Precalculus Examples

$p (x) = 12 x^{3} + 16 x^{2} - x - 5$

Step 1

Set $12 x^{3} + 16 x^{2} - x - 5$ equal to $0$ .

$12 x^{3} + 16 x^{2} - x - 5 = 0$

Step 2

Solve for $x$ .

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Step 2.1

Factor the left side of the equation.

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Step 2.1.1

Factor $12 x^{3} + 16 x^{2} - x - 5$ using the rational roots test.

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Step 2.1.1.1

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.

$p = \pm 1, \pm 5$

$q = \pm 1, \pm 12, \pm 2, \pm 6, \pm 3, \pm 4$

Step 2.1.1.2

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

$\pm 1, \pm 0.08\overline{3}, \pm 0.5, \pm 0.1\overline{6}, \pm 0.\overline{3}, \pm 0.25, \pm 5, \pm 0.41\overline{6}, \pm 2.5, \pm 0.8\overline{3}, \pm 1.\overline{6}, \pm 1.25$

Step 2.1.1.3

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

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Step 2.1.1.3.1

Substitute $- 1$ into the polynomial.

$12 {(- 1)}^{3} + 16 {(- 1)}^{2} + 1 - 5$

Step 2.1.1.3.2

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

$12 \cdot - 1 + 16 {(- 1)}^{2} + 1 - 5$

Step 2.1.1.3.3

Multiply $12$ by $- 1$ .

$- 12 + 16 {(- 1)}^{2} + 1 - 5$

Step 2.1.1.3.4

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

$- 12 + 16 \cdot 1 + 1 - 5$

Step 2.1.1.3.5

Multiply $16$ by $1$ .

$- 12 + 16 + 1 - 5$

Step 2.1.1.3.6

Add $- 12$ and $16$ .

$4 + 1 - 5$

Step 2.1.1.3.7

Add $4$ and $1$ .

$5 - 5$

Step 2.1.1.3.8

Subtract $5$ from $5$ .

$0$

Step 2.1.1.4

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.

$\frac{12 x^{3} + 16 x^{2} - x - 5}{x + 1}$

Step 2.1.1.5

Divide $12 x^{3} + 16 x^{2} - x - 5$ by $x + 1$ .

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Step 2.1.1.5.1

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

$x$

$1$

$12 x^{3}$

$16 x^{2}$

$x$

$5$

Step 2.1.1.5.2

Divide the highest order term in the dividend $12 x^{3}$ by the highest order term in divisor $x$ .

					$12 x^{2}$
	$x$	+	$1$		$12 x^{3}$	+	$16 x^{2}$	-	$x$	-	$5$

Step 2.1.1.5.3

Multiply the new quotient term by the divisor.

				$12 x^{2}$
$x$	+	$1$		$12 x^{3}$	+	$16 x^{2}$	-	$x$	-	$5$
			+	$12 x^{3}$	+	$12 x^{2}$

Step 2.1.1.5.4

The expression needs to be subtracted from the dividend, so change all the signs in $12 x^{3} + 12 x^{2}$

				$12 x^{2}$
$x$	+	$1$		$12 x^{3}$	+	$16 x^{2}$	-	$x$	-	$5$
			-	$12 x^{3}$	-	$12 x^{2}$

Step 2.1.1.5.5

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

				$12 x^{2}$
$x$	+	$1$		$12 x^{3}$	+	$16 x^{2}$	-	$x$	-	$5$
			-	$12 x^{3}$	-	$12 x^{2}$
					+	$4 x^{2}$

Step 2.1.1.5.6

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

				$12 x^{2}$
$x$	+	$1$		$12 x^{3}$	+	$16 x^{2}$	-	$x$	-	$5$
			-	$12 x^{3}$	-	$12 x^{2}$
					+	$4 x^{2}$	-	$x$

Step 2.1.1.5.7

Divide the highest order term in the dividend $4 x^{2}$ by the highest order term in divisor $x$ .

				$12 x^{2}$	+	$4 x$
$x$	+	$1$		$12 x^{3}$	+	$16 x^{2}$	-	$x$	-	$5$
			-	$12 x^{3}$	-	$12 x^{2}$
					+	$4 x^{2}$	-	$x$

Step 2.1.1.5.8

Multiply the new quotient term by the divisor.

				$12 x^{2}$	+	$4 x$
$x$	+	$1$		$12 x^{3}$	+	$16 x^{2}$	-	$x$	-	$5$
			-	$12 x^{3}$	-	$12 x^{2}$
					+	$4 x^{2}$	-	$x$
					+	$4 x^{2}$	+	$4 x$

Step 2.1.1.5.9

The expression needs to be subtracted from the dividend, so change all the signs in $4 x^{2} + 4 x$

				$12 x^{2}$	+	$4 x$
$x$	+	$1$		$12 x^{3}$	+	$16 x^{2}$	-	$x$	-	$5$
			-	$12 x^{3}$	-	$12 x^{2}$
					+	$4 x^{2}$	-	$x$
					-	$4 x^{2}$	-	$4 x$

Step 2.1.1.5.10

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

				$12 x^{2}$	+	$4 x$
$x$	+	$1$		$12 x^{3}$	+	$16 x^{2}$	-	$x$	-	$5$
			-	$12 x^{3}$	-	$12 x^{2}$
					+	$4 x^{2}$	-	$x$
					-	$4 x^{2}$	-	$4 x$
							-	$5 x$

Step 2.1.1.5.11

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

				$12 x^{2}$	+	$4 x$
$x$	+	$1$		$12 x^{3}$	+	$16 x^{2}$	-	$x$	-	$5$
			-	$12 x^{3}$	-	$12 x^{2}$
					+	$4 x^{2}$	-	$x$
					-	$4 x^{2}$	-	$4 x$
							-	$5 x$	-	$5$

Step 2.1.1.5.12

Divide the highest order term in the dividend $- 5 x$ by the highest order term in divisor $x$ .

				$12 x^{2}$	+	$4 x$	-	$5$
$x$	+	$1$		$12 x^{3}$	+	$16 x^{2}$	-	$x$	-	$5$
			-	$12 x^{3}$	-	$12 x^{2}$
					+	$4 x^{2}$	-	$x$
					-	$4 x^{2}$	-	$4 x$
							-	$5 x$	-	$5$

Step 2.1.1.5.13

Multiply the new quotient term by the divisor.

				$12 x^{2}$	+	$4 x$	-	$5$
$x$	+	$1$		$12 x^{3}$	+	$16 x^{2}$	-	$x$	-	$5$
			-	$12 x^{3}$	-	$12 x^{2}$
					+	$4 x^{2}$	-	$x$
					-	$4 x^{2}$	-	$4 x$
							-	$5 x$	-	$5$
							-	$5 x$	-	$5$

Step 2.1.1.5.14

The expression needs to be subtracted from the dividend, so change all the signs in $- 5 x - 5$

				$12 x^{2}$	+	$4 x$	-	$5$
$x$	+	$1$		$12 x^{3}$	+	$16 x^{2}$	-	$x$	-	$5$
			-	$12 x^{3}$	-	$12 x^{2}$
					+	$4 x^{2}$	-	$x$
					-	$4 x^{2}$	-	$4 x$
							-	$5 x$	-	$5$
							+	$5 x$	+	$5$

Step 2.1.1.5.15

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

				$12 x^{2}$	+	$4 x$	-	$5$
$x$	+	$1$		$12 x^{3}$	+	$16 x^{2}$	-	$x$	-	$5$
			-	$12 x^{3}$	-	$12 x^{2}$
					+	$4 x^{2}$	-	$x$
					-	$4 x^{2}$	-	$4 x$
							-	$5 x$	-	$5$
							+	$5 x$	+	$5$
										$0$

Step 2.1.1.5.16

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

$12 x^{2} + 4 x - 5$

Step 2.1.1.6

Write $12 x^{3} + 16 x^{2} - x - 5$ as a set of factors.

$(x + 1) (12 x^{2} + 4 x - 5) = 0$

Step 2.1.2

Factor by grouping.

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Step 2.1.2.1

Factor by grouping.

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Step 2.1.2.1.1

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 $a \cdot c = 12 \cdot - 5 = - 60$ and whose sum is $b = 4$ .

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Step 2.1.2.1.1.1

Factor $4$ out of $4 x$ .

$(x + 1) (12 x^{2} + 4 (x) - 5) = 0$

Step 2.1.2.1.1.2

Rewrite $4$ as $- 6$ plus $10$

$(x + 1) (12 x^{2} + (- 6 + 10) x - 5) = 0$

Step 2.1.2.1.1.3

Apply the distributive property.

$(x + 1) (12 x^{2} - 6 x + 10 x - 5) = 0$

Step 2.1.2.1.2

Factor out the greatest common factor from each group.

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Step 2.1.2.1.2.1

Group the first two terms and the last two terms.

$(x + 1) ((12 x^{2} - 6 x) + 10 x - 5) = 0$

Step 2.1.2.1.2.2

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

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

Step 2.1.2.1.3

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

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

Step 2.1.2.2

Remove unnecessary parentheses.

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

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 $0$ .

$x + 1 = 0$

$2 x - 1 = 0$

$6 x + 5 = 0$

Step 2.3

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

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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 $2 x - 1$ equal to $0$ and solve for $x$ .

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Step 2.4.1

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

$2 x - 1 = 0$

Step 2.4.2

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

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Step 2.4.2.1

Add $1$ to both sides of the equation.

$2 x = 1$

Step 2.4.2.2

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

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Step 2.4.2.2.1

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

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

Step 2.4.2.2.2

Simplify the left side.

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Step 2.4.2.2.2.1

Cancel the common factor of $2$ .

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Step 2.4.2.2.2.1.1

Cancel the common factor.

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

Step 2.4.2.2.2.1.2

Divide $x$ by $1$ .

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

Step 2.5

Set $6 x + 5$ equal to $0$ and solve for $x$ .

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Step 2.5.1

Set $6 x + 5$ equal to $0$ .

$6 x + 5 = 0$

Step 2.5.2

Solve $6 x + 5 = 0$ for $x$ .

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Step 2.5.2.1

Subtract $5$ from both sides of the equation.

$6 x = - 5$

Step 2.5.2.2

Divide each term in $6 x = - 5$ by $6$ and simplify.

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Step 2.5.2.2.1

Divide each term in $6 x = - 5$ by $6$ .

$\frac{6 x}{6} = \frac{- 5}{6}$

Step 2.5.2.2.2

Simplify the left side.

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Step 2.5.2.2.2.1

Cancel the common factor of $6$ .

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Step 2.5.2.2.2.1.1

Cancel the common factor.

$\frac{6 x}{6} = \frac{- 5}{6}$

Step 2.5.2.2.2.1.2

Divide $x$ by $1$ .

$x = \frac{- 5}{6}$

Step 2.5.2.2.3

Simplify the right side.

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Step 2.5.2.2.3.1

Move the negative in front of the fraction.

$x = - \frac{5}{6}$

Step 2.6

The final solution is all the values that make $(x + 1) (2 x - 1) (6 x + 5) = 0$ true.

$x = - 1, \frac{1}{2}, - \frac{5}{6}$

Step 3