Find the Roots (Zeros) f(x)=x^3-4x^2+9x-10

$f (x) = x^{3} - 4 x^{2} + 9 x - 10$

Set $x^{3} - 4 x^{2} + 9 x - 10$ equal to $0$ .

$x^{3} - 4 x^{2} + 9 x - 10 = 0$

Solve for $x$ .

Factor $x^{3} - 4 x^{2} + 9 x - 10$ 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.

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

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

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

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.

$2^{3} - 4 \cdot 2^{2} + 9 \cdot 2 - 10$

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

$8 - 4 \cdot 2^{2} + 9 \cdot 2 - 10$

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

$8 - 4 \cdot 4 + 9 \cdot 2 - 10$

Multiply $- 4$ by $4$ .

$8 - 16 + 9 \cdot 2 - 10$

Subtract $16$ from $8$ .

$- 8 + 9 \cdot 2 - 10$

Multiply $9$ by $2$ .

$- 8 + 18 - 10$

Add $- 8$ and $18$ .

$10 - 10$

Subtract $10$ from $10$ .

$0$

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.

$\frac{x^{3} - 4 x^{2} + 9 x - 10}{x - 2}$

Divide $x^{3} - 4 x^{2} + 9 x - 10$ 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 $0$ .

$x$

$2$

$x^{3}$

$4 x^{2}$

$9 x$

$10$

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

					$x^{2}$
	$x$	-	$2$		$x^{3}$	-	$4 x^{2}$	+	$9 x$	-	$10$

Multiply the new quotient term by the divisor.

				$x^{2}$
$x$	-	$2$		$x^{3}$	-	$4 x^{2}$	+	$9 x$	-	$10$
			+	$x^{3}$	-	$2 x^{2}$

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

				$x^{2}$
$x$	-	$2$		$x^{3}$	-	$4 x^{2}$	+	$9 x$	-	$10$
			-	$x^{3}$	+	$2 x^{2}$

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

				$x^{2}$
$x$	-	$2$		$x^{3}$	-	$4 x^{2}$	+	$9 x$	-	$10$
			-	$x^{3}$	+	$2 x^{2}$
					-	$2 x^{2}$

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

				$x^{2}$
$x$	-	$2$		$x^{3}$	-	$4 x^{2}$	+	$9 x$	-	$10$
			-	$x^{3}$	+	$2 x^{2}$
					-	$2 x^{2}$	+	$9 x$

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

				$x^{2}$	-	$2 x$
$x$	-	$2$		$x^{3}$	-	$4 x^{2}$	+	$9 x$	-	$10$
			-	$x^{3}$	+	$2 x^{2}$
					-	$2 x^{2}$	+	$9 x$

Multiply the new quotient term by the divisor.

				$x^{2}$	-	$2 x$
$x$	-	$2$		$x^{3}$	-	$4 x^{2}$	+	$9 x$	-	$10$
			-	$x^{3}$	+	$2 x^{2}$
					-	$2 x^{2}$	+	$9 x$
					-	$2 x^{2}$	+	$4 x$

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

				$x^{2}$	-	$2 x$
$x$	-	$2$		$x^{3}$	-	$4 x^{2}$	+	$9 x$	-	$10$
			-	$x^{3}$	+	$2 x^{2}$
					-	$2 x^{2}$	+	$9 x$
					+	$2 x^{2}$	-	$4 x$

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

				$x^{2}$	-	$2 x$
$x$	-	$2$		$x^{3}$	-	$4 x^{2}$	+	$9 x$	-	$10$
			-	$x^{3}$	+	$2 x^{2}$
					-	$2 x^{2}$	+	$9 x$
					+	$2 x^{2}$	-	$4 x$
							+	$5 x$

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

				$x^{2}$	-	$2 x$
$x$	-	$2$		$x^{3}$	-	$4 x^{2}$	+	$9 x$	-	$10$
			-	$x^{3}$	+	$2 x^{2}$
					-	$2 x^{2}$	+	$9 x$
					+	$2 x^{2}$	-	$4 x$
							+	$5 x$	-	$10$

Divide the highest order term in the dividend $5 x$ by the highest order term in divisor $x$ .

				$x^{2}$	-	$2 x$	+	$5$
$x$	-	$2$		$x^{3}$	-	$4 x^{2}$	+	$9 x$	-	$10$
			-	$x^{3}$	+	$2 x^{2}$
					-	$2 x^{2}$	+	$9 x$
					+	$2 x^{2}$	-	$4 x$
							+	$5 x$	-	$10$

Multiply the new quotient term by the divisor.

				$x^{2}$	-	$2 x$	+	$5$
$x$	-	$2$		$x^{3}$	-	$4 x^{2}$	+	$9 x$	-	$10$
			-	$x^{3}$	+	$2 x^{2}$
					-	$2 x^{2}$	+	$9 x$
					+	$2 x^{2}$	-	$4 x$
							+	$5 x$	-	$10$
							+	$5 x$	-	$10$

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

				$x^{2}$	-	$2 x$	+	$5$
$x$	-	$2$		$x^{3}$	-	$4 x^{2}$	+	$9 x$	-	$10$
			-	$x^{3}$	+	$2 x^{2}$
					-	$2 x^{2}$	+	$9 x$
					+	$2 x^{2}$	-	$4 x$
							+	$5 x$	-	$10$
							-	$5 x$	+	$10$

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

				$x^{2}$	-	$2 x$	+	$5$
$x$	-	$2$		$x^{3}$	-	$4 x^{2}$	+	$9 x$	-	$10$
			-	$x^{3}$	+	$2 x^{2}$
					-	$2 x^{2}$	+	$9 x$
					+	$2 x^{2}$	-	$4 x$
							+	$5 x$	-	$10$
							-	$5 x$	+	$10$
										$0$

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

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

Write $x^{3} - 4 x^{2} + 9 x - 10$ as a set of factors.

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

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

$x - 2 = 0$

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

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

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

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

Solve $x^{2} - 2 x + 5 = 0$ 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 $c = 5$ 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 $- 4$ 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.

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

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

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

Find the Roots (Zeros) f(x)=x^3-4x^2+9x-10

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