Precalculus Examples

$2 x^{3} + 11 x^{2} - 7 x - 6$

Step 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.

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

$q = \pm 1, \pm 2$

Step 2

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

$\pm 1, \pm \frac{1}{2}, \pm 2, \pm 3, \pm \frac{3}{2}, \pm 6$

Step 3

Substitute the possible roots one by one into the polynomial to find the actual roots. Simplify to check if the value is $0$ , which means it is a root.

$2 {(1)}^{3} + 11 {(1)}^{2} - 7 \cdot 1 - 6$

Step 4

Simplify the expression. In this case, the expression is equal to $0$ so $x = 1$ is a root of the polynomial.

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Step 4.1

Simplify each term.

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Step 4.1.1

One to any power is one.

$2 \cdot 1 + 11 {(1)}^{2} - 7 \cdot 1 - 6$

Step 4.1.2

Multiply $2$ by $1$ .

$2 + 11 {(1)}^{2} - 7 \cdot 1 - 6$

Step 4.1.3

One to any power is one.

$2 + 11 \cdot 1 - 7 \cdot 1 - 6$

Step 4.1.4

Multiply $11$ by $1$ .

$2 + 11 - 7 \cdot 1 - 6$

Step 4.1.5

Multiply $- 7$ by $1$ .

$2 + 11 - 7 - 6$

Step 4.2

Simplify by adding and subtracting.

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Step 4.2.1

Add $2$ and $11$ .

$13 - 7 - 6$

Step 4.2.2

Subtract $7$ from $13$ .

$6 - 6$

Step 4.2.3

Subtract $6$ from $6$ .

$0$

Step 5

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.

$\frac{2 x^{3} + 11 x^{2} - 7 x - 6}{x - 1}$

Step 6

Next, find the roots of the remaining polynomial. The order of the polynomial has been reduced by $1$ .

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Step 6.1

Place the numbers representing the divisor and the dividend into a division-like configuration.

$1$	$2$	$11$	$- 7$	$- 6$

Step 6.2

The first number in the dividend $(2)$ is put into the first position of the result area (below the horizontal line).

$1$	$2$	$11$	$- 7$	$- 6$

	$2$

Step 6.3

Multiply the newest entry in the result $(2)$ by the divisor $(1)$ and place the result of $(2)$ under the next term in the dividend $(11)$ .

$1$	$2$	$11$	$- 7$	$- 6$
		$2$
	$2$

Step 6.4

Add the product of the multiplication and the number from the dividend and put the result in the next position on the result line.

$1$	$2$	$11$	$- 7$	$- 6$
		$2$
	$2$	$13$

Step 6.5

Multiply the newest entry in the result $(13)$ by the divisor $(1)$ and place the result of $(13)$ under the next term in the dividend $(- 7)$ .

$1$	$2$	$11$	$- 7$	$- 6$
		$2$	$13$
	$2$	$13$

Step 6.6

Add the product of the multiplication and the number from the dividend and put the result in the next position on the result line.

$1$	$2$	$11$	$- 7$	$- 6$
		$2$	$13$
	$2$	$13$	$6$

Step 6.7

Multiply the newest entry in the result $(6)$ by the divisor $(1)$ and place the result of $(6)$ under the next term in the dividend $(- 6)$ .

$1$	$2$	$11$	$- 7$	$- 6$
		$2$	$13$	$6$
	$2$	$13$	$6$

Step 6.8

Add the product of the multiplication and the number from the dividend and put the result in the next position on the result line.

$1$	$2$	$11$	$- 7$	$- 6$
		$2$	$13$	$6$
	$2$	$13$	$6$	$0$

Step 6.9

All numbers except the last become the coefficients of the quotient polynomial. The last value in the result line is the remainder.

$2 x^{2} + 13 x + 6$

Step 7

Factor by grouping.

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Step 7.1

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 $a \cdot c = 2 \cdot 6 = 12$ and whose sum is $b = 13$ .

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Step 7.1.1

Factor $13$ out of $13 x$ .

$(x - 1) + 2 x^{2} + 13 (x) + 6$

Step 7.1.2

Rewrite $13$ as $1$ plus $12$

$(x - 1) + 2 x^{2} + (1 + 12) x + 6$

Step 7.1.3

Apply the distributive property.

$(x - 1) + 2 x^{2} + 1 x + 12 x + 6$

Step 7.1.4

Multiply $x$ by $1$ .

$(x - 1) + 2 x^{2} + x + 12 x + 6$

Step 7.2

Factor out the greatest common factor from each group.

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Step 7.2.1

Group the first two terms and the last two terms.

$(x - 1) (2 x^{2} + x) + 12 x + 6$

Step 7.2.2

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

$(x - 1) + x (2 x + 1) + 6 (2 x + 1)$

Step 7.3

Factor the polynomial by factoring out the greatest common factor, $2 x + 1$ .

$(x - 1) + (2 x + 1) (x + 6)$

$(x - 1) (2 x + 1) (x + 6)$

Step 8

Factor the left side of the equation.

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Step 8.1

Factor $2 x^{3} + 11 x^{2} - 7 x - 6$ using the rational roots test.

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Step 8.1.1

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.

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

$q = \pm 1, \pm 2$

Step 8.1.2

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

$\pm 1, \pm 0.5, \pm 6, \pm 3, \pm 2, \pm 1.5$

Step 8.1.3

Substitute $- 0.5$ and simplify the expression. In this case, the expression is equal to $0$ so $- 0.5$ is a root of the polynomial.

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Step 8.1.3.1

Substitute $- 0.5$ into the polynomial.

$2 {(- 0.5)}^{3} + 11 {(- 0.5)}^{2} - 7 \cdot - 0.5 - 6$

Step 8.1.3.2

Raise $- 0.5$ to the power of $3$ .

$2 \cdot - 0.125 + 11 {(- 0.5)}^{2} - 7 \cdot - 0.5 - 6$

Step 8.1.3.3

Multiply $2$ by $- 0.125$ .

$- 0.25 + 11 {(- 0.5)}^{2} - 7 \cdot - 0.5 - 6$

Step 8.1.3.4

Raise $- 0.5$ to the power of $2$ .

$- 0.25 + 11 \cdot 0.25 - 7 \cdot - 0.5 - 6$

Step 8.1.3.5

Multiply $11$ by $0.25$ .

$- 0.25 + 2.75 - 7 \cdot - 0.5 - 6$

Step 8.1.3.6

Add $- 0.25$ and $2.75$ .

$2.5 - 7 \cdot - 0.5 - 6$

Step 8.1.3.7

Multiply $- 7$ by $- 0.5$ .

$2.5 + 3.5 - 6$

Step 8.1.3.8

Add $2.5$ and $3.5$ .

$6 - 6$

Step 8.1.3.9

Subtract $6$ from $6$ .

$0$

Step 8.1.4

Since $- 0.5$ is a known root, divide the polynomial by $2 x + 1$ to find the quotient polynomial. This polynomial can then be used to find the remaining roots.

$\frac{2 x^{3} + 11 x^{2} - 7 x - 6}{2 x + 1}$

Step 8.1.5

Divide $2 x^{3} + 11 x^{2} - 7 x - 6$ by $2 x + 1$ .

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Step 8.1.5.1

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

$2 x$

$1$

$2 x^{3}$

$11 x^{2}$

$7 x$

$6$

Step 8.1.5.2

Divide the highest order term in the dividend $2 x^{3}$ by the highest order term in divisor $2 x$ .

					$x^{2}$
	$2 x$	+	$1$		$2 x^{3}$	+	$11 x^{2}$	-	$7 x$	-	$6$

Step 8.1.5.3

Multiply the new quotient term by the divisor.

				$x^{2}$
$2 x$	+	$1$		$2 x^{3}$	+	$11 x^{2}$	-	$7 x$	-	$6$
			+	$2 x^{3}$	+	$x^{2}$

Step 8.1.5.4

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

				$x^{2}$
$2 x$	+	$1$		$2 x^{3}$	+	$11 x^{2}$	-	$7 x$	-	$6$
			-	$2 x^{3}$	-	$x^{2}$

Step 8.1.5.5

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

				$x^{2}$
$2 x$	+	$1$		$2 x^{3}$	+	$11 x^{2}$	-	$7 x$	-	$6$
			-	$2 x^{3}$	-	$x^{2}$
					+	$10 x^{2}$

Step 8.1.5.6

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

				$x^{2}$
$2 x$	+	$1$		$2 x^{3}$	+	$11 x^{2}$	-	$7 x$	-	$6$
			-	$2 x^{3}$	-	$x^{2}$
					+	$10 x^{2}$	-	$7 x$

Step 8.1.5.7

Divide the highest order term in the dividend $10 x^{2}$ by the highest order term in divisor $2 x$ .

				$x^{2}$	+	$5 x$
$2 x$	+	$1$		$2 x^{3}$	+	$11 x^{2}$	-	$7 x$	-	$6$
			-	$2 x^{3}$	-	$x^{2}$
					+	$10 x^{2}$	-	$7 x$

Step 8.1.5.8

Multiply the new quotient term by the divisor.

				$x^{2}$	+	$5 x$
$2 x$	+	$1$		$2 x^{3}$	+	$11 x^{2}$	-	$7 x$	-	$6$
			-	$2 x^{3}$	-	$x^{2}$
					+	$10 x^{2}$	-	$7 x$
					+	$10 x^{2}$	+	$5 x$

Step 8.1.5.9

The expression needs to be subtracted from the dividend, so change all the signs in $10 x^{2} + 5 x$

				$x^{2}$	+	$5 x$
$2 x$	+	$1$		$2 x^{3}$	+	$11 x^{2}$	-	$7 x$	-	$6$
			-	$2 x^{3}$	-	$x^{2}$
					+	$10 x^{2}$	-	$7 x$
					-	$10 x^{2}$	-	$5 x$

Step 8.1.5.10

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

				$x^{2}$	+	$5 x$
$2 x$	+	$1$		$2 x^{3}$	+	$11 x^{2}$	-	$7 x$	-	$6$
			-	$2 x^{3}$	-	$x^{2}$
					+	$10 x^{2}$	-	$7 x$
					-	$10 x^{2}$	-	$5 x$
							-	$12 x$

Step 8.1.5.11

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

				$x^{2}$	+	$5 x$
$2 x$	+	$1$		$2 x^{3}$	+	$11 x^{2}$	-	$7 x$	-	$6$
			-	$2 x^{3}$	-	$x^{2}$
					+	$10 x^{2}$	-	$7 x$
					-	$10 x^{2}$	-	$5 x$
							-	$12 x$	-	$6$

Step 8.1.5.12

Divide the highest order term in the dividend $- 12 x$ by the highest order term in divisor $2 x$ .

				$x^{2}$	+	$5 x$	-	$6$
$2 x$	+	$1$		$2 x^{3}$	+	$11 x^{2}$	-	$7 x$	-	$6$
			-	$2 x^{3}$	-	$x^{2}$
					+	$10 x^{2}$	-	$7 x$
					-	$10 x^{2}$	-	$5 x$
							-	$12 x$	-	$6$

Step 8.1.5.13

Multiply the new quotient term by the divisor.

				$x^{2}$	+	$5 x$	-	$6$
$2 x$	+	$1$		$2 x^{3}$	+	$11 x^{2}$	-	$7 x$	-	$6$
			-	$2 x^{3}$	-	$x^{2}$
					+	$10 x^{2}$	-	$7 x$
					-	$10 x^{2}$	-	$5 x$
							-	$12 x$	-	$6$
							-	$12 x$	-	$6$

Step 8.1.5.14

The expression needs to be subtracted from the dividend, so change all the signs in $- 12 x - 6$

				$x^{2}$	+	$5 x$	-	$6$
$2 x$	+	$1$		$2 x^{3}$	+	$11 x^{2}$	-	$7 x$	-	$6$
			-	$2 x^{3}$	-	$x^{2}$
					+	$10 x^{2}$	-	$7 x$
					-	$10 x^{2}$	-	$5 x$
							-	$12 x$	-	$6$
							+	$12 x$	+	$6$

Step 8.1.5.15

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

				$x^{2}$	+	$5 x$	-	$6$
$2 x$	+	$1$		$2 x^{3}$	+	$11 x^{2}$	-	$7 x$	-	$6$
			-	$2 x^{3}$	-	$x^{2}$
					+	$10 x^{2}$	-	$7 x$
					-	$10 x^{2}$	-	$5 x$
							-	$12 x$	-	$6$
							+	$12 x$	+	$6$
										$0$

Step 8.1.5.16

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

$x^{2} + 5 x - 6$

Step 8.1.6

Write $2 x^{3} + 11 x^{2} - 7 x - 6$ as a set of factors.

$(2 x + 1) (x^{2} + 5 x - 6) = 0$

Step 8.2

Factor $x^{2} + 5 x - 6$ using the AC method.

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Step 8.2.1

Factor $x^{2} + 5 x - 6$ using the AC method.

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Step 8.2.1.1

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 $5$ .

$- 1, 6$

Step 8.2.1.2

Write the factored form using these integers.

$(2 x + 1) ((x - 1) (x + 6)) = 0$

Step 8.2.2

Remove unnecessary parentheses.

$(2 x + 1) (x - 1) (x + 6) = 0$

Step 9

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

$2 x + 1 = 0$

$x - 1 = 0$

$x + 6 = 0$

Step 10

Set $2 x + 1$ equal to $0$ and solve for $x$ .

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Step 10.1

Set $2 x + 1$ equal to $0$ .

$2 x + 1 = 0$

Step 10.2

Solve $2 x + 1 = 0$ for $x$ .

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Step 10.2.1

Subtract $1$ from both sides of the equation.

$2 x = - 1$

Step 10.2.2

Divide each term in $2 x = - 1$ by $2$ and simplify.

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Step 10.2.2.1

Divide each term in $2 x = - 1$ by $2$ .

$\frac{2 x}{2} = \frac{- 1}{2}$

Step 10.2.2.2

Simplify the left side.

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Step 10.2.2.2.1

Cancel the common factor of $2$ .

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Step 10.2.2.2.1.1

Cancel the common factor.

$\frac{2 x}{2} = \frac{- 1}{2}$

Step 10.2.2.2.1.2

Divide $x$ by $1$ .

$x = \frac{- 1}{2}$

Step 10.2.2.3

Simplify the right side.

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Step 10.2.2.3.1

Move the negative in front of the fraction.

$x = - \frac{1}{2}$

Step 11

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

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Step 11.1

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

$x - 1 = 0$

Step 11.2

Add $1$ to both sides of the equation.

$x = 1$

Step 12

Set $x + 6$ equal to $0$ and solve for $x$ .

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Step 12.1

Set $x + 6$ equal to $0$ .

$x + 6 = 0$

Step 12.2

Subtract $6$ from both sides of the equation.

$x = - 6$

Step 13

The final solution is all the values that make $(2 x + 1) (x - 1) (x + 6) = 0$ true.

$x = - \frac{1}{2}, 1, - 6$

Step 14