Find the Roots (Zeros) -x^3-3x^2=6x+4

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

Step 1

Subtract $6 x$ from both sides of the equation.

$- x^{3} - 3 x^{2} - 6 x = 4$

Step 2

Subtract $4$ from both sides of the equation.

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

Step 3

Factor the left side of the equation.

Factor $- 1$ out of $- x^{3} - 3 x^{2} - 6 x - 4$ .

Factor $- 1$ out of $- x^{3}$ .

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

Factor $- 1$ out of $- 3 x^{2}$ .

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

Factor $- 1$ out of $- 6 x$ .

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

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

$- (x^{3}) - (3 x^{2}) - (6 x) - 1 \cdot 4 = 0$

Factor $- 1$ out of $- (x^{3}) - (3 x^{2})$ .

$- (x^{3} + 3 x^{2}) - (6 x) - 1 \cdot 4 = 0$

Factor $- 1$ out of $- (x^{3} + 3 x^{2}) - (6 x)$ .

$- (x^{3} + 3 x^{2} + 6 x) - 1 \cdot 4 = 0$

Factor $- 1$ out of $- (x^{3} + 3 x^{2} + 6 x) - 1 (4)$ .

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

Factor.

Factor $x^{3} + 3 x^{2} + 6 x + 4$ 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.

$p = \pm 1, \pm 4, \pm 2 q = \pm 1$

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

$\pm 1, \pm 4, \pm 2$

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.

${(- 1)}^{3} + 3 {(- 1)}^{2} + 6 \cdot - 1 + 4$

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

$- 1 + 3 {(- 1)}^{2} + 6 \cdot - 1 + 4$

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

$- 1 + 3 \cdot 1 + 6 \cdot - 1 + 4$

Multiply $3$ by $1$ .

$- 1 + 3 + 6 \cdot - 1 + 4$

Add $- 1$ and $3$ .

$2 + 6 \cdot - 1 + 4$

Multiply $6$ by $- 1$ .

$2 - 6 + 4$

Subtract $6$ from $2$ .

$- 4 + 4$

Add $- 4$ and $4$ .

$0$

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.

$\frac{x^{3} + 3 x^{2} + 6 x + 4}{x + 1}$

Divide $x^{3} + 3 x^{2} + 6 x + 4$ 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 $0$ .

$x$

$1$

$x^{3}$

$3 x^{2}$

$6 x$

$4$

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

					$x^{2}$
	$x$	+	$1$		$x^{3}$	+	$3 x^{2}$	+	$6 x$	+	$4$

Multiply the new quotient term by the divisor.

				$x^{2}$
$x$	+	$1$		$x^{3}$	+	$3 x^{2}$	+	$6 x$	+	$4$
			+	$x^{3}$	+	$x^{2}$

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

				$x^{2}$
$x$	+	$1$		$x^{3}$	+	$3 x^{2}$	+	$6 x$	+	$4$
			-	$x^{3}$	-	$x^{2}$

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

				$x^{2}$
$x$	+	$1$		$x^{3}$	+	$3 x^{2}$	+	$6 x$	+	$4$
			-	$x^{3}$	-	$x^{2}$
					+	$2 x^{2}$

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

				$x^{2}$
$x$	+	$1$		$x^{3}$	+	$3 x^{2}$	+	$6 x$	+	$4$
			-	$x^{3}$	-	$x^{2}$
					+	$2 x^{2}$	+	$6 x$

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

				$x^{2}$	+	$2 x$
$x$	+	$1$		$x^{3}$	+	$3 x^{2}$	+	$6 x$	+	$4$
			-	$x^{3}$	-	$x^{2}$
					+	$2 x^{2}$	+	$6 x$

Multiply the new quotient term by the divisor.

				$x^{2}$	+	$2 x$
$x$	+	$1$		$x^{3}$	+	$3 x^{2}$	+	$6 x$	+	$4$
			-	$x^{3}$	-	$x^{2}$
					+	$2 x^{2}$	+	$6 x$
					+	$2 x^{2}$	+	$2 x$

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

				$x^{2}$	+	$2 x$
$x$	+	$1$		$x^{3}$	+	$3 x^{2}$	+	$6 x$	+	$4$
			-	$x^{3}$	-	$x^{2}$
					+	$2 x^{2}$	+	$6 x$
					-	$2 x^{2}$	-	$2 x$

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

				$x^{2}$	+	$2 x$
$x$	+	$1$		$x^{3}$	+	$3 x^{2}$	+	$6 x$	+	$4$
			-	$x^{3}$	-	$x^{2}$
					+	$2 x^{2}$	+	$6 x$
					-	$2 x^{2}$	-	$2 x$
							+	$4 x$

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

				$x^{2}$	+	$2 x$
$x$	+	$1$		$x^{3}$	+	$3 x^{2}$	+	$6 x$	+	$4$
			-	$x^{3}$	-	$x^{2}$
					+	$2 x^{2}$	+	$6 x$
					-	$2 x^{2}$	-	$2 x$
							+	$4 x$	+	$4$

Divide the highest order term in the dividend $4 x$ by the highest order term in divisor $x$ .

				$x^{2}$	+	$2 x$	+	$4$
$x$	+	$1$		$x^{3}$	+	$3 x^{2}$	+	$6 x$	+	$4$
			-	$x^{3}$	-	$x^{2}$
					+	$2 x^{2}$	+	$6 x$
					-	$2 x^{2}$	-	$2 x$
							+	$4 x$	+	$4$

Multiply the new quotient term by the divisor.

				$x^{2}$	+	$2 x$	+	$4$
$x$	+	$1$		$x^{3}$	+	$3 x^{2}$	+	$6 x$	+	$4$
			-	$x^{3}$	-	$x^{2}$
					+	$2 x^{2}$	+	$6 x$
					-	$2 x^{2}$	-	$2 x$
							+	$4 x$	+	$4$
							+	$4 x$	+	$4$

The expression needs to be subtracted from the dividend, so change all the signs in $4 x + 4$

				$x^{2}$	+	$2 x$	+	$4$
$x$	+	$1$		$x^{3}$	+	$3 x^{2}$	+	$6 x$	+	$4$
			-	$x^{3}$	-	$x^{2}$
					+	$2 x^{2}$	+	$6 x$
					-	$2 x^{2}$	-	$2 x$
							+	$4 x$	+	$4$
							-	$4 x$	-	$4$

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

				$x^{2}$	+	$2 x$	+	$4$
$x$	+	$1$		$x^{3}$	+	$3 x^{2}$	+	$6 x$	+	$4$
			-	$x^{3}$	-	$x^{2}$
					+	$2 x^{2}$	+	$6 x$
					-	$2 x^{2}$	-	$2 x$
							+	$4 x$	+	$4$
							-	$4 x$	-	$4$
										$0$

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

$x^{2} + 2 x + 4$

Write $x^{3} + 3 x^{2} + 6 x + 4$ as a set of factors.

$- ((x + 1) (x^{2} + 2 x + 4)) = 0$

Remove unnecessary parentheses.

$- (x + 1) (x^{2} + 2 x + 4) = 0$

Step 4

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

$x + 1 = 0$

$x^{2} + 2 x + 4 = 0$

Step 5

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

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

$x + 1 = 0$

Subtract $1$ from both sides of the equation.

$x = - 1$

Step 6

Set $x^{2} + 2 x + 4$ equal to $0$ and solve for $x$ .

Set $x^{2} + 2 x + 4$ equal to $0$ .

$x^{2} + 2 x + 4 = 0$

Solve $x^{2} + 2 x + 4 = 0$ 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 $c = 4$ into the quadratic formula and solve for $x$ .

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

Simplify.

Simplify the numerator.

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

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

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

Multiply $- 4$ by $1$ .

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

Multiply $- 4$ by $4$ .

$x = \frac{- 2 \pm \sqrt{4 - 16}}{2 \cdot 1}$

Subtract $16$ from $4$ .

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

Rewrite $- 12$ as $- 1 (12)$ .

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

Rewrite $\sqrt{- 1 (12)}$ as $\sqrt{- 1} \cdot \sqrt{12}$ .

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

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

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

Rewrite $12$ as $2^{2} \cdot 3$ .

Factor $4$ out of $12$ .

$x = \frac{- 2 \pm i \cdot \sqrt{4 (3)}}{2 \cdot 1}$

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

$x = \frac{- 2 \pm i \cdot \sqrt{2^{2} \cdot 3}}{2 \cdot 1}$

Pull terms out from under the radical.

$x = \frac{- 2 \pm i \cdot (2 \sqrt{3})}{2 \cdot 1}$

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

$x = \frac{- 2 \pm 2 i \sqrt{3}}{2 \cdot 1}$

Multiply $2$ by $1$ .

$x = \frac{- 2 \pm 2 i \sqrt{3}}{2}$

Simplify $\frac{- 2 \pm 2 i \sqrt{3}}{2}$ .

$x = - 1 \pm i \sqrt{3}$

The final answer is the combination of both solutions.

$x = - 1 + i \sqrt{3}, - 1 - i \sqrt{3}$

Step 7

The final solution is all the values that make $- (x + 1) (x^{2} + 2 x + 4) = 0$ true.

$x = - 1, - 1 + i \sqrt{3}, - 1 - i \sqrt{3}$

Step 8

Find the Roots (Zeros) -x^3-3x^2=6x+4

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