Find the Roots (Zeros) f(x)=x^3-10x^2+27x-12

$f (x) = x^{3} - 10 x^{2} + 27 x - 12$

Step 1

Set $x^{3} - 10 x^{2} + 27 x - 12$ equal to $0$ .

$x^{3} - 10 x^{2} + 27 x - 12 = 0$

Step 2

Solve for $x$ .

Factor $x^{3} - 10 x^{2} + 27 x - 12$ 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.

$p = \pm 1, \pm 12, \pm 2, \pm 6, \pm 3, \pm 4 q = \pm 1$

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

$\pm 1, \pm 12, \pm 2, \pm 6, \pm 3, \pm 4$

Substitute $4$ and simplify the expression. In this case, the expression is equal to $0$ so $4$ is a root of the polynomial.

Substitute $4$ into the polynomial.

$4^{3} - 10 \cdot 4^{2} + 27 \cdot 4 - 12$

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

$64 - 10 \cdot 4^{2} + 27 \cdot 4 - 12$

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

$64 - 10 \cdot 16 + 27 \cdot 4 - 12$

Multiply $- 10$ by $16$ .

$64 - 160 + 27 \cdot 4 - 12$

Subtract $160$ from $64$ .

$- 96 + 27 \cdot 4 - 12$

Multiply $27$ by $4$ .

$- 96 + 108 - 12$

Add $- 96$ and $108$ .

$12 - 12$

Subtract $12$ from $12$ .

$0$

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.

$\frac{x^{3} - 10 x^{2} + 27 x - 12}{x - 4}$

Divide $x^{3} - 10 x^{2} + 27 x - 12$ by $x - 4$ .

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

$x$

$4$

$x^{3}$

$10 x^{2}$

$27 x$

$12$

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

					$x^{2}$
	$x$	-	$4$		$x^{3}$	-	$10 x^{2}$	+	$27 x$	-	$12$

Multiply the new quotient term by the divisor.

				$x^{2}$
$x$	-	$4$		$x^{3}$	-	$10 x^{2}$	+	$27 x$	-	$12$
			+	$x^{3}$	-	$4 x^{2}$

The expression needs to be subtracted from the dividend, so change all the signs in $x^{3} - 4 x^{2}$

				$x^{2}$
$x$	-	$4$		$x^{3}$	-	$10 x^{2}$	+	$27 x$	-	$12$
			-	$x^{3}$	+	$4 x^{2}$

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

				$x^{2}$
$x$	-	$4$		$x^{3}$	-	$10 x^{2}$	+	$27 x$	-	$12$
			-	$x^{3}$	+	$4 x^{2}$
					-	$6 x^{2}$

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

				$x^{2}$
$x$	-	$4$		$x^{3}$	-	$10 x^{2}$	+	$27 x$	-	$12$
			-	$x^{3}$	+	$4 x^{2}$
					-	$6 x^{2}$	+	$27 x$

Divide the highest order term in the dividend $- 6 x^{2}$ by the highest order term in divisor $x$ .

				$x^{2}$	-	$6 x$
$x$	-	$4$		$x^{3}$	-	$10 x^{2}$	+	$27 x$	-	$12$
			-	$x^{3}$	+	$4 x^{2}$
					-	$6 x^{2}$	+	$27 x$

Multiply the new quotient term by the divisor.

				$x^{2}$	-	$6 x$
$x$	-	$4$		$x^{3}$	-	$10 x^{2}$	+	$27 x$	-	$12$
			-	$x^{3}$	+	$4 x^{2}$
					-	$6 x^{2}$	+	$27 x$
					-	$6 x^{2}$	+	$24 x$

The expression needs to be subtracted from the dividend, so change all the signs in $- 6 x^{2} + 24 x$

				$x^{2}$	-	$6 x$
$x$	-	$4$		$x^{3}$	-	$10 x^{2}$	+	$27 x$	-	$12$
			-	$x^{3}$	+	$4 x^{2}$
					-	$6 x^{2}$	+	$27 x$
					+	$6 x^{2}$	-	$24 x$

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

				$x^{2}$	-	$6 x$
$x$	-	$4$		$x^{3}$	-	$10 x^{2}$	+	$27 x$	-	$12$
			-	$x^{3}$	+	$4 x^{2}$
					-	$6 x^{2}$	+	$27 x$
					+	$6 x^{2}$	-	$24 x$
							+	$3 x$

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

				$x^{2}$	-	$6 x$
$x$	-	$4$		$x^{3}$	-	$10 x^{2}$	+	$27 x$	-	$12$
			-	$x^{3}$	+	$4 x^{2}$
					-	$6 x^{2}$	+	$27 x$
					+	$6 x^{2}$	-	$24 x$
							+	$3 x$	-	$12$

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

				$x^{2}$	-	$6 x$	+	$3$
$x$	-	$4$		$x^{3}$	-	$10 x^{2}$	+	$27 x$	-	$12$
			-	$x^{3}$	+	$4 x^{2}$
					-	$6 x^{2}$	+	$27 x$
					+	$6 x^{2}$	-	$24 x$
							+	$3 x$	-	$12$

Multiply the new quotient term by the divisor.

				$x^{2}$	-	$6 x$	+	$3$
$x$	-	$4$		$x^{3}$	-	$10 x^{2}$	+	$27 x$	-	$12$
			-	$x^{3}$	+	$4 x^{2}$
					-	$6 x^{2}$	+	$27 x$
					+	$6 x^{2}$	-	$24 x$
							+	$3 x$	-	$12$
							+	$3 x$	-	$12$

The expression needs to be subtracted from the dividend, so change all the signs in $3 x - 12$

				$x^{2}$	-	$6 x$	+	$3$
$x$	-	$4$		$x^{3}$	-	$10 x^{2}$	+	$27 x$	-	$12$
			-	$x^{3}$	+	$4 x^{2}$
					-	$6 x^{2}$	+	$27 x$
					+	$6 x^{2}$	-	$24 x$
							+	$3 x$	-	$12$
							-	$3 x$	+	$12$

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

				$x^{2}$	-	$6 x$	+	$3$
$x$	-	$4$		$x^{3}$	-	$10 x^{2}$	+	$27 x$	-	$12$
			-	$x^{3}$	+	$4 x^{2}$
					-	$6 x^{2}$	+	$27 x$
					+	$6 x^{2}$	-	$24 x$
							+	$3 x$	-	$12$
							-	$3 x$	+	$12$
										$0$

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

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

Write $x^{3} - 10 x^{2} + 27 x - 12$ as a set of factors.

$(x - 4) (x^{2} - 6 x + 3) = 0$

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

$x - 4 = 0$

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

Set $x - 4$ equal to $0$ and solve for $x$ .

Set $x - 4$ equal to $0$ .

$x - 4 = 0$

Add $4$ to both sides of the equation.

$x = 4$

Set $x^{2} - 6 x + 3$ equal to $0$ and solve for $x$ .

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

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

Solve $x^{2} - 6 x + 3 = 0$ for $x$ .

Use the quadratic formula to find the solutions.

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

Substitute the values $a = 1$ , $b = - 6$ , and $c = 3$ into the quadratic formula and solve for $x$ .

$\frac{6 \pm \sqrt{{(- 6)}^{2} - 4 \cdot (1 \cdot 3)}}{2 \cdot 1}$

Simplify.

Simplify the numerator.

Raise $- 6$ to the power of $2$ .

$x = \frac{6 \pm \sqrt{36 - 4 \cdot 1 \cdot 3}}{2 \cdot 1}$

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

Multiply $- 4$ by $1$ .

$x = \frac{6 \pm \sqrt{36 - 4 \cdot 3}}{2 \cdot 1}$

Multiply $- 4$ by $3$ .

$x = \frac{6 \pm \sqrt{36 - 12}}{2 \cdot 1}$

Subtract $12$ from $36$ .

$x = \frac{6 \pm \sqrt{24}}{2 \cdot 1}$

Rewrite $24$ as $2^{2} \cdot 6$ .

Factor $4$ out of $24$ .

$x = \frac{6 \pm \sqrt{4 (6)}}{2 \cdot 1}$

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

$x = \frac{6 \pm \sqrt{2^{2} \cdot 6}}{2 \cdot 1}$

Pull terms out from under the radical.

$x = \frac{6 \pm 2 \sqrt{6}}{2 \cdot 1}$

Multiply $2$ by $1$ .

$x = \frac{6 \pm 2 \sqrt{6}}{2}$

Simplify $\frac{6 \pm 2 \sqrt{6}}{2}$ .

$x = 3 \pm \sqrt{6}$

The final answer is the combination of both solutions.

$x = 3 + \sqrt{6}, 3 - \sqrt{6}$

The final solution is all the values that make $(x - 4) (x^{2} - 6 x + 3) = 0$ true.

$x = 4, 3 + \sqrt{6}, 3 - \sqrt{6}$

Step 3

The result can be shown in multiple forms.

Exact Form:

$x = 4, 3 + \sqrt{6}, 3 - \sqrt{6}$

Decimal Form:

$x = 4, 5.44948974 \dots, 0.55051025 \dots$

Step 4

Find the Roots (Zeros) f(x)=x^3-10x^2+27x-12

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