Find the Roots (Zeros) f(x)=x^3-6x^2-9x+14

$f (x) = x^{3} - 6 x^{2} - 9 x + 14$

Set $x^{3} - 6 x^{2} - 9 x + 14$ equal to $0$ .

$x^{3} - 6 x^{2} - 9 x + 14 = 0$

Solve for $x$ .

Factor the left side of the equation.

Factor $x^{3} - 6 x^{2} - 9 x + 14$ 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.

$p = \pm 1, \pm 14, \pm 2, \pm 7 q = \pm 1$

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

$\pm 1, \pm 14, \pm 2, \pm 7$

Substitute $1$ and simplify the expression. In this case, the expression is equal to $0$ so $1$ is a root of the polynomial.

Substitute $1$ into the polynomial.

$1^{3} - 6 \cdot 1^{2} - 9 \cdot 1 + 14$

Raise $1$ to the power of $3$ .

$1 - 6 \cdot 1^{2} - 9 \cdot 1 + 14$

Raise $1$ to the power of $2$ .

$1 - 6 \cdot 1 - 9 \cdot 1 + 14$

Multiply $- 6$ by $1$ .

$1 - 6 - 9 \cdot 1 + 14$

Subtract $6$ from $1$ .

$- 5 - 9 \cdot 1 + 14$

Multiply $- 9$ by $1$ .

$- 5 - 9 + 14$

Subtract $9$ from $- 5$ .

$- 14 + 14$

Add $- 14$ and $14$ .

$0$

Since $1$ is a known root, divide the polynomial by $x - 1$ to find the quotient polynomial. This polynomial can then be used to find the remaining roots.

$\frac{x^{3} - 6 x^{2} - 9 x + 14}{x - 1}$

Divide $x^{3} - 6 x^{2} - 9 x + 14$ by $x - 1$ .

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

$x$

$1$

$x^{3}$

$6 x^{2}$

$9 x$

$14$

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

					$x^{2}$
	$x$	-	$1$		$x^{3}$	-	$6 x^{2}$	-	$9 x$	+	$14$

Multiply the new quotient term by the divisor.

				$x^{2}$
$x$	-	$1$		$x^{3}$	-	$6 x^{2}$	-	$9 x$	+	$14$
			+	$x^{3}$	-	$x^{2}$

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

				$x^{2}$
$x$	-	$1$		$x^{3}$	-	$6 x^{2}$	-	$9 x$	+	$14$
			-	$x^{3}$	+	$x^{2}$

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

				$x^{2}$
$x$	-	$1$		$x^{3}$	-	$6 x^{2}$	-	$9 x$	+	$14$
			-	$x^{3}$	+	$x^{2}$
					-	$5 x^{2}$

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

				$x^{2}$
$x$	-	$1$		$x^{3}$	-	$6 x^{2}$	-	$9 x$	+	$14$
			-	$x^{3}$	+	$x^{2}$
					-	$5 x^{2}$	-	$9 x$

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

				$x^{2}$	-	$5 x$
$x$	-	$1$		$x^{3}$	-	$6 x^{2}$	-	$9 x$	+	$14$
			-	$x^{3}$	+	$x^{2}$
					-	$5 x^{2}$	-	$9 x$

Multiply the new quotient term by the divisor.

				$x^{2}$	-	$5 x$
$x$	-	$1$		$x^{3}$	-	$6 x^{2}$	-	$9 x$	+	$14$
			-	$x^{3}$	+	$x^{2}$
					-	$5 x^{2}$	-	$9 x$
					-	$5 x^{2}$	+	$5 x$

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

				$x^{2}$	-	$5 x$
$x$	-	$1$		$x^{3}$	-	$6 x^{2}$	-	$9 x$	+	$14$
			-	$x^{3}$	+	$x^{2}$
					-	$5 x^{2}$	-	$9 x$
					+	$5 x^{2}$	-	$5 x$

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

				$x^{2}$	-	$5 x$
$x$	-	$1$		$x^{3}$	-	$6 x^{2}$	-	$9 x$	+	$14$
			-	$x^{3}$	+	$x^{2}$
					-	$5 x^{2}$	-	$9 x$
					+	$5 x^{2}$	-	$5 x$
							-	$14 x$

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

				$x^{2}$	-	$5 x$
$x$	-	$1$		$x^{3}$	-	$6 x^{2}$	-	$9 x$	+	$14$
			-	$x^{3}$	+	$x^{2}$
					-	$5 x^{2}$	-	$9 x$
					+	$5 x^{2}$	-	$5 x$
							-	$14 x$	+	$14$

Divide the highest order term in the dividend $- 14 x$ by the highest order term in divisor $x$ .

				$x^{2}$	-	$5 x$	-	$14$
$x$	-	$1$		$x^{3}$	-	$6 x^{2}$	-	$9 x$	+	$14$
			-	$x^{3}$	+	$x^{2}$
					-	$5 x^{2}$	-	$9 x$
					+	$5 x^{2}$	-	$5 x$
							-	$14 x$	+	$14$

Multiply the new quotient term by the divisor.

				$x^{2}$	-	$5 x$	-	$14$
$x$	-	$1$		$x^{3}$	-	$6 x^{2}$	-	$9 x$	+	$14$
			-	$x^{3}$	+	$x^{2}$
					-	$5 x^{2}$	-	$9 x$
					+	$5 x^{2}$	-	$5 x$
							-	$14 x$	+	$14$
							-	$14 x$	+	$14$

The expression needs to be subtracted from the dividend, so change all the signs in $- 14 x + 14$

				$x^{2}$	-	$5 x$	-	$14$
$x$	-	$1$		$x^{3}$	-	$6 x^{2}$	-	$9 x$	+	$14$
			-	$x^{3}$	+	$x^{2}$
					-	$5 x^{2}$	-	$9 x$
					+	$5 x^{2}$	-	$5 x$
							-	$14 x$	+	$14$
							+	$14 x$	-	$14$

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

				$x^{2}$	-	$5 x$	-	$14$
$x$	-	$1$		$x^{3}$	-	$6 x^{2}$	-	$9 x$	+	$14$
			-	$x^{3}$	+	$x^{2}$
					-	$5 x^{2}$	-	$9 x$
					+	$5 x^{2}$	-	$5 x$
							-	$14 x$	+	$14$
							+	$14 x$	-	$14$
										$0$

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

$x^{2} - 5 x - 14$

Write $x^{3} - 6 x^{2} - 9 x + 14$ as a set of factors.

$(x - 1) (x^{2} - 5 x - 14) = 0$

Factor $x^{2} - 5 x - 14$ using the AC method.

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 14$ and whose sum is $- 5$ .

$- 7, 2$

Write the factored form using these integers.

$(x - 1) ((x - 7) (x + 2)) = 0$

Remove unnecessary parentheses.

$(x - 1) (x - 7) (x + 2) = 0$

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

$x - 1 = 0$

$x - 7 = 0$

$x + 2 = 0$

Set $x - 1$ equal to $0$ and solve for $x$ .

Set $x - 1$ equal to $0$ .

$x - 1 = 0$

Add $1$ to both sides of the equation.

$x = 1$

Set $x - 7$ equal to $0$ and solve for $x$ .

Set $x - 7$ equal to $0$ .

$x - 7 = 0$

Add $7$ to both sides of the equation.

$x = 7$

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

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

$x + 2 = 0$

Subtract $2$ from both sides of the equation.

$x = - 2$

The final solution is all the values that make $(x - 1) (x - 7) (x + 2) = 0$ true.

$x = 1, 7, - 2$

Find the Roots (Zeros) f(x)=x^3-6x^2-9x+14

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