Find the Roots (Zeros) x^3-7x^2+13x-6

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

Step 1

Set $x^{3} - 7 x^{2} + 13 x - 6$ equal to $0$ .

$x^{3} - 7 x^{2} + 13 x - 6 = 0$

Step 2

Solve for $x$ .

Factor $x^{3} - 7 x^{2} + 13 x - 6$ 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.

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

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

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

Substitute $2$ and simplify the expression. In this case, the expression is equal to $0$ so $2$ is a root of the polynomial.

Substitute $2$ into the polynomial.

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

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

$8 - 7 \cdot 2^{2} + 13 \cdot 2 - 6$

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

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

Multiply $- 7$ by $4$ .

$8 - 28 + 13 \cdot 2 - 6$

Subtract $28$ from $8$ .

$- 20 + 13 \cdot 2 - 6$

Multiply $13$ by $2$ .

$- 20 + 26 - 6$

Add $- 20$ and $26$ .

$6 - 6$

Subtract $6$ from $6$ .

$0$

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.

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

Divide $x^{3} - 7 x^{2} + 13 x - 6$ by $x - 2$ .

Set up the polynomials to be divided. If there is not a term for every exponent, insert one with a value of $0$ .

$x$

$2$

$x^{3}$

$7 x^{2}$

$13 x$

$6$

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

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

Multiply the new quotient term by the divisor.

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

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

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

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

				$x^{2}$
$x$	-	$2$		$x^{3}$	-	$7 x^{2}$	+	$13 x$	-	$6$
			-	$x^{3}$	+	$2 x^{2}$
					-	$5 x^{2}$

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

				$x^{2}$
$x$	-	$2$		$x^{3}$	-	$7 x^{2}$	+	$13 x$	-	$6$
			-	$x^{3}$	+	$2 x^{2}$
					-	$5 x^{2}$	+	$13 x$

Divide the highest order term in the dividend $- 5 x^{2}$ by the highest order term in divisor $x$ .

				$x^{2}$	-	$5 x$
$x$	-	$2$		$x^{3}$	-	$7 x^{2}$	+	$13 x$	-	$6$
			-	$x^{3}$	+	$2 x^{2}$
					-	$5 x^{2}$	+	$13 x$

Multiply the new quotient term by the divisor.

				$x^{2}$	-	$5 x$
$x$	-	$2$		$x^{3}$	-	$7 x^{2}$	+	$13 x$	-	$6$
			-	$x^{3}$	+	$2 x^{2}$
					-	$5 x^{2}$	+	$13 x$
					-	$5 x^{2}$	+	$10 x$

The expression needs to be subtracted from the dividend, so change all the signs in $- 5 x^{2} + 10 x$

				$x^{2}$	-	$5 x$
$x$	-	$2$		$x^{3}$	-	$7 x^{2}$	+	$13 x$	-	$6$
			-	$x^{3}$	+	$2 x^{2}$
					-	$5 x^{2}$	+	$13 x$
					+	$5 x^{2}$	-	$10 x$

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

				$x^{2}$	-	$5 x$
$x$	-	$2$		$x^{3}$	-	$7 x^{2}$	+	$13 x$	-	$6$
			-	$x^{3}$	+	$2 x^{2}$
					-	$5 x^{2}$	+	$13 x$
					+	$5 x^{2}$	-	$10 x$
							+	$3 x$

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

				$x^{2}$	-	$5 x$
$x$	-	$2$		$x^{3}$	-	$7 x^{2}$	+	$13 x$	-	$6$
			-	$x^{3}$	+	$2 x^{2}$
					-	$5 x^{2}$	+	$13 x$
					+	$5 x^{2}$	-	$10 x$
							+	$3 x$	-	$6$

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

				$x^{2}$	-	$5 x$	+	$3$
$x$	-	$2$		$x^{3}$	-	$7 x^{2}$	+	$13 x$	-	$6$
			-	$x^{3}$	+	$2 x^{2}$
					-	$5 x^{2}$	+	$13 x$
					+	$5 x^{2}$	-	$10 x$
							+	$3 x$	-	$6$

Multiply the new quotient term by the divisor.

				$x^{2}$	-	$5 x$	+	$3$
$x$	-	$2$		$x^{3}$	-	$7 x^{2}$	+	$13 x$	-	$6$
			-	$x^{3}$	+	$2 x^{2}$
					-	$5 x^{2}$	+	$13 x$
					+	$5 x^{2}$	-	$10 x$
							+	$3 x$	-	$6$
							+	$3 x$	-	$6$

The expression needs to be subtracted from the dividend, so change all the signs in $3 x - 6$

				$x^{2}$	-	$5 x$	+	$3$
$x$	-	$2$		$x^{3}$	-	$7 x^{2}$	+	$13 x$	-	$6$
			-	$x^{3}$	+	$2 x^{2}$
					-	$5 x^{2}$	+	$13 x$
					+	$5 x^{2}$	-	$10 x$
							+	$3 x$	-	$6$
							-	$3 x$	+	$6$

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

				$x^{2}$	-	$5 x$	+	$3$
$x$	-	$2$		$x^{3}$	-	$7 x^{2}$	+	$13 x$	-	$6$
			-	$x^{3}$	+	$2 x^{2}$
					-	$5 x^{2}$	+	$13 x$
					+	$5 x^{2}$	-	$10 x$
							+	$3 x$	-	$6$
							-	$3 x$	+	$6$
										$0$

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

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

Write $x^{3} - 7 x^{2} + 13 x - 6$ as a set of factors.

$(x - 2) (x^{2} - 5 x + 3) = 0$

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

$x - 2 = 0$

$x^{2} - 5 x + 3 = 0$

Set $x - 2$ equal to $0$ and solve for $x$ .

Set $x - 2$ equal to $0$ .

$x - 2 = 0$

Add $2$ to both sides of the equation.

$x = 2$

Set $x^{2} - 5 x + 3$ equal to $0$ and solve for $x$ .

Set $x^{2} - 5 x + 3$ equal to $0$ .

$x^{2} - 5 x + 3 = 0$

Solve $x^{2} - 5 x + 3 = 0$ for $x$ .

Use the quadratic formula to find the solutions.

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

Substitute the values $a = 1$ , $b = - 5$ , 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 $1$ .

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

The final answer is the combination of both solutions.

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

The final solution is all the values that make $(x - 2) (x^{2} - 5 x + 3) = 0$ true.

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

Step 3

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 4

Find the Roots (Zeros) x^3-7x^2+13x-6

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