Precalculus Examples

$x^{4} + x^{3} - 7 x^{2} - x + 6$

Step 1

Set $x^{4} + x^{3} - 7 x^{2} - x + 6$ equal to $0$ .

$x^{4} + x^{3} - 7 x^{2} - x + 6 = 0$

Step 2

Solve for $x$ .

Tap for more steps...

Step 2.1

Factor the left side of the equation.

Tap for more steps...

Step 2.1.1

Regroup terms.

$x^{4} + x^{3} - 7 x^{2} - x + 6 = 0$

Step 2.1.2

Factor $x^{3}$ out of $x^{4} + x^{3}$ .

Tap for more steps...

Step 2.1.2.1

Factor $x^{3}$ out of $x^{4}$ .

$x^{3} x + x^{3} - 7 x^{2} - x + 6 = 0$

Step 2.1.2.2

Multiply by $1$ .

$x^{3} x + x^{3} \cdot 1 - 7 x^{2} - x + 6 = 0$

Step 2.1.2.3

Factor $x^{3}$ out of $x^{3} x + x^{3} \cdot 1$ .

$x^{3} (x + 1) - 7 x^{2} - x + 6 = 0$

Step 2.1.3

Factor by grouping.

Tap for more steps...

Step 2.1.3.1

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = - 7 \cdot 6 = - 42$ and whose sum is $b = - 1$ .

Tap for more steps...

Step 2.1.3.1.1

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

$x^{3} (x + 1) - 7 x^{2} - (x) + 6 = 0$

Step 2.1.3.1.2

Rewrite $- 1$ as $6$ plus $- 7$

$x^{3} (x + 1) - 7 x^{2} + (6 - 7) x + 6 = 0$

Step 2.1.3.1.3

Apply the distributive property.

$x^{3} (x + 1) - 7 x^{2} + 6 x - 7 x + 6 = 0$

Step 2.1.3.2

Factor out the greatest common factor from each group.

Tap for more steps...

Step 2.1.3.2.1

Group the first two terms and the last two terms.

$x^{3} (x + 1) - 7 x^{2} + 6 x - 7 x + 6 = 0$

Step 2.1.3.2.2

Factor out the greatest common factor (GCF) from each group.

$x^{3} (x + 1) + x (- 7 x + 6) + 1 (- 7 x + 6) = 0$

Step 2.1.3.3

Factor the polynomial by factoring out the greatest common factor, $- 7 x + 6$ .

$x^{3} (x + 1) + (- 7 x + 6) (x + 1) = 0$

Step 2.1.4

Factor $x + 1$ out of $x^{3} (x + 1) + (- 7 x + 6) (x + 1)$ .

Tap for more steps...

Step 2.1.4.1

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

$(x + 1) x^{3} + (- 7 x + 6) (x + 1) = 0$

Step 2.1.4.2

Factor $x + 1$ out of $(- 7 x + 6) (x + 1)$ .

$(x + 1) x^{3} + (x + 1) (- 7 x + 6) = 0$

Step 2.1.4.3

Factor $x + 1$ out of $(x + 1) x^{3} + (x + 1) (- 7 x + 6)$ .

$(x + 1) (x^{3} - 7 x + 6) = 0$

Step 2.1.5

Factor.

Tap for more steps...

Step 2.1.5.1

Rewrite $x^{3} - 7 x + 6$ in a factored form.

Tap for more steps...

Step 2.1.5.1.1

Factor $x^{3} - 7 x + 6$ using the rational roots test.

Tap for more steps...

Step 2.1.5.1.1.1

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.

$p = \pm 1, \pm 6, \pm 2, \pm 3$

$q = \pm 1$

Step 2.1.5.1.1.2

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

$\pm 1, \pm 6, \pm 2, \pm 3$

Step 2.1.5.1.1.3

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

Tap for more steps...

Step 2.1.5.1.1.3.1

Substitute $1$ into the polynomial.

$1^{3} - 7 \cdot 1 + 6$

Step 2.1.5.1.1.3.2

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

$1 - 7 \cdot 1 + 6$

Step 2.1.5.1.1.3.3

Multiply $- 7$ by $1$ .

$1 - 7 + 6$

Step 2.1.5.1.1.3.4

Subtract $7$ from $1$ .

$- 6 + 6$

Step 2.1.5.1.1.3.5

Add $- 6$ and $6$ .

$0$

Step 2.1.5.1.1.4

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.

$\frac{x^{3} - 7 x + 6}{x - 1}$

Step 2.1.5.1.1.5

Divide $x^{3} - 7 x + 6$ by $x - 1$ .

Tap for more steps...

Step 2.1.5.1.1.5.1

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

$x$

$1$

$x^{3}$

$0 x^{2}$

$7 x$

$6$

Step 2.1.5.1.1.5.2

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

					$x^{2}$
	$x$	-	$1$		$x^{3}$	+	$0 x^{2}$	-	$7 x$	+	$6$

Step 2.1.5.1.1.5.3

Multiply the new quotient term by the divisor.

				$x^{2}$
$x$	-	$1$		$x^{3}$	+	$0 x^{2}$	-	$7 x$	+	$6$
			+	$x^{3}$	-	$x^{2}$

Step 2.1.5.1.1.5.4

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

				$x^{2}$
$x$	-	$1$		$x^{3}$	+	$0 x^{2}$	-	$7 x$	+	$6$
			-	$x^{3}$	+	$x^{2}$

Step 2.1.5.1.1.5.5

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

				$x^{2}$
$x$	-	$1$		$x^{3}$	+	$0 x^{2}$	-	$7 x$	+	$6$
			-	$x^{3}$	+	$x^{2}$
					+	$x^{2}$

Step 2.1.5.1.1.5.6

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

				$x^{2}$
$x$	-	$1$		$x^{3}$	+	$0 x^{2}$	-	$7 x$	+	$6$
			-	$x^{3}$	+	$x^{2}$
					+	$x^{2}$	-	$7 x$

Step 2.1.5.1.1.5.7

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

				$x^{2}$	+	$x$
$x$	-	$1$		$x^{3}$	+	$0 x^{2}$	-	$7 x$	+	$6$
			-	$x^{3}$	+	$x^{2}$
					+	$x^{2}$	-	$7 x$

Step 2.1.5.1.1.5.8

Multiply the new quotient term by the divisor.

				$x^{2}$	+	$x$
$x$	-	$1$		$x^{3}$	+	$0 x^{2}$	-	$7 x$	+	$6$
			-	$x^{3}$	+	$x^{2}$
					+	$x^{2}$	-	$7 x$
					+	$x^{2}$	-	$x$

Step 2.1.5.1.1.5.9

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

				$x^{2}$	+	$x$
$x$	-	$1$		$x^{3}$	+	$0 x^{2}$	-	$7 x$	+	$6$
			-	$x^{3}$	+	$x^{2}$
					+	$x^{2}$	-	$7 x$
					-	$x^{2}$	+	$x$

Step 2.1.5.1.1.5.10

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

				$x^{2}$	+	$x$
$x$	-	$1$		$x^{3}$	+	$0 x^{2}$	-	$7 x$	+	$6$
			-	$x^{3}$	+	$x^{2}$
					+	$x^{2}$	-	$7 x$
					-	$x^{2}$	+	$x$
							-	$6 x$

Step 2.1.5.1.1.5.11

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

				$x^{2}$	+	$x$
$x$	-	$1$		$x^{3}$	+	$0 x^{2}$	-	$7 x$	+	$6$
			-	$x^{3}$	+	$x^{2}$
					+	$x^{2}$	-	$7 x$
					-	$x^{2}$	+	$x$
							-	$6 x$	+	$6$

Step 2.1.5.1.1.5.12

Divide the highest order term in the dividend $- 6 x$ by the highest order term in divisor $x$ .

				$x^{2}$	+	$x$	-	$6$
$x$	-	$1$		$x^{3}$	+	$0 x^{2}$	-	$7 x$	+	$6$
			-	$x^{3}$	+	$x^{2}$
					+	$x^{2}$	-	$7 x$
					-	$x^{2}$	+	$x$
							-	$6 x$	+	$6$

Step 2.1.5.1.1.5.13

Multiply the new quotient term by the divisor.

				$x^{2}$	+	$x$	-	$6$
$x$	-	$1$		$x^{3}$	+	$0 x^{2}$	-	$7 x$	+	$6$
			-	$x^{3}$	+	$x^{2}$
					+	$x^{2}$	-	$7 x$
					-	$x^{2}$	+	$x$
							-	$6 x$	+	$6$
							-	$6 x$	+	$6$

Step 2.1.5.1.1.5.14

The expression needs to be subtracted from the dividend, so change all the signs in $- 6 x + 6$

				$x^{2}$	+	$x$	-	$6$
$x$	-	$1$		$x^{3}$	+	$0 x^{2}$	-	$7 x$	+	$6$
			-	$x^{3}$	+	$x^{2}$
					+	$x^{2}$	-	$7 x$
					-	$x^{2}$	+	$x$
							-	$6 x$	+	$6$
							+	$6 x$	-	$6$

Step 2.1.5.1.1.5.15

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

				$x^{2}$	+	$x$	-	$6$
$x$	-	$1$		$x^{3}$	+	$0 x^{2}$	-	$7 x$	+	$6$
			-	$x^{3}$	+	$x^{2}$
					+	$x^{2}$	-	$7 x$
					-	$x^{2}$	+	$x$
							-	$6 x$	+	$6$
							+	$6 x$	-	$6$
										$0$

Step 2.1.5.1.1.5.16

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

$x^{2} + x - 6$

Step 2.1.5.1.1.6

Write $x^{3} - 7 x + 6$ as a set of factors.

$(x + 1) ((x - 1) (x^{2} + x - 6)) = 0$

Step 2.1.5.1.2

Factor $x^{2} + x - 6$ using the AC method.

Tap for more steps...

Step 2.1.5.1.2.1

Factor $x^{2} + x - 6$ using the AC method.

Tap for more steps...

Step 2.1.5.1.2.1.1

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 $- 6$ and whose sum is $1$ .

$- 2, 3$

Step 2.1.5.1.2.1.2

Write the factored form using these integers.

$(x + 1) ((x - 1) ((x - 2) (x + 3))) = 0$

Step 2.1.5.1.2.2

Remove unnecessary parentheses.

$(x + 1) ((x - 1) (x - 2) (x + 3)) = 0$

Step 2.1.5.2

Remove unnecessary parentheses.

$(x + 1) (x - 1) (x - 2) (x + 3) = 0$

Step 2.2

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

$x + 1 = 0$

$x - 1 = 0$

$x - 2 = 0$

$x + 3 = 0$

Step 2.3

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

Tap for more steps...

Step 2.3.1

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

$x + 1 = 0$

Step 2.3.2

Subtract $1$ from both sides of the equation.

$x = - 1$

Step 2.4

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

Tap for more steps...

Step 2.4.1

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

$x - 1 = 0$

Step 2.4.2

Add $1$ to both sides of the equation.

$x = 1$

Step 2.5

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

Tap for more steps...

Step 2.5.1

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

$x - 2 = 0$

Step 2.5.2

Add $2$ to both sides of the equation.

$x = 2$

Step 2.6

Set $x + 3$ equal to $0$ and solve for $x$ .

Tap for more steps...

Step 2.6.1

Set $x + 3$ equal to $0$ .

$x + 3 = 0$

Step 2.6.2

Subtract $3$ from both sides of the equation.

$x = - 3$

Step 2.7

The final solution is all the values that make $(x + 1) (x - 1) (x - 2) (x + 3) = 0$ true.

$x = - 1, 1, 2, - 3$

Step 3