Divide (3x^3-3-6x)÷(3+3x)

$(3 x^{3} - 3 - 6 x) \div (3 + 3 x)$

Step 1

Rewrite the division as a fraction.

$\frac{3 x^{3} - 3 - 6 x}{3 + 3 x}$

Step 2

Cancel the common factor of $3 x^{3} - 3 - 6 x$ and $3 + 3 x$ .

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

$\frac{3 (x^{3}) - 3 - 6 x}{3 + 3 x}$

Factor $3$ out of $- 3$ .

$\frac{3 (x^{3}) + 3 \cdot - 1 - 6 x}{3 + 3 x}$

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

$\frac{3 (x^{3} - 1) - 6 x}{3 + 3 x}$

Factor $3$ out of $- 6 x$ .

$\frac{3 (x^{3} - 1) + 3 (- 2 x)}{3 + 3 x}$

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

$\frac{3 (x^{3} - 1 - 2 x)}{3 + 3 x}$

Cancel the common factors.

Factor $3$ out of $3$ .

$\frac{3 (x^{3} - 1 - 2 x)}{3 \cdot 1 + 3 x}$

Factor $3$ out of $3 x$ .

$\frac{3 (x^{3} - 1 - 2 x)}{3 \cdot 1 + 3 (x)}$

Factor $3$ out of $3 \cdot 1 + 3 (x)$ .

$\frac{3 (x^{3} - 1 - 2 x)}{3 \cdot (1 + x)}$

Cancel the common factor.

$\frac{3 (x^{3} - 1 - 2 x)}{3 \cdot (1 + x)}$

Rewrite the expression.

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

Step 3

Simplify the numerator.

Reorder terms.

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

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

$p = \pm 1 q = \pm 1$

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

$\pm 1$

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.

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

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

$- 1 - 2 \cdot - 1 - 1$

Multiply $- 2$ by $- 1$ .

$- 1 + 2 - 1$

Add $- 1$ and $2$ .

$1 - 1$

Subtract $1$ from $1$ .

$0$

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.

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

Divide $x^{3} - 2 x - 1$ 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 $0$ .

$x$

$1$

$x^{3}$

$0 x^{2}$

$2 x$

$1$

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

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

Multiply the new quotient term by the divisor.

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

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

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

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

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

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

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

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

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

Multiply the new quotient term by the divisor.

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

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

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

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

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

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

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

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

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

Multiply the new quotient term by the divisor.

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

The expression needs to be subtracted from the dividend, so change all the signs in $- x - 1$

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

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

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

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

$x^{2} - x - 1$

Write $x^{3} - 2 x - 1$ as a set of factors.

$\frac{(x + 1) (x^{2} - x - 1)}{1 + x}$

Step 4

Cancel the common factor of $x + 1$ and $1 + x$ .

Reorder terms.

$\frac{(x + 1) (x^{2} - x - 1)}{x + 1}$

Cancel the common factor.

$\frac{(x + 1) (x^{2} - x - 1)}{x + 1}$

Divide $x^{2} - x - 1$ by $1$ .

$x^{2} - x - 1$

Divide (3x^3-3-6x)÷(3+3x)

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