Precalculus Examples

$x^{4} - 2 x^{2} - 16 x - 15$

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.

$p = \pm 1, \pm 3, \pm 5, \pm 15$

$q = \pm 1$

Step 2

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

$\pm 1, \pm 3, \pm 5, \pm 15$

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.

${(- 1)}^{4} - 2 {(- 1)}^{2} - 16 \cdot - 1 - 15$

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

Raise $- 1$ to the power of $4$ .

$1 - 2 {(- 1)}^{2} - 16 \cdot - 1 - 15$

Step 4.1.2

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

$1 - 2 \cdot 1 - 16 \cdot - 1 - 15$

Step 4.1.3

Multiply $- 2$ by $1$ .

$1 - 2 - 16 \cdot - 1 - 15$

Step 4.1.4

Multiply $- 16$ by $- 1$ .

$1 - 2 + 16 - 15$

Step 4.2

Simplify by adding and subtracting.

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Step 4.2.1

Subtract $2$ from $1$ .

$- 1 + 16 - 15$

Step 4.2.2

Add $- 1$ and $16$ .

$15 - 15$

Step 4.2.3

Subtract $15$ from $15$ .

$0$

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.

$\frac{x^{4} - 2 x^{2} - 16 x - 15}{x + 1}$

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.

$- 1$	$1$	$0$	$- 2$	$- 16$	$- 15$

Step 6.2

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

$- 1$	$1$	$0$	$- 2$	$- 16$	$- 15$

	$1$

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 $(0)$ .

$- 1$	$1$	$0$	$- 2$	$- 16$	$- 15$
		$- 1$
	$1$

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.

$- 1$	$1$	$0$	$- 2$	$- 16$	$- 15$
		$- 1$
	$1$	$- 1$

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 $(- 2)$ .

$- 1$	$1$	$0$	$- 2$	$- 16$	$- 15$
		$- 1$	$1$
	$1$	$- 1$

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.

$- 1$	$1$	$0$	$- 2$	$- 16$	$- 15$
		$- 1$	$1$
	$1$	$- 1$	$- 1$

Step 6.7

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 $(- 16)$ .

$- 1$	$1$	$0$	$- 2$	$- 16$	$- 15$
		$- 1$	$1$	$1$
	$1$	$- 1$	$- 1$

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.

$- 1$	$1$	$0$	$- 2$	$- 16$	$- 15$
		$- 1$	$1$	$1$
	$1$	$- 1$	$- 1$	$- 15$

Step 6.9

Multiply the newest entry in the result $(- 15)$ by the divisor $(- 1)$ and place the result of $(15)$ under the next term in the dividend $(- 15)$ .

$- 1$	$1$	$0$	$- 2$	$- 16$	$- 15$
		$- 1$	$1$	$1$	$15$
	$1$	$- 1$	$- 1$	$- 15$

Step 6.10

Add the product of the multiplication and the number from the dividend and put the result in the next position on the result line.

$- 1$	$1$	$0$	$- 2$	$- 16$	$- 15$
		$- 1$	$1$	$1$	$15$
	$1$	$- 1$	$- 1$	$- 15$	$0$

Step 6.11

All numbers except the last become the coefficients of the quotient polynomial. The last value in the result line is the remainder.

$1 x^{3} + - 1 x^{2} + (- 1) x - 15$

Step 6.12

Simplify the quotient polynomial.

$x^{3} - x^{2} - x - 15$

Step 7

Solve the equation to find any remaining roots.

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Step 7.1

Factor $x^{3} - x^{2} - x - 15$ using the rational roots test.

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Step 7.1.1

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.

$p = \pm 1, \pm 15, \pm 3, \pm 5$

$q = \pm 1$

Step 7.1.2

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

$\pm 1, \pm 15, \pm 3, \pm 5$

Step 7.1.3

Substitute $3$ and simplify the expression. In this case, the expression is equal to $0$ so $3$ is a root of the polynomial.

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Step 7.1.3.1

Substitute $3$ into the polynomial.

$3^{3} - 3^{2} - 3 - 15$

Step 7.1.3.2

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

$27 - 3^{2} - 3 - 15$

Step 7.1.3.3

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

$27 - 1 \cdot 9 - 3 - 15$

Step 7.1.3.4

Multiply $- 1$ by $9$ .

$27 - 9 - 3 - 15$

Step 7.1.3.5

Subtract $9$ from $27$ .

$18 - 3 - 15$

Step 7.1.3.6

Subtract $3$ from $18$ .

$15 - 15$

Step 7.1.3.7

Subtract $15$ from $15$ .

$0$

Step 7.1.4

Since $3$ is a known root, divide the polynomial by $x - 3$ to find the quotient polynomial. This polynomial can then be used to find the remaining roots.

$\frac{x^{3} - x^{2} - x - 15}{x - 3}$

Step 7.1.5

Divide $x^{3} - x^{2} - x - 15$ by $x - 3$ .

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Step 7.1.5.1

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

$x$

$3$

$x^{3}$

$x^{2}$

$x$

$15$

Step 7.1.5.2

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

					$x^{2}$
	$x$	-	$3$		$x^{3}$	-	$x^{2}$	-	$x$	-	$15$

Step 7.1.5.3

Multiply the new quotient term by the divisor.

				$x^{2}$
$x$	-	$3$		$x^{3}$	-	$x^{2}$	-	$x$	-	$15$
			+	$x^{3}$	-	$3 x^{2}$

Step 7.1.5.4

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

				$x^{2}$
$x$	-	$3$		$x^{3}$	-	$x^{2}$	-	$x$	-	$15$
			-	$x^{3}$	+	$3 x^{2}$

Step 7.1.5.5

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

				$x^{2}$
$x$	-	$3$		$x^{3}$	-	$x^{2}$	-	$x$	-	$15$
			-	$x^{3}$	+	$3 x^{2}$
					+	$2 x^{2}$

Step 7.1.5.6

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

				$x^{2}$
$x$	-	$3$		$x^{3}$	-	$x^{2}$	-	$x$	-	$15$
			-	$x^{3}$	+	$3 x^{2}$
					+	$2 x^{2}$	-	$x$

Step 7.1.5.7

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

				$x^{2}$	+	$2 x$
$x$	-	$3$		$x^{3}$	-	$x^{2}$	-	$x$	-	$15$
			-	$x^{3}$	+	$3 x^{2}$
					+	$2 x^{2}$	-	$x$

Step 7.1.5.8

Multiply the new quotient term by the divisor.

				$x^{2}$	+	$2 x$
$x$	-	$3$		$x^{3}$	-	$x^{2}$	-	$x$	-	$15$
			-	$x^{3}$	+	$3 x^{2}$
					+	$2 x^{2}$	-	$x$
					+	$2 x^{2}$	-	$6 x$

Step 7.1.5.9

The expression needs to be subtracted from the dividend, so change all the signs in $2 x^{2} - 6 x$

				$x^{2}$	+	$2 x$
$x$	-	$3$		$x^{3}$	-	$x^{2}$	-	$x$	-	$15$
			-	$x^{3}$	+	$3 x^{2}$
					+	$2 x^{2}$	-	$x$
					-	$2 x^{2}$	+	$6 x$

Step 7.1.5.10

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

				$x^{2}$	+	$2 x$
$x$	-	$3$		$x^{3}$	-	$x^{2}$	-	$x$	-	$15$
			-	$x^{3}$	+	$3 x^{2}$
					+	$2 x^{2}$	-	$x$
					-	$2 x^{2}$	+	$6 x$
							+	$5 x$

Step 7.1.5.11

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

				$x^{2}$	+	$2 x$
$x$	-	$3$		$x^{3}$	-	$x^{2}$	-	$x$	-	$15$
			-	$x^{3}$	+	$3 x^{2}$
					+	$2 x^{2}$	-	$x$
					-	$2 x^{2}$	+	$6 x$
							+	$5 x$	-	$15$

Step 7.1.5.12

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

				$x^{2}$	+	$2 x$	+	$5$
$x$	-	$3$		$x^{3}$	-	$x^{2}$	-	$x$	-	$15$
			-	$x^{3}$	+	$3 x^{2}$
					+	$2 x^{2}$	-	$x$
					-	$2 x^{2}$	+	$6 x$
							+	$5 x$	-	$15$

Step 7.1.5.13

Multiply the new quotient term by the divisor.

				$x^{2}$	+	$2 x$	+	$5$
$x$	-	$3$		$x^{3}$	-	$x^{2}$	-	$x$	-	$15$
			-	$x^{3}$	+	$3 x^{2}$
					+	$2 x^{2}$	-	$x$
					-	$2 x^{2}$	+	$6 x$
							+	$5 x$	-	$15$
							+	$5 x$	-	$15$

Step 7.1.5.14

The expression needs to be subtracted from the dividend, so change all the signs in $5 x - 15$

				$x^{2}$	+	$2 x$	+	$5$
$x$	-	$3$		$x^{3}$	-	$x^{2}$	-	$x$	-	$15$
			-	$x^{3}$	+	$3 x^{2}$
					+	$2 x^{2}$	-	$x$
					-	$2 x^{2}$	+	$6 x$
							+	$5 x$	-	$15$
							-	$5 x$	+	$15$

Step 7.1.5.15

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

				$x^{2}$	+	$2 x$	+	$5$
$x$	-	$3$		$x^{3}$	-	$x^{2}$	-	$x$	-	$15$
			-	$x^{3}$	+	$3 x^{2}$
					+	$2 x^{2}$	-	$x$
					-	$2 x^{2}$	+	$6 x$
							+	$5 x$	-	$15$
							-	$5 x$	+	$15$
										$0$

Step 7.1.5.16

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

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

Step 7.1.6

Write $x^{3} - x^{2} - x - 15$ as a set of factors.

$(x - 3) (x^{2} + 2 x + 5) = 0$

Step 7.2

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$x - 3 = 0$

$x^{2} + 2 x + 5 = 0$

Step 7.3

Set $x - 3$ equal to $0$ and solve for $x$ .

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Step 7.3.1

Set $x - 3$ equal to $0$ .

$x - 3 = 0$

Step 7.3.2

Add $3$ to both sides of the equation.

$x = 3$

Step 7.4

Set $x^{2} + 2 x + 5$ equal to $0$ and solve for $x$ .

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Step 7.4.1

Set $x^{2} + 2 x + 5$ equal to $0$ .

$x^{2} + 2 x + 5 = 0$

Step 7.4.2

Solve $x^{2} + 2 x + 5 = 0$ for $x$ .

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Step 7.4.2.1

Use the quadratic formula to find the solutions.

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

Step 7.4.2.2

Substitute the values $a = 1$ , $b = 2$ , and $c = 5$ into the quadratic formula and solve for $x$ .

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

Step 7.4.2.3

Simplify.

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Step 7.4.2.3.1

Simplify the numerator.

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Step 7.4.2.3.1.1

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

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

Step 7.4.2.3.1.2

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

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Step 7.4.2.3.1.2.1

Multiply $- 4$ by $1$ .

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

Step 7.4.2.3.1.2.2

Multiply $- 4$ by $5$ .

$x = \frac{- 2 \pm \sqrt{4 - 20}}{2 \cdot 1}$

Step 7.4.2.3.1.3

Subtract $20$ from $4$ .

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

Step 7.4.2.3.1.4

Rewrite $- 16$ as $- 1 (16)$ .

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

Step 7.4.2.3.1.5

Rewrite $\sqrt{- 1 (16)}$ as $\sqrt{- 1} \cdot \sqrt{16}$ .

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

Step 7.4.2.3.1.6

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

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

Step 7.4.2.3.1.7

Rewrite $16$ as $4^{2}$ .

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

Step 7.4.2.3.1.8

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

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

Step 7.4.2.3.1.9

Move $4$ to the left of $i$ .

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

Step 7.4.2.3.2

Multiply $2$ by $1$ .

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

Step 7.4.2.3.3

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

$x = - 1 \pm 2 i$

Step 7.4.2.4

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

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Step 7.4.2.4.1

Simplify the numerator.

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Step 7.4.2.4.1.1

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

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

Step 7.4.2.4.1.2

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

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Step 7.4.2.4.1.2.1

Multiply $- 4$ by $1$ .

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

Step 7.4.2.4.1.2.2

Multiply $- 4$ by $5$ .

$x = \frac{- 2 \pm \sqrt{4 - 20}}{2 \cdot 1}$

Step 7.4.2.4.1.3

Subtract $20$ from $4$ .

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

Step 7.4.2.4.1.4

Rewrite $- 16$ as $- 1 (16)$ .

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

Step 7.4.2.4.1.5

Rewrite $\sqrt{- 1 (16)}$ as $\sqrt{- 1} \cdot \sqrt{16}$ .

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

Step 7.4.2.4.1.6

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

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

Step 7.4.2.4.1.7

Rewrite $16$ as $4^{2}$ .

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

Step 7.4.2.4.1.8

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

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

Step 7.4.2.4.1.9

Move $4$ to the left of $i$ .

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

Step 7.4.2.4.2

Multiply $2$ by $1$ .

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

Step 7.4.2.4.3

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

$x = - 1 \pm 2 i$

Step 7.4.2.4.4

Change the $\pm$ to $+$ .

$x = - 1 + 2 i$

Step 7.4.2.5

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

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Step 7.4.2.5.1

Simplify the numerator.

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Step 7.4.2.5.1.1

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

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

Step 7.4.2.5.1.2

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

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Step 7.4.2.5.1.2.1

Multiply $- 4$ by $1$ .

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

Step 7.4.2.5.1.2.2

Multiply $- 4$ by $5$ .

$x = \frac{- 2 \pm \sqrt{4 - 20}}{2 \cdot 1}$

Step 7.4.2.5.1.3

Subtract $20$ from $4$ .

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

Step 7.4.2.5.1.4

Rewrite $- 16$ as $- 1 (16)$ .

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

Step 7.4.2.5.1.5

Rewrite $\sqrt{- 1 (16)}$ as $\sqrt{- 1} \cdot \sqrt{16}$ .

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

Step 7.4.2.5.1.6

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

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

Step 7.4.2.5.1.7

Rewrite $16$ as $4^{2}$ .

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

Step 7.4.2.5.1.8

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

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

Step 7.4.2.5.1.9

Move $4$ to the left of $i$ .

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

Step 7.4.2.5.2

Multiply $2$ by $1$ .

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

Step 7.4.2.5.3

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

$x = - 1 \pm 2 i$

Step 7.4.2.5.4

Change the $\pm$ to $-$ .

$x = - 1 - 2 i$

Step 7.4.2.6

The final answer is the combination of both solutions.

$x = - 1 + 2 i, - 1 - 2 i$

Step 7.5

The final solution is all the values that make $(x - 3) (x^{2} + 2 x + 5) = 0$ true.

$x = 3, - 1 + 2 i, - 1 - 2 i$

Step 8

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

$(x + 1) (x - 3) (x + 1 - 2 i) (x + 1 + 2 i)$

Step 9

These are the roots (zeros) of the polynomial $x^{4} - 2 x^{2} - 16 x - 15$ .

$x = - 1, 3, - 1 + 2 i, - 1 - 2 i$

Step 10