Find the Roots (Zeros) f(X)=2x^3-24x^2+44x-40

$f (X) = 2 x^{3} - 24 x^{2} + 44 x - 40$

Set $2 x^{3} - 24 x^{2} + 44 x - 40$ equal to $0$ .

$2 x^{3} - 24 x^{2} + 44 x - 40 = 0$

Solve for $x$ .

Factor the left side of the equation.

Factor $2$ out of $2 x^{3} - 24 x^{2} + 44 x - 40$ .

Factor $2$ out of $2 x^{3}$ .

$2 (x^{3}) - 24 x^{2} + 44 x - 40 = 0$

Factor $2$ out of $- 24 x^{2}$ .

$2 (x^{3}) + 2 (- 12 x^{2}) + 44 x - 40 = 0$

Factor $2$ out of $44 x$ .

$2 (x^{3}) + 2 (- 12 x^{2}) + 2 (22 x) - 40 = 0$

Factor $2$ out of $- 40$ .

$2 x^{3} + 2 (- 12 x^{2}) + 2 (22 x) + 2 \cdot - 20 = 0$

Factor $2$ out of $2 x^{3} + 2 (- 12 x^{2})$ .

$2 (x^{3} - 12 x^{2}) + 2 (22 x) + 2 \cdot - 20 = 0$

Factor $2$ out of $2 (x^{3} - 12 x^{2}) + 2 (22 x)$ .

$2 (x^{3} - 12 x^{2} + 22 x) + 2 \cdot - 20 = 0$

Factor $2$ out of $2 (x^{3} - 12 x^{2} + 22 x) + 2 \cdot - 20$ .

$2 (x^{3} - 12 x^{2} + 22 x - 20) = 0$

Factor.

Factor $x^{3} - 12 x^{2} + 22 x - 20$ 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.

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

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

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

Substitute $10$ and simplify the expression. In this case, the expression is equal to $0$ so $10$ is a root of the polynomial.

Substitute $10$ into the polynomial.

$10^{3} - 12 \cdot 10^{2} + 22 \cdot 10 - 20$

Raise $10$ to the power of $3$ .

$1000 - 12 \cdot 10^{2} + 22 \cdot 10 - 20$

Raise $10$ to the power of $2$ .

$1000 - 12 \cdot 100 + 22 \cdot 10 - 20$

Multiply $- 12$ by $100$ .

$1000 - 1200 + 22 \cdot 10 - 20$

Subtract $1200$ from $1000$ .

$- 200 + 22 \cdot 10 - 20$

Multiply $22$ by $10$ .

$- 200 + 220 - 20$

Add $- 200$ and $220$ .

$20 - 20$

Subtract $20$ from $20$ .

$0$

Since $10$ is a known root, divide the polynomial by $x - 10$ to find the quotient polynomial. This polynomial can then be used to find the remaining roots.

$\frac{x^{3} - 12 x^{2} + 22 x - 20}{x - 10}$

Divide $x^{3} - 12 x^{2} + 22 x - 20$ by $x - 10$ .

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

$x$

$10$

$x^{3}$

$12 x^{2}$

$22 x$

$20$

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

					$x^{2}$
	$x$	-	$10$		$x^{3}$	-	$12 x^{2}$	+	$22 x$	-	$20$

Multiply the new quotient term by the divisor.

				$x^{2}$
$x$	-	$10$		$x^{3}$	-	$12 x^{2}$	+	$22 x$	-	$20$
			+	$x^{3}$	-	$10 x^{2}$

The expression needs to be subtracted from the dividend, so change all the signs in $x^{3} - 10 x^{2}$

				$x^{2}$
$x$	-	$10$		$x^{3}$	-	$12 x^{2}$	+	$22 x$	-	$20$
			-	$x^{3}$	+	$10 x^{2}$

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

				$x^{2}$
$x$	-	$10$		$x^{3}$	-	$12 x^{2}$	+	$22 x$	-	$20$
			-	$x^{3}$	+	$10 x^{2}$
					-	$2 x^{2}$

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

				$x^{2}$
$x$	-	$10$		$x^{3}$	-	$12 x^{2}$	+	$22 x$	-	$20$
			-	$x^{3}$	+	$10 x^{2}$
					-	$2 x^{2}$	+	$22 x$

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

				$x^{2}$	-	$2 x$
$x$	-	$10$		$x^{3}$	-	$12 x^{2}$	+	$22 x$	-	$20$
			-	$x^{3}$	+	$10 x^{2}$
					-	$2 x^{2}$	+	$22 x$

Multiply the new quotient term by the divisor.

				$x^{2}$	-	$2 x$
$x$	-	$10$		$x^{3}$	-	$12 x^{2}$	+	$22 x$	-	$20$
			-	$x^{3}$	+	$10 x^{2}$
					-	$2 x^{2}$	+	$22 x$
					-	$2 x^{2}$	+	$20 x$

The expression needs to be subtracted from the dividend, so change all the signs in $- 2 x^{2} + 20 x$

				$x^{2}$	-	$2 x$
$x$	-	$10$		$x^{3}$	-	$12 x^{2}$	+	$22 x$	-	$20$
			-	$x^{3}$	+	$10 x^{2}$
					-	$2 x^{2}$	+	$22 x$
					+	$2 x^{2}$	-	$20 x$

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

				$x^{2}$	-	$2 x$
$x$	-	$10$		$x^{3}$	-	$12 x^{2}$	+	$22 x$	-	$20$
			-	$x^{3}$	+	$10 x^{2}$
					-	$2 x^{2}$	+	$22 x$
					+	$2 x^{2}$	-	$20 x$
							+	$2 x$

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

				$x^{2}$	-	$2 x$
$x$	-	$10$		$x^{3}$	-	$12 x^{2}$	+	$22 x$	-	$20$
			-	$x^{3}$	+	$10 x^{2}$
					-	$2 x^{2}$	+	$22 x$
					+	$2 x^{2}$	-	$20 x$
							+	$2 x$	-	$20$

Divide the highest order term in the dividend $2 x$ by the highest order term in divisor $x$ .

				$x^{2}$	-	$2 x$	+	$2$
$x$	-	$10$		$x^{3}$	-	$12 x^{2}$	+	$22 x$	-	$20$
			-	$x^{3}$	+	$10 x^{2}$
					-	$2 x^{2}$	+	$22 x$
					+	$2 x^{2}$	-	$20 x$
							+	$2 x$	-	$20$

Multiply the new quotient term by the divisor.

				$x^{2}$	-	$2 x$	+	$2$
$x$	-	$10$		$x^{3}$	-	$12 x^{2}$	+	$22 x$	-	$20$
			-	$x^{3}$	+	$10 x^{2}$
					-	$2 x^{2}$	+	$22 x$
					+	$2 x^{2}$	-	$20 x$
							+	$2 x$	-	$20$
							+	$2 x$	-	$20$

The expression needs to be subtracted from the dividend, so change all the signs in $2 x - 20$

				$x^{2}$	-	$2 x$	+	$2$
$x$	-	$10$		$x^{3}$	-	$12 x^{2}$	+	$22 x$	-	$20$
			-	$x^{3}$	+	$10 x^{2}$
					-	$2 x^{2}$	+	$22 x$
					+	$2 x^{2}$	-	$20 x$
							+	$2 x$	-	$20$
							-	$2 x$	+	$20$

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

				$x^{2}$	-	$2 x$	+	$2$
$x$	-	$10$		$x^{3}$	-	$12 x^{2}$	+	$22 x$	-	$20$
			-	$x^{3}$	+	$10 x^{2}$
					-	$2 x^{2}$	+	$22 x$
					+	$2 x^{2}$	-	$20 x$
							+	$2 x$	-	$20$
							-	$2 x$	+	$20$
										$0$

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

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

Write $x^{3} - 12 x^{2} + 22 x - 20$ as a set of factors.

$2 ((x - 10) (x^{2} - 2 x + 2)) = 0$

Remove unnecessary parentheses.

$2 (x - 10) (x^{2} - 2 x + 2) = 0$

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

$x - 10 = 0$

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

Set $x - 10$ equal to $0$ and solve for $x$ .

Set $x - 10$ equal to $0$ .

$x - 10 = 0$

Add $10$ to both sides of the equation.

$x = 10$

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

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

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

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

Use the quadratic formula to find the solutions.

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

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

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

Simplify.

Simplify the numerator.

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

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

Multiply $- 4$ by $1$ .

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

Multiply $- 4$ by $2$ .

$x = \frac{2 \pm \sqrt{4 - 8}}{2 \cdot 1}$

Subtract $8$ from $4$ .

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

Rewrite $- 4$ as $- 1 (4)$ .

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

Rewrite $\sqrt{- 1 (4)}$ as $\sqrt{- 1} \cdot \sqrt{4}$ .

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

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

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

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

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

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

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

Move $2$ to the left of $i$ .

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

Multiply $2$ by $1$ .

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

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

$x = 1 \pm i$

The final answer is the combination of both solutions.

$x = 1 + i, 1 - i$

The final solution is all the values that make $2 (x - 10) (x^{2} - 2 x + 2) = 0$ true.

$x = 10, 1 + i, 1 - i$

Find the Roots (Zeros) f(X)=2x^3-24x^2+44x-40

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