Precalculus Examples

$f (x) = 7 x^{3} + 12 x^{2} - 2 x - 5$

Step 1

Set $7 x^{3} + 12 x^{2} - 2 x - 5$ equal to $0$ .

$7 x^{3} + 12 x^{2} - 2 x - 5 = 0$

Step 2

Solve for $x$ .

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Step 2.1

Factor $7 x^{3} + 12 x^{2} - 2 x - 5$ using the rational roots test.

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Step 2.1.1

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.

$p = \pm 1, \pm 5$

$q = \pm 1, \pm 7$

Step 2.1.2

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

$\pm 1, \pm 0.\overline{142857}, \pm 5, \pm 0.\overline{714285}$

Step 2.1.3

Substitute $- 0.\overline{714285}$ and simplify the expression. In this case, the expression is equal to $0$ so $- 0.\overline{714285}$ is a root of the polynomial.

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Step 2.1.3.1

Substitute $- 0.\overline{714285}$ into the polynomial.

$7 {(- 0.\overline{714285})}^{3} + 12 {(- 0.\overline{714285})}^{2} - 2 \cdot - 0.\overline{714285} - 5$

Step 2.1.3.2

Raise $- 0.\overline{714285}$ to the power of $3$ .

$7 \cdot - 0.36443148 + 12 {(- 0.\overline{714285})}^{2} - 2 \cdot - 0.\overline{714285} - 5$

Step 2.1.3.3

Multiply $7$ by $- 0.36443148$ .

$- 2.5510204 + 12 {(- 0.\overline{714285})}^{2} - 2 \cdot - 0.\overline{714285} - 5$

Step 2.1.3.4

Raise $- 0.\overline{714285}$ to the power of $2$ .

$- 2.5510204 + 12 \cdot 0.51020408 - 2 \cdot - 0.\overline{714285} - 5$

Step 2.1.3.5

Multiply $12$ by $0.51020408$ .

$- 2.5510204 + 6.12244897 - 2 \cdot - 0.\overline{714285} - 5$

Step 2.1.3.6

Add $- 2.5510204$ and $6.12244897$ .

$3.\overline{571428} - 2 \cdot - 0.\overline{714285} - 5$

Step 2.1.3.7

Multiply $- 2$ by $- 0.\overline{714285}$ .

$3.\overline{571428} + 1.\overline{428571} - 5$

Step 2.1.3.8

Add $3.\overline{571428}$ and $1.\overline{428571}$ .

$5 - 5$

Step 2.1.3.9

Subtract $5$ from $5$ .

$0$

Step 2.1.4

Since $- 0.\overline{714285}$ is a known root, divide the polynomial by $7 x + 5$ to find the quotient polynomial. This polynomial can then be used to find the remaining roots.

$\frac{7 x^{3} + 12 x^{2} - 2 x - 5}{7 x + 5}$

Step 2.1.5

Divide $7 x^{3} + 12 x^{2} - 2 x - 5$ by $7 x + 5$ .

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Step 2.1.5.1

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

$7 x$

$5$

$7 x^{3}$

$12 x^{2}$

$2 x$

$5$

Step 2.1.5.2

Divide the highest order term in the dividend $7 x^{3}$ by the highest order term in divisor $7 x$ .

					$x^{2}$
	$7 x$	+	$5$		$7 x^{3}$	+	$12 x^{2}$	-	$2 x$	-	$5$

Step 2.1.5.3

Multiply the new quotient term by the divisor.

				$x^{2}$
$7 x$	+	$5$		$7 x^{3}$	+	$12 x^{2}$	-	$2 x$	-	$5$
			+	$7 x^{3}$	+	$5 x^{2}$

Step 2.1.5.4

The expression needs to be subtracted from the dividend, so change all the signs in $7 x^{3} + 5 x^{2}$

				$x^{2}$
$7 x$	+	$5$		$7 x^{3}$	+	$12 x^{2}$	-	$2 x$	-	$5$
			-	$7 x^{3}$	-	$5 x^{2}$

Step 2.1.5.5

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

				$x^{2}$
$7 x$	+	$5$		$7 x^{3}$	+	$12 x^{2}$	-	$2 x$	-	$5$
			-	$7 x^{3}$	-	$5 x^{2}$
					+	$7 x^{2}$

Step 2.1.5.6

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

				$x^{2}$
$7 x$	+	$5$		$7 x^{3}$	+	$12 x^{2}$	-	$2 x$	-	$5$
			-	$7 x^{3}$	-	$5 x^{2}$
					+	$7 x^{2}$	-	$2 x$

Step 2.1.5.7

Divide the highest order term in the dividend $7 x^{2}$ by the highest order term in divisor $7 x$ .

				$x^{2}$	+	$x$
$7 x$	+	$5$		$7 x^{3}$	+	$12 x^{2}$	-	$2 x$	-	$5$
			-	$7 x^{3}$	-	$5 x^{2}$
					+	$7 x^{2}$	-	$2 x$

Step 2.1.5.8

Multiply the new quotient term by the divisor.

				$x^{2}$	+	$x$
$7 x$	+	$5$		$7 x^{3}$	+	$12 x^{2}$	-	$2 x$	-	$5$
			-	$7 x^{3}$	-	$5 x^{2}$
					+	$7 x^{2}$	-	$2 x$
					+	$7 x^{2}$	+	$5 x$

Step 2.1.5.9

The expression needs to be subtracted from the dividend, so change all the signs in $7 x^{2} + 5 x$

				$x^{2}$	+	$x$
$7 x$	+	$5$		$7 x^{3}$	+	$12 x^{2}$	-	$2 x$	-	$5$
			-	$7 x^{3}$	-	$5 x^{2}$
					+	$7 x^{2}$	-	$2 x$
					-	$7 x^{2}$	-	$5 x$

Step 2.1.5.10

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

				$x^{2}$	+	$x$
$7 x$	+	$5$		$7 x^{3}$	+	$12 x^{2}$	-	$2 x$	-	$5$
			-	$7 x^{3}$	-	$5 x^{2}$
					+	$7 x^{2}$	-	$2 x$
					-	$7 x^{2}$	-	$5 x$
							-	$7 x$

Step 2.1.5.11

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

				$x^{2}$	+	$x$
$7 x$	+	$5$		$7 x^{3}$	+	$12 x^{2}$	-	$2 x$	-	$5$
			-	$7 x^{3}$	-	$5 x^{2}$
					+	$7 x^{2}$	-	$2 x$
					-	$7 x^{2}$	-	$5 x$
							-	$7 x$	-	$5$

Step 2.1.5.12

Divide the highest order term in the dividend $- 7 x$ by the highest order term in divisor $7 x$ .

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

Step 2.1.5.13

Multiply the new quotient term by the divisor.

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

Step 2.1.5.14

The expression needs to be subtracted from the dividend, so change all the signs in $- 7 x - 5$

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

Step 2.1.5.15

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

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

Step 2.1.5.16

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

$x^{2} + x - 1$

Step 2.1.6

Write $7 x^{3} + 12 x^{2} - 2 x - 5$ as a set of factors.

$(7 x + 5) (x^{2} + x - 1) = 0$

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 $0$ .

$7 x + 5 = 0$

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

Step 2.3

Set $7 x + 5$ equal to $0$ and solve for $x$ .

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Step 2.3.1

Set $7 x + 5$ equal to $0$ .

$7 x + 5 = 0$

Step 2.3.2

Solve $7 x + 5 = 0$ for $x$ .

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Step 2.3.2.1

Subtract $5$ from both sides of the equation.

$7 x = - 5$

Step 2.3.2.2

Divide each term in $7 x = - 5$ by $7$ and simplify.

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Step 2.3.2.2.1

Divide each term in $7 x = - 5$ by $7$ .

$\frac{7 x}{7} = \frac{- 5}{7}$

Step 2.3.2.2.2

Simplify the left side.

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Step 2.3.2.2.2.1

Cancel the common factor of $7$ .

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Step 2.3.2.2.2.1.1

Cancel the common factor.

$\frac{7 x}{7} = \frac{- 5}{7}$

Step 2.3.2.2.2.1.2

Divide $x$ by $1$ .

$x = \frac{- 5}{7}$

Step 2.3.2.2.3

Simplify the right side.

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Step 2.3.2.2.3.1

Move the negative in front of the fraction.

$x = - \frac{5}{7}$

Step 2.4

Set $x^{2} + x - 1$ equal to $0$ and solve for $x$ .

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Step 2.4.1

Set $x^{2} + x - 1$ equal to $0$ .

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

Step 2.4.2

Solve $x^{2} + x - 1 = 0$ for $x$ .

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Step 2.4.2.1

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Step 2.4.2.2

Substitute the values $a = 1$ , $b = 1$ , and $c = - 1$ into the quadratic formula and solve for $x$ .

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

Step 2.4.2.3

Simplify.

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Step 2.4.2.3.1

Simplify the numerator.

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Step 2.4.2.3.1.1

One to any power is one.

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

Step 2.4.2.3.1.2

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

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Step 2.4.2.3.1.2.1

Multiply $- 4$ by $1$ .

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

Step 2.4.2.3.1.2.2

Multiply $- 4$ by $- 1$ .

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

Step 2.4.2.3.1.3

Add $1$ and $4$ .

$x = \frac{- 1 \pm \sqrt{5}}{2 \cdot 1}$

Step 2.4.2.3.2

Multiply $2$ by $1$ .

$x = \frac{- 1 \pm \sqrt{5}}{2}$

Step 2.4.2.4

Simplify the expression to solve for the $+$ portion of the $\pm$ .

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Step 2.4.2.4.1

Simplify the numerator.

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Step 2.4.2.4.1.1

One to any power is one.

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

Step 2.4.2.4.1.2

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

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Step 2.4.2.4.1.2.1

Multiply $- 4$ by $1$ .

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

Step 2.4.2.4.1.2.2

Multiply $- 4$ by $- 1$ .

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

Step 2.4.2.4.1.3

Add $1$ and $4$ .

$x = \frac{- 1 \pm \sqrt{5}}{2 \cdot 1}$

Step 2.4.2.4.2

Multiply $2$ by $1$ .

$x = \frac{- 1 \pm \sqrt{5}}{2}$

Step 2.4.2.4.3

Change the $\pm$ to $+$ .

$x = \frac{- 1 + \sqrt{5}}{2}$

Step 2.4.2.4.4

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

$x = \frac{- 1 \cdot 1 + \sqrt{5}}{2}$

Step 2.4.2.4.5

Factor $- 1$ out of $\sqrt{5}$ .

$x = \frac{- 1 \cdot 1 - 1 (- \sqrt{5})}{2}$

Step 2.4.2.4.6

Factor $- 1$ out of $- 1 (1) - 1 (- \sqrt{5})$ .

$x = \frac{- 1 (1 - \sqrt{5})}{2}$

Step 2.4.2.4.7

Move the negative in front of the fraction.

$x = - \frac{1 - \sqrt{5}}{2}$

Step 2.4.2.5

Simplify the expression to solve for the $-$ portion of the $\pm$ .

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Step 2.4.2.5.1

Simplify the numerator.

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Step 2.4.2.5.1.1

One to any power is one.

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

Step 2.4.2.5.1.2

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

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Step 2.4.2.5.1.2.1

Multiply $- 4$ by $1$ .

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

Step 2.4.2.5.1.2.2

Multiply $- 4$ by $- 1$ .

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

Step 2.4.2.5.1.3

Add $1$ and $4$ .

$x = \frac{- 1 \pm \sqrt{5}}{2 \cdot 1}$

Step 2.4.2.5.2

Multiply $2$ by $1$ .

$x = \frac{- 1 \pm \sqrt{5}}{2}$

Step 2.4.2.5.3

Change the $\pm$ to $-$ .

$x = \frac{- 1 - \sqrt{5}}{2}$

Step 2.4.2.5.4

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

$x = \frac{- 1 \cdot 1 - \sqrt{5}}{2}$

Step 2.4.2.5.5

Factor $- 1$ out of $- \sqrt{5}$ .

$x = \frac{- 1 \cdot 1 - (\sqrt{5})}{2}$

Step 2.4.2.5.6

Factor $- 1$ out of $- 1 (1) - (\sqrt{5})$ .

$x = \frac{- 1 (1 + \sqrt{5})}{2}$

Step 2.4.2.5.7

Move the negative in front of the fraction.

$x = - \frac{1 + \sqrt{5}}{2}$

Step 2.4.2.6

The final answer is the combination of both solutions.

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

Step 2.5

The final solution is all the values that make $(7 x + 5) (x^{2} + x - 1) = 0$ true. The multiplicity of a root is the number of times the root appears.

$x = - \frac{5}{7}$ (Multiplicity of $1$ )

$x = - \frac{1 - \sqrt{5}}{2}$ (Multiplicity of $1$ )

$x = - \frac{1 + \sqrt{5}}{2}$ (Multiplicity of $1$ )

$x = - \frac{5}{7}$ (Multiplicity of $1$ )

$x = - \frac{1 - \sqrt{5}}{2}$ (Multiplicity of $1$ )

$x = - \frac{1 + \sqrt{5}}{2}$ (Multiplicity of $1$ )

Step 3