \sqrt{13}}{2 \cdot 1}$

Multiply $2$ by $1$ .

$x = \frac{5 \pm \sqrt{13}}{2}$

Change the $\pm$ to $-$ .

$x = \frac{5 - \sqrt{13}}{2}$

The final answer is the combination of both solutions.

$x = \frac{5 + \sqrt{13}}{2}, \frac{5 - \sqrt{13}}{2}$

Step 8

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

$(x - 2) (x - \frac{5 + \sqrt{13}}{2}) (x - \frac{5 - \sqrt{13}}{2})$

Step 9

These are the roots (zeros) of the polynomial $x^{3} - 7 x^{2} + 13 x - 6$ .

$x = 2, \frac{5 + \sqrt{13}}{2}, \frac{5 - \sqrt{13}}{2}$

Step 10

The result can be shown in multiple forms.

Exact Form:

$x = 2, \frac{5 + \sqrt{13}}{2}, \frac{5 - \sqrt{13}}{2}$

Decimal Form:

$x = 2, 4.30277563 \dots, 0.69722436 \dots$

Step 11

Find the Roots/Zeros Using the Rational Roots Test f(x)=x^3-7x^2+13x-6

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