Precalculus Examples

$2 x^{3} + 6 x^{2} + 5 x + 2$

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.

$p = \pm 1, \pm 2$

$q = \pm 1, \pm 2$

Step 2

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

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

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.

$2 {(- 2)}^{3} + 6 {(- 2)}^{2} + 5 (- 2) + 2$

Step 4

Simplify the expression. In this case, the expression is equal to $0$ so $x = - 2$ is a root of the polynomial.

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Step 4.1

Simplify each term.

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Step 4.1.1

Raise $- 2$ to the power of $3$ .

$2 \cdot - 8 + 6 {(- 2)}^{2} + 5 (- 2) + 2$

Step 4.1.2

Multiply $2$ by $- 8$ .

$- 16 + 6 {(- 2)}^{2} + 5 (- 2) + 2$

Step 4.1.3

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

$- 16 + 6 \cdot 4 + 5 (- 2) + 2$

Step 4.1.4

Multiply $6$ by $4$ .

$- 16 + 24 + 5 (- 2) + 2$

Step 4.1.5

Multiply $5$ by $- 2$ .

$- 16 + 24 - 10 + 2$

Step 4.2

Simplify by adding and subtracting.

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Step 4.2.1

Add $- 16$ and $24$ .

$8 - 10 + 2$

Step 4.2.2

Subtract $10$ from $8$ .

$- 2 + 2$

Step 4.2.3

Add $- 2$ and $2$ .

$0$

Step 5

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

$\frac{2 x^{3} + 6 x^{2} + 5 x + 2}{x + 2}$

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.

$- 2$	$2$	$6$	$5$	$2$

Step 6.2

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

$- 2$	$2$	$6$	$5$	$2$

	$2$

Step 6.3

Multiply the newest entry in the result $(2)$ by the divisor $(- 2)$ and place the result of $(- 4)$ under the next term in the dividend $(6)$ .

$- 2$	$2$	$6$	$5$	$2$
		$- 4$
	$2$

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.

$- 2$	$2$	$6$	$5$	$2$
		$- 4$
	$2$	$2$

Step 6.5

Multiply the newest entry in the result $(2)$ by the divisor $(- 2)$ and place the result of $(- 4)$ under the next term in the dividend $(5)$ .

$- 2$	$2$	$6$	$5$	$2$
		$- 4$	$- 4$
	$2$	$2$

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.

$- 2$	$2$	$6$	$5$	$2$
		$- 4$	$- 4$
	$2$	$2$	$1$

Step 6.7

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

$- 2$	$2$	$6$	$5$	$2$
		$- 4$	$- 4$	$- 2$
	$2$	$2$	$1$

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.

$- 2$	$2$	$6$	$5$	$2$
		$- 4$	$- 4$	$- 2$
	$2$	$2$	$1$	$0$

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.

$2 x^{2} + 2 x + 1$

Step 7

Solve the equation to find any remaining roots.

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Step 7.1

Use the quadratic formula to find the solutions.

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

Step 7.2

Substitute the values $a = 2$ , $b = 2$ , and $c = 1$ into the quadratic formula and solve for $x$ .

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

Step 7.3

Simplify.

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Step 7.3.1

Simplify the numerator.

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Step 7.3.1.1

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

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

Step 7.3.1.2

Multiply $- 4 \cdot 2 \cdot 1$ .

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Step 7.3.1.2.1

Multiply $- 4$ by $2$ .

$x = \frac{- 2 \pm \sqrt{4 - 8 \cdot 1}}{2 \cdot 2}$

Step 7.3.1.2.2

Multiply $- 8$ by $1$ .

$x = \frac{- 2 \pm \sqrt{4 - 8}}{2 \cdot 2}$

Step 7.3.1.3

Subtract $8$ from $4$ .

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

Step 7.3.1.4

Rewrite $- 4$ as $- 1 (4)$ .

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

Step 7.3.1.5

Rewrite $\sqrt{- 1 (4)}$ as $\sqrt{- 1} \cdot \sqrt{4}$ .

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

Step 7.3.1.6

Rewrite $\sqrt{- 1}$ as $i$ .

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

Step 7.3.1.7

Rewrite $4$ as $2^{2}$ .

$x = \frac{- 2 \pm i \cdot \sqrt{2^{2}}}{2 \cdot 2}$

Step 7.3.1.8

Pull terms out from under the radical, assuming positive real numbers.

$x = \frac{- 2 \pm i \cdot 2}{2 \cdot 2}$

Step 7.3.1.9

Move $2$ to the left of $i$ .

$x = \frac{- 2 \pm 2 i}{2 \cdot 2}$

Step 7.3.2

Multiply $2$ by $2$ .

$x = \frac{- 2 \pm 2 i}{4}$

Step 7.3.3

Simplify $\frac{- 2 \pm 2 i}{4}$ .

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

Step 7.4

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

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Step 7.4.1

Simplify the numerator.

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Step 7.4.1.1

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

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

Step 7.4.1.2

Multiply $- 4 \cdot 2 \cdot 1$ .

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Step 7.4.1.2.1

Multiply $- 4$ by $2$ .

$x = \frac{- 2 \pm \sqrt{4 - 8 \cdot 1}}{2 \cdot 2}$

Step 7.4.1.2.2

Multiply $- 8$ by $1$ .

$x = \frac{- 2 \pm \sqrt{4 - 8}}{2 \cdot 2}$

Step 7.4.1.3

Subtract $8$ from $4$ .

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

Step 7.4.1.4

Rewrite $- 4$ as $- 1 (4)$ .

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

Step 7.4.1.5

Rewrite $\sqrt{- 1 (4)}$ as $\sqrt{- 1} \cdot \sqrt{4}$ .

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

Step 7.4.1.6

Rewrite $\sqrt{- 1}$ as $i$ .

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

Step 7.4.1.7

Rewrite $4$ as $2^{2}$ .

$x = \frac{- 2 \pm i \cdot \sqrt{2^{2}}}{2 \cdot 2}$

Step 7.4.1.8

Pull terms out from under the radical, assuming positive real numbers.

$x = \frac{- 2 \pm i \cdot 2}{2 \cdot 2}$

Step 7.4.1.9

Move $2$ to the left of $i$ .

$x = \frac{- 2 \pm 2 i}{2 \cdot 2}$

Step 7.4.2

Multiply $2$ by $2$ .

$x = \frac{- 2 \pm 2 i}{4}$

Step 7.4.3

Simplify $\frac{- 2 \pm 2 i}{4}$ .

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

Step 7.4.4

Change the $\pm$ to $+$ .

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

Step 7.4.5

Split the fraction $\frac{- 1 + i}{2}$ into two fractions.

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

Step 7.4.6

Move the negative in front of the fraction.

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

Step 7.5

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

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Step 7.5.1

Simplify the numerator.

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Step 7.5.1.1

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

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

Step 7.5.1.2

Multiply $- 4 \cdot 2 \cdot 1$ .

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Step 7.5.1.2.1

Multiply $- 4$ by $2$ .

$x = \frac{- 2 \pm \sqrt{4 - 8 \cdot 1}}{2 \cdot 2}$

Step 7.5.1.2.2

Multiply $- 8$ by $1$ .

$x = \frac{- 2 \pm \sqrt{4 - 8}}{2 \cdot 2}$

Step 7.5.1.3

Subtract $8$ from $4$ .

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

Step 7.5.1.4

Rewrite $- 4$ as $- 1 (4)$ .

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

Step 7.5.1.5

Rewrite $\sqrt{- 1 (4)}$ as $\sqrt{- 1} \cdot \sqrt{4}$ .

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

Step 7.5.1.6

Rewrite $\sqrt{- 1}$ as $i$ .

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

Step 7.5.1.7

Rewrite $4$ as $2^{2}$ .

$x = \frac{- 2 \pm i \cdot \sqrt{2^{2}}}{2 \cdot 2}$

Step 7.5.1.8

Pull terms out from under the radical, assuming positive real numbers.

$x = \frac{- 2 \pm i \cdot 2}{2 \cdot 2}$

Step 7.5.1.9

Move $2$ to the left of $i$ .

$x = \frac{- 2 \pm 2 i}{2 \cdot 2}$

Step 7.5.2

Multiply $2$ by $2$ .

$x = \frac{- 2 \pm 2 i}{4}$

Step 7.5.3

Simplify $\frac{- 2 \pm 2 i}{4}$ .

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

Step 7.5.4

Change the $\pm$ to $-$ .

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

Step 7.5.5

Split the fraction $\frac{- 1 - i}{2}$ into two fractions.

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

Step 7.5.6

Simplify each term.

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Step 7.5.6.1

Move the negative in front of the fraction.

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

Step 7.5.6.2

Move the negative in front of the fraction.

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

Step 7.6

The final answer is the combination of both solutions.

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

Step 8

The polynomial can be written as a set of linear factors.

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

Step 9

These are the roots (zeros) of the polynomial $2 x^{3} + 6 x^{2} + 5 x + 2$ .

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

Step 10