Precalculus Examples

$x^{3} - 15 x - 4 = 0$

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.

$p = \pm 1, \pm 2, \pm 4$

$q = \pm 1$

Step 2

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

$\pm 1, \pm 2, \pm 4$

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.

${(4)}^{3} - 15 \cdot 4 - 4$

Step 4

Simplify the expression. In this case, the expression is equal to $0$ so $x = 4$ 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 $4$ to the power of $3$ .

$64 - 15 \cdot 4 - 4$

Step 4.1.2

Multiply $- 15$ by $4$ .

$64 - 60 - 4$

Step 4.2

Simplify by subtracting numbers.

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Step 4.2.1

Subtract $60$ from $64$ .

$4 - 4$

Step 4.2.2

Subtract $4$ from $4$ .

$0$

Step 5

Since $4$ is a known root, divide the polynomial by $x - 4$ to find the quotient polynomial. This polynomial can then be used to find the remaining roots.

$\frac{x^{3} - 15 x - 4}{x - 4}$

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.

$4$	$1$	$0$	$- 15$	$- 4$

Step 6.2

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

$4$	$1$	$0$	$- 15$	$- 4$

	$1$

Step 6.3

Multiply the newest entry in the result $(1)$ by the divisor $(4)$ and place the result of $(4)$ under the next term in the dividend $(0)$ .

$4$	$1$	$0$	$- 15$	$- 4$
		$4$
	$1$

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.

$4$	$1$	$0$	$- 15$	$- 4$
		$4$
	$1$	$4$

Step 6.5

Multiply the newest entry in the result $(4)$ by the divisor $(4)$ and place the result of $(16)$ under the next term in the dividend $(- 15)$ .

$4$	$1$	$0$	$- 15$	$- 4$
		$4$	$16$
	$1$	$4$

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.

$4$	$1$	$0$	$- 15$	$- 4$
		$4$	$16$
	$1$	$4$	$1$

Step 6.7

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

$4$	$1$	$0$	$- 15$	$- 4$
		$4$	$16$	$4$
	$1$	$4$	$1$

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.

$4$	$1$	$0$	$- 15$	$- 4$
		$4$	$16$	$4$
	$1$	$4$	$1$	$0$

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.

$1 x^{2} + 4 x + 1$

Step 6.10

Simplify the quotient polynomial.

$x^{2} + 4 x + 1$

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 $a = 1$ , $b = 4$ , and $c = 1$ into the quadratic formula and solve for $x$ .

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

Step 7.3

Simplify.

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Step 7.3.1

Simplify the numerator.

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Step 7.3.1.1

Raise $4$ to the power of $2$ .

$x = \frac{- 4 \pm \sqrt{16 - 4 \cdot 1 \cdot 1}}{2 \cdot 1}$

Step 7.3.1.2

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

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Step 7.3.1.2.1

Multiply $- 4$ by $1$ .

$x = \frac{- 4 \pm \sqrt{16 - 4 \cdot 1}}{2 \cdot 1}$

Step 7.3.1.2.2

Multiply $- 4$ by $1$ .

$x = \frac{- 4 \pm \sqrt{16 - 4}}{2 \cdot 1}$

Step 7.3.1.3

Subtract $4$ from $16$ .

$x = \frac{- 4 \pm \sqrt{12}}{2 \cdot 1}$

Step 7.3.1.4

Rewrite $12$ as $2^{2} \cdot 3$ .

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Step 7.3.1.4.1

Factor $4$ out of $12$ .

$x = \frac{- 4 \pm \sqrt{4 (3)}}{2 \cdot 1}$

Step 7.3.1.4.2

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

$x = \frac{- 4 \pm \sqrt{2^{2} \cdot 3}}{2 \cdot 1}$

Step 7.3.1.5

Pull terms out from under the radical.

$x = \frac{- 4 \pm 2 \sqrt{3}}{2 \cdot 1}$

Step 7.3.2

Multiply $2$ by $1$ .

$x = \frac{- 4 \pm 2 \sqrt{3}}{2}$

Step 7.3.3

Simplify $\frac{- 4 \pm 2 \sqrt{3}}{2}$ .

$x = - 2 \pm \sqrt{3}$

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 $4$ to the power of $2$ .

$x = \frac{- 4 \pm \sqrt{16 - 4 \cdot 1 \cdot 1}}{2 \cdot 1}$

Step 7.4.1.2

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

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Step 7.4.1.2.1

Multiply $- 4$ by $1$ .

$x = \frac{- 4 \pm \sqrt{16 - 4 \cdot 1}}{2 \cdot 1}$

Step 7.4.1.2.2

Multiply $- 4$ by $1$ .

$x = \frac{- 4 \pm \sqrt{16 - 4}}{2 \cdot 1}$

Step 7.4.1.3

Subtract $4$ from $16$ .

$x = \frac{- 4 \pm \sqrt{12}}{2 \cdot 1}$

Step 7.4.1.4

Rewrite $12$ as $2^{2} \cdot 3$ .

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Step 7.4.1.4.1

Factor $4$ out of $12$ .

$x = \frac{- 4 \pm \sqrt{4 (3)}}{2 \cdot 1}$

Step 7.4.1.4.2

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

$x = \frac{- 4 \pm \sqrt{2^{2} \cdot 3}}{2 \cdot 1}$

Step 7.4.1.5

Pull terms out from under the radical.

$x = \frac{- 4 \pm 2 \sqrt{3}}{2 \cdot 1}$

Step 7.4.2

Multiply $2$ by $1$ .

$x = \frac{- 4 \pm 2 \sqrt{3}}{2}$

Step 7.4.3

Simplify $\frac{- 4 \pm 2 \sqrt{3}}{2}$ .

$x = - 2 \pm \sqrt{3}$

Step 7.4.4

Change the $\pm$ to $+$ .

$x = - 2 + \sqrt{3}$

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 $4$ to the power of $2$ .

$x = \frac{- 4 \pm \sqrt{16 - 4 \cdot 1 \cdot 1}}{2 \cdot 1}$

Step 7.5.1.2

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

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Step 7.5.1.2.1

Multiply $- 4$ by $1$ .

$x = \frac{- 4 \pm \sqrt{16 - 4 \cdot 1}}{2 \cdot 1}$

Step 7.5.1.2.2

Multiply $- 4$ by $1$ .

$x = \frac{- 4 \pm \sqrt{16 - 4}}{2 \cdot 1}$

Step 7.5.1.3

Subtract $4$ from $16$ .

$x = \frac{- 4 \pm \sqrt{12}}{2 \cdot 1}$

Step 7.5.1.4

Rewrite $12$ as $2^{2} \cdot 3$ .

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Step 7.5.1.4.1

Factor $4$ out of $12$ .

$x = \frac{- 4 \pm \sqrt{4 (3)}}{2 \cdot 1}$

Step 7.5.1.4.2

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

$x = \frac{- 4 \pm \sqrt{2^{2} \cdot 3}}{2 \cdot 1}$

Step 7.5.1.5

Pull terms out from under the radical.

$x = \frac{- 4 \pm 2 \sqrt{3}}{2 \cdot 1}$

Step 7.5.2

Multiply $2$ by $1$ .

$x = \frac{- 4 \pm 2 \sqrt{3}}{2}$

Step 7.5.3

Simplify $\frac{- 4 \pm 2 \sqrt{3}}{2}$ .

$x = - 2 \pm \sqrt{3}$

Step 7.5.4

Change the $\pm$ to $-$ .

$x = - 2 - \sqrt{3}$

Step 7.6

The final answer is the combination of both solutions.

$x = - 2 + \sqrt{3}, - 2 - \sqrt{3}$

Step 8

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

$(x - 4) (x + 2 - \sqrt{3}) (x + 2 + \sqrt{3})$

Step 9

These are the roots (zeros) of the polynomial $x^{3} - 15 x - 4$ .

$x = 4, - 2 + \sqrt{3}, - 2 - \sqrt{3}$

Step 10

The result can be shown in multiple forms.

Exact Form:

$x = 4, - 2 + \sqrt{3}, - 2 - \sqrt{3}$

Decimal Form:

$x = 4, - 0.26794919 \dots, - 3.73205080 \dots$

Step 11