Precalculus Examples

$x^{3} + 4 x - 5$

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.

$p = \pm 1, \pm 5$

$q = \pm 1$

Step 2

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

$\pm 1, \pm 5$

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.

${(1)}^{3} + 4 (1) - 5$

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.

$1 + 4 (1) - 5$

Step 4.1.2

Multiply $4$ by $1$ .

$1 + 4 - 5$

Step 4.2

Simplify by adding and subtracting.

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Step 4.2.1

Add $1$ and $4$ .

$5 - 5$

Step 4.2.2

Subtract $5$ from $5$ .

$0$

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.

$\frac{x^{3} + 4 x - 5}{x - 1}$

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.

$1$	$1$	$0$	$4$	$- 5$

Step 6.2

The first number in the dividend $(1)$ is put into the first position of the result area (below the horizontal line).

$1$	$1$	$0$	$4$	$- 5$

	$1$

Step 6.3

Multiply the newest entry in the result $(1)$ by the divisor $(1)$ and place the result of $(1)$ under the next term in the dividend $(0)$ .

$1$	$1$	$0$	$4$	$- 5$
		$1$
	$1$

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.

$1$	$1$	$0$	$4$	$- 5$
		$1$
	$1$	$1$

Step 6.5

Multiply the newest entry in the result $(1)$ by the divisor $(1)$ and place the result of $(1)$ under the next term in the dividend $(4)$ .

$1$	$1$	$0$	$4$	$- 5$
		$1$	$1$
	$1$	$1$

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.

$1$	$1$	$0$	$4$	$- 5$
		$1$	$1$
	$1$	$1$	$5$

Step 6.7

Multiply the newest entry in the result $(5)$ by the divisor $(1)$ and place the result of $(5)$ under the next term in the dividend $(- 5)$ .

$1$	$1$	$0$	$4$	$- 5$
		$1$	$1$	$5$
	$1$	$1$	$5$

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.

$1$	$1$	$0$	$4$	$- 5$
		$1$	$1$	$5$
	$1$	$1$	$5$	$0$

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.

$1 x^{2} + 1 x + 5$

Step 6.10

Simplify the quotient polynomial.

$x^{2} + x + 5$

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 $a = 1$ , $b = 1$ , and $c = 5$ into the quadratic formula and solve for $x$ .

$\frac{- 1 \pm \sqrt{1^{2} - 4 \cdot (1 \cdot 5)}}{2 \cdot 1}$

Step 7.3

Simplify.

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Step 7.3.1

Simplify the numerator.

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Step 7.3.1.1

One to any power is one.

$x = \frac{- 1 \pm \sqrt{1 - 4 \cdot 1 \cdot 5}}{2 \cdot 1}$

Step 7.3.1.2

Multiply $- 4 \cdot 1 \cdot 5$ .

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Step 7.3.1.2.1

Multiply $- 4$ by $1$ .

$x = \frac{- 1 \pm \sqrt{1 - 4 \cdot 5}}{2 \cdot 1}$

Step 7.3.1.2.2

Multiply $- 4$ by $5$ .

$x = \frac{- 1 \pm \sqrt{1 - 20}}{2 \cdot 1}$

Step 7.3.1.3

Subtract $20$ from $1$ .

$x = \frac{- 1 \pm \sqrt{- 19}}{2 \cdot 1}$

Step 7.3.1.4

Rewrite $- 19$ as $- 1 (19)$ .

$x = \frac{- 1 \pm \sqrt{- 1 \cdot 19}}{2 \cdot 1}$

Step 7.3.1.5

Rewrite $\sqrt{- 1 (19)}$ as $\sqrt{- 1} \cdot \sqrt{19}$ .

$x = \frac{- 1 \pm \sqrt{- 1} \cdot \sqrt{19}}{2 \cdot 1}$

Step 7.3.1.6

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

$x = \frac{- 1 \pm i \sqrt{19}}{2 \cdot 1}$

Step 7.3.2

Multiply $2$ by $1$ .

$x = \frac{- 1 \pm i \sqrt{19}}{2}$

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

One to any power is one.

$x = \frac{- 1 \pm \sqrt{1 - 4 \cdot 1 \cdot 5}}{2 \cdot 1}$

Step 7.4.1.2

Multiply $- 4 \cdot 1 \cdot 5$ .

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Step 7.4.1.2.1

Multiply $- 4$ by $1$ .

$x = \frac{- 1 \pm \sqrt{1 - 4 \cdot 5}}{2 \cdot 1}$

Step 7.4.1.2.2

Multiply $- 4$ by $5$ .

$x = \frac{- 1 \pm \sqrt{1 - 20}}{2 \cdot 1}$

Step 7.4.1.3

Subtract $20$ from $1$ .

$x = \frac{- 1 \pm \sqrt{- 19}}{2 \cdot 1}$

Step 7.4.1.4

Rewrite $- 19$ as $- 1 (19)$ .

$x = \frac{- 1 \pm \sqrt{- 1 \cdot 19}}{2 \cdot 1}$

Step 7.4.1.5

Rewrite $\sqrt{- 1 (19)}$ as $\sqrt{- 1} \cdot \sqrt{19}$ .

$x = \frac{- 1 \pm \sqrt{- 1} \cdot \sqrt{19}}{2 \cdot 1}$

Step 7.4.1.6

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

$x = \frac{- 1 \pm i \sqrt{19}}{2 \cdot 1}$

Step 7.4.2

Multiply $2$ by $1$ .

$x = \frac{- 1 \pm i \sqrt{19}}{2}$

Step 7.4.3

Change the $\pm$ to $+$ .

$x = \frac{- 1 + i \sqrt{19}}{2}$

Step 7.4.4

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

$x = \frac{- 1 \cdot 1 + i \sqrt{19}}{2}$

Step 7.4.5

Factor $- 1$ out of $i \sqrt{19}$ .

$x = \frac{- 1 \cdot 1 - (- i \sqrt{19})}{2}$

Step 7.4.6

Factor $- 1$ out of $- 1 (1) - (- i \sqrt{19})$ .

$x = \frac{- 1 (1 - i \sqrt{19})}{2}$

Step 7.4.7

Move the negative in front of the fraction.

$x = - \frac{1 - i \sqrt{19}}{2}$

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

One to any power is one.

$x = \frac{- 1 \pm \sqrt{1 - 4 \cdot 1 \cdot 5}}{2 \cdot 1}$

Step 7.5.1.2

Multiply $- 4 \cdot 1 \cdot 5$ .

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Step 7.5.1.2.1

Multiply $- 4$ by $1$ .

$x = \frac{- 1 \pm \sqrt{1 - 4 \cdot 5}}{2 \cdot 1}$

Step 7.5.1.2.2

Multiply $- 4$ by $5$ .

$x = \frac{- 1 \pm \sqrt{1 - 20}}{2 \cdot 1}$

Step 7.5.1.3

Subtract $20$ from $1$ .

$x = \frac{- 1 \pm \sqrt{- 19}}{2 \cdot 1}$

Step 7.5.1.4

Rewrite $- 19$ as $- 1 (19)$ .

$x = \frac{- 1 \pm \sqrt{- 1 \cdot 19}}{2 \cdot 1}$

Step 7.5.1.5

Rewrite $\sqrt{- 1 (19)}$ as $\sqrt{- 1} \cdot \sqrt{19}$ .

$x = \frac{- 1 \pm \sqrt{- 1} \cdot \sqrt{19}}{2 \cdot 1}$

Step 7.5.1.6

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

$x = \frac{- 1 \pm i \sqrt{19}}{2 \cdot 1}$

Step 7.5.2

Multiply $2$ by $1$ .

$x = \frac{- 1 \pm i \sqrt{19}}{2}$

Step 7.5.3

Change the $\pm$ to $-$ .

$x = \frac{- 1 - i \sqrt{19}}{2}$

Step 7.5.4

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

$x = \frac{- 1 \cdot 1 - i \sqrt{19}}{2}$

Step 7.5.5

Factor $- 1$ out of $- i \sqrt{19}$ .

$x = \frac{- 1 \cdot 1 - (i \sqrt{19})}{2}$

Step 7.5.6

Factor $- 1$ out of $- 1 (1) - (i \sqrt{19})$ .

$x = \frac{- 1 (1 + i \sqrt{19})}{2}$

Step 7.5.7

Move the negative in front of the fraction.

$x = - \frac{1 + i \sqrt{19}}{2}$

Step 7.6

The final answer is the combination of both solutions.

$x = - \frac{1 - i \sqrt{19}}{2}, - \frac{1 + i \sqrt{19}}{2}$

Step 8

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

$(x - 1) (x + \frac{1 - i \sqrt{19}}{2}) (x + \frac{1 + i \sqrt{19}}{2})$

Step 9

These are the roots (zeros) of the polynomial $x^{3} + 4 x - 5$ .

$x = 1, - \frac{1 - i \sqrt{19}}{2}, - \frac{1 + i \sqrt{19}}{2}$

Step 10