Precalculus Examples

$x^{4} - 6 x^{3} + 10 x^{2} + 2 x - 15 = 0$

Step 1

Factor the left side of the equation.

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Step 1.1

Factor $x^{4} - 6 x^{3} + 10 x^{2} + 2 x - 15$ using the rational roots test.

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Step 1.1.1

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 1.1.2

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 1.1.3

Substitute $- 1$ and simplify the expression. In this case, the expression is equal to $0$ so $- 1$ is a root of the polynomial.

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Step 1.1.3.1

Substitute $- 1$ into the polynomial.

${(- 1)}^{4} - 6 {(- 1)}^{3} + 10 {(- 1)}^{2} + 2 \cdot - 1 - 15$

Step 1.1.3.2

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

$1 - 6 {(- 1)}^{3} + 10 {(- 1)}^{2} + 2 \cdot - 1 - 15$

Step 1.1.3.3

Raise $- 1$ to the power of $3$ .

$1 - 6 \cdot - 1 + 10 {(- 1)}^{2} + 2 \cdot - 1 - 15$

Step 1.1.3.4

Multiply $- 6$ by $- 1$ .

$1 + 6 + 10 {(- 1)}^{2} + 2 \cdot - 1 - 15$

Step 1.1.3.5

Add $1$ and $6$ .

$7 + 10 {(- 1)}^{2} + 2 \cdot - 1 - 15$

Step 1.1.3.6

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

$7 + 10 \cdot 1 + 2 \cdot - 1 - 15$

Step 1.1.3.7

Multiply $10$ by $1$ .

$7 + 10 + 2 \cdot - 1 - 15$

Step 1.1.3.8

Add $7$ and $10$ .

$17 + 2 \cdot - 1 - 15$

Step 1.1.3.9

Multiply $2$ by $- 1$ .

$17 - 2 - 15$

Step 1.1.3.10

Subtract $2$ from $17$ .

$15 - 15$

Step 1.1.3.11

Subtract $15$ from $15$ .

$0$

Step 1.1.4

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.

$\frac{x^{4} - 6 x^{3} + 10 x^{2} + 2 x - 15}{x + 1}$

Step 1.1.5

Divide $x^{4} - 6 x^{3} + 10 x^{2} + 2 x - 15$ by $x + 1$ .

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Step 1.1.5.1

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

$x$

$1$

$x^{4}$

$6 x^{3}$

$10 x^{2}$

$2 x$

$15$

Step 1.1.5.2

Divide the highest order term in the dividend $x^{4}$ by the highest order term in divisor $x$ .

$x^{3}$

$x$

$1$

$x^{4}$

$6 x^{3}$

$10 x^{2}$

$2 x$

$15$

Step 1.1.5.3

Multiply the new quotient term by the divisor.

$x^{3}$

$x$

$1$

$x^{4}$

$6 x^{3}$

$10 x^{2}$

$2 x$

$15$

$x^{4}$

$x^{3}$

Step 1.1.5.4

The expression needs to be subtracted from the dividend, so change all the signs in $x^{4} + x^{3}$

$x^{3}$

$x$

$1$

$x^{4}$

$6 x^{3}$

$10 x^{2}$

$2 x$

$15$

$x^{4}$

$x^{3}$

Step 1.1.5.5

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

$x^{3}$

$x$

$1$

$x^{4}$

$6 x^{3}$

$10 x^{2}$

$2 x$

$15$

$x^{4}$

$x^{3}$

$7 x^{3}$

Step 1.1.5.6

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

$x^{3}$

$x$

$1$

$x^{4}$

$6 x^{3}$

$10 x^{2}$

$2 x$

$15$

$x^{4}$

$x^{3}$

$7 x^{3}$

$10 x^{2}$

Step 1.1.5.7

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

$x^{3}$

$7 x^{2}$

$x$

$1$

$x^{4}$

$6 x^{3}$

$10 x^{2}$

$2 x$

$15$

$x^{4}$

$x^{3}$

$7 x^{3}$

$10 x^{2}$

Step 1.1.5.8

Multiply the new quotient term by the divisor.

$x^{3}$

$7 x^{2}$

$x$

$1$

$x^{4}$

$6 x^{3}$

$10 x^{2}$

$2 x$

$15$

$x^{4}$

$x^{3}$

$7 x^{3}$

$10 x^{2}$

$7 x^{3}$

$7 x^{2}$

Step 1.1.5.9

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

$x^{3}$

$7 x^{2}$

$x$

$1$

$x^{4}$

$6 x^{3}$

$10 x^{2}$

$2 x$

$15$

$x^{4}$

$x^{3}$

$7 x^{3}$

$10 x^{2}$

$7 x^{3}$

$7 x^{2}$

Step 1.1.5.10

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

$x^{3}$

$7 x^{2}$

$x$

$1$

$x^{4}$

$6 x^{3}$

$10 x^{2}$

$2 x$

$15$

$x^{4}$

$x^{3}$

$7 x^{3}$

$10 x^{2}$

$7 x^{3}$

$7 x^{2}$

$17 x^{2}$

Step 1.1.5.11

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

$x^{3}$

$7 x^{2}$

$x$

$1$

$x^{4}$

$6 x^{3}$

$10 x^{2}$

$2 x$

$15$

$x^{4}$

$x^{3}$

$7 x^{3}$

$10 x^{2}$

$7 x^{3}$

$7 x^{2}$

$17 x^{2}$

$2 x$

Step 1.1.5.12

Divide the highest order term in the dividend $17 x^{2}$ by the highest order term in divisor $x$ .

$x^{3}$

$7 x^{2}$

$17 x$

$x$

$1$

$x^{4}$

$6 x^{3}$

$10 x^{2}$

$2 x$

$15$

$x^{4}$

$x^{3}$

$7 x^{3}$

$10 x^{2}$

$7 x^{3}$

$7 x^{2}$

$17 x^{2}$

$2 x$

Step 1.1.5.13

Multiply the new quotient term by the divisor.

$x^{3}$

$7 x^{2}$

$17 x$

$x$

$1$

$x^{4}$

$6 x^{3}$

$10 x^{2}$

$2 x$

$15$

$x^{4}$

$x^{3}$

$7 x^{3}$

$10 x^{2}$

$7 x^{3}$

$7 x^{2}$

$17 x^{2}$

$2 x$

$17 x^{2}$

$17 x$

Step 1.1.5.14

The expression needs to be subtracted from the dividend, so change all the signs in $17 x^{2} + 17 x$

$x^{3}$

$7 x^{2}$

$17 x$

$x$

$1$

$x^{4}$

$6 x^{3}$

$10 x^{2}$

$2 x$

$15$

$x^{4}$

$x^{3}$

$7 x^{3}$

$10 x^{2}$

$7 x^{3}$

$7 x^{2}$

$17 x^{2}$

$2 x$

$17 x^{2}$

$17 x$

Step 1.1.5.15

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

$x^{3}$

$7 x^{2}$

$17 x$

$x$

$1$

$x^{4}$

$6 x^{3}$

$10 x^{2}$

$2 x$

$15$

$x^{4}$

$x^{3}$

$7 x^{3}$

$10 x^{2}$

$7 x^{3}$

$7 x^{2}$

$17 x^{2}$

$2 x$

$17 x^{2}$

$17 x$

$15 x$

Step 1.1.5.16

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

$x^{3}$

$7 x^{2}$

$17 x$

$x$

$1$

$x^{4}$

$6 x^{3}$

$10 x^{2}$

$2 x$

$15$

$x^{4}$

$x^{3}$

$7 x^{3}$

$10 x^{2}$

$7 x^{3}$

$7 x^{2}$

$17 x^{2}$

$2 x$

$17 x^{2}$

$17 x$

$15 x$

$15$

Step 1.1.5.17

Divide the highest order term in the dividend $- 15 x$ by the highest order term in divisor $x$ .

$x^{3}$

$7 x^{2}$

$17 x$

$15$

$x$

$1$

$x^{4}$

$6 x^{3}$

$10 x^{2}$

$2 x$

$15$

$x^{4}$

$x^{3}$

$7 x^{3}$

$10 x^{2}$

$7 x^{3}$

$7 x^{2}$

$17 x^{2}$

$2 x$

$17 x^{2}$

$17 x$

$15 x$

$15$

Step 1.1.5.18

Multiply the new quotient term by the divisor.

$x^{3}$

$7 x^{2}$

$17 x$

$15$

$x$

$1$

$x^{4}$

$6 x^{3}$

$10 x^{2}$

$2 x$

$15$

$x^{4}$

$x^{3}$

$7 x^{3}$

$10 x^{2}$

$7 x^{3}$

$7 x^{2}$

$17 x^{2}$

$2 x$

$17 x^{2}$

$17 x$

$15 x$

$15$

$15 x$

$15$

Step 1.1.5.19

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

$x^{3}$

$7 x^{2}$

$17 x$

$15$

$x$

$1$

$x^{4}$

$6 x^{3}$

$10 x^{2}$

$2 x$

$15$

$x^{4}$

$x^{3}$

$7 x^{3}$

$10 x^{2}$

$7 x^{3}$

$7 x^{2}$

$17 x^{2}$

$2 x$

$17 x^{2}$

$17 x$

$15 x$

$15$

$15 x$

$15$

Step 1.1.5.20

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

$x^{3}$

$7 x^{2}$

$17 x$

$15$

$x$

$1$

$x^{4}$

$6 x^{3}$

$10 x^{2}$

$2 x$

$15$

$x^{4}$

$x^{3}$

$7 x^{3}$

$10 x^{2}$

$7 x^{3}$

$7 x^{2}$

$17 x^{2}$

$2 x$

$17 x^{2}$

$17 x$

$15 x$

$15$

$15 x$

$15$

$0$

Step 1.1.5.21

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

$x^{3} - 7 x^{2} + 17 x - 15$

Step 1.1.6

Write $x^{4} - 6 x^{3} + 10 x^{2} + 2 x - 15$ as a set of factors.

$(x + 1) (x^{3} - 7 x^{2} + 17 x - 15) = 0$

Step 1.2

Factor $x^{3} - 7 x^{2} + 17 x - 15$ using the rational roots test.

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Step 1.2.1

Factor $x^{3} - 7 x^{2} + 17 x - 15$ using the rational roots test.

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Step 1.2.1.1

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.

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

$q = \pm 1$

Step 1.2.1.2

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 1.2.1.3

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 1.2.1.3.1

Substitute $3$ into the polynomial.

$3^{3} - 7 \cdot 3^{2} + 17 \cdot 3 - 15$

Step 1.2.1.3.2

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

$27 - 7 \cdot 3^{2} + 17 \cdot 3 - 15$

Step 1.2.1.3.3

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

$27 - 7 \cdot 9 + 17 \cdot 3 - 15$

Step 1.2.1.3.4

Multiply $- 7$ by $9$ .

$27 - 63 + 17 \cdot 3 - 15$

Step 1.2.1.3.5

Subtract $63$ from $27$ .

$- 36 + 17 \cdot 3 - 15$

Step 1.2.1.3.6

Multiply $17$ by $3$ .

$- 36 + 51 - 15$

Step 1.2.1.3.7

Add $- 36$ and $51$ .

$15 - 15$

Step 1.2.1.3.8

Subtract $15$ from $15$ .

$0$

Step 1.2.1.4

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.

$\frac{x^{3} - 7 x^{2} + 17 x - 15}{x - 3}$

Step 1.2.1.5

Divide $x^{3} - 7 x^{2} + 17 x - 15$ by $x - 3$ .

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Step 1.2.1.5.1

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

$x$

$3$

$x^{3}$

$7 x^{2}$

$17 x$

$15$

Step 1.2.1.5.2

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

					$x^{2}$
	$x$	-	$3$		$x^{3}$	-	$7 x^{2}$	+	$17 x$	-	$15$

Step 1.2.1.5.3

Multiply the new quotient term by the divisor.

				$x^{2}$
$x$	-	$3$		$x^{3}$	-	$7 x^{2}$	+	$17 x$	-	$15$
			+	$x^{3}$	-	$3 x^{2}$

Step 1.2.1.5.4

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

				$x^{2}$
$x$	-	$3$		$x^{3}$	-	$7 x^{2}$	+	$17 x$	-	$15$
			-	$x^{3}$	+	$3 x^{2}$

Step 1.2.1.5.5

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

				$x^{2}$
$x$	-	$3$		$x^{3}$	-	$7 x^{2}$	+	$17 x$	-	$15$
			-	$x^{3}$	+	$3 x^{2}$
					-	$4 x^{2}$

Step 1.2.1.5.6

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

				$x^{2}$
$x$	-	$3$		$x^{3}$	-	$7 x^{2}$	+	$17 x$	-	$15$
			-	$x^{3}$	+	$3 x^{2}$
					-	$4 x^{2}$	+	$17 x$

Step 1.2.1.5.7

Divide the highest order term in the dividend $- 4 x^{2}$ by the highest order term in divisor $x$ .

				$x^{2}$	-	$4 x$
$x$	-	$3$		$x^{3}$	-	$7 x^{2}$	+	$17 x$	-	$15$
			-	$x^{3}$	+	$3 x^{2}$
					-	$4 x^{2}$	+	$17 x$

Step 1.2.1.5.8

Multiply the new quotient term by the divisor.

				$x^{2}$	-	$4 x$
$x$	-	$3$		$x^{3}$	-	$7 x^{2}$	+	$17 x$	-	$15$
			-	$x^{3}$	+	$3 x^{2}$
					-	$4 x^{2}$	+	$17 x$
					-	$4 x^{2}$	+	$12 x$

Step 1.2.1.5.9

The expression needs to be subtracted from the dividend, so change all the signs in $- 4 x^{2} + 12 x$

				$x^{2}$	-	$4 x$
$x$	-	$3$		$x^{3}$	-	$7 x^{2}$	+	$17 x$	-	$15$
			-	$x^{3}$	+	$3 x^{2}$
					-	$4 x^{2}$	+	$17 x$
					+	$4 x^{2}$	-	$12 x$

Step 1.2.1.5.10

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

				$x^{2}$	-	$4 x$
$x$	-	$3$		$x^{3}$	-	$7 x^{2}$	+	$17 x$	-	$15$
			-	$x^{3}$	+	$3 x^{2}$
					-	$4 x^{2}$	+	$17 x$
					+	$4 x^{2}$	-	$12 x$
							+	$5 x$

Step 1.2.1.5.11

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

				$x^{2}$	-	$4 x$
$x$	-	$3$		$x^{3}$	-	$7 x^{2}$	+	$17 x$	-	$15$
			-	$x^{3}$	+	$3 x^{2}$
					-	$4 x^{2}$	+	$17 x$
					+	$4 x^{2}$	-	$12 x$
							+	$5 x$	-	$15$

Step 1.2.1.5.12

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

				$x^{2}$	-	$4 x$	+	$5$
$x$	-	$3$		$x^{3}$	-	$7 x^{2}$	+	$17 x$	-	$15$
			-	$x^{3}$	+	$3 x^{2}$
					-	$4 x^{2}$	+	$17 x$
					+	$4 x^{2}$	-	$12 x$
							+	$5 x$	-	$15$

Step 1.2.1.5.13

Multiply the new quotient term by the divisor.

				$x^{2}$	-	$4 x$	+	$5$
$x$	-	$3$		$x^{3}$	-	$7 x^{2}$	+	$17 x$	-	$15$
			-	$x^{3}$	+	$3 x^{2}$
					-	$4 x^{2}$	+	$17 x$
					+	$4 x^{2}$	-	$12 x$
							+	$5 x$	-	$15$
							+	$5 x$	-	$15$

Step 1.2.1.5.14

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

				$x^{2}$	-	$4 x$	+	$5$
$x$	-	$3$		$x^{3}$	-	$7 x^{2}$	+	$17 x$	-	$15$
			-	$x^{3}$	+	$3 x^{2}$
					-	$4 x^{2}$	+	$17 x$
					+	$4 x^{2}$	-	$12 x$
							+	$5 x$	-	$15$
							-	$5 x$	+	$15$

Step 1.2.1.5.15

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

				$x^{2}$	-	$4 x$	+	$5$
$x$	-	$3$		$x^{3}$	-	$7 x^{2}$	+	$17 x$	-	$15$
			-	$x^{3}$	+	$3 x^{2}$
					-	$4 x^{2}$	+	$17 x$
					+	$4 x^{2}$	-	$12 x$
							+	$5 x$	-	$15$
							-	$5 x$	+	$15$
										$0$

Step 1.2.1.5.16

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

$x^{2} - 4 x + 5$

Step 1.2.1.6

Write $x^{3} - 7 x^{2} + 17 x - 15$ as a set of factors.

$(x + 1) ((x - 3) (x^{2} - 4 x + 5)) = 0$

Step 1.2.2

Remove unnecessary parentheses.

$(x + 1) (x - 3) (x^{2} - 4 x + 5) = 0$

Step 2

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

$x + 1 = 0$

$x - 3 = 0$

$x^{2} - 4 x + 5 = 0$

Step 3

Set $x + 1$ equal to $0$ and solve for $x$ .

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Step 3.1

Set $x + 1$ equal to $0$ .

$x + 1 = 0$

Step 3.2

Subtract $1$ from both sides of the equation.

$x = - 1$

Step 4

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

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Step 4.1

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

$x - 3 = 0$

Step 4.2

Add $3$ to both sides of the equation.

$x = 3$

Step 5

Set $x^{2} - 4 x + 5$ equal to $0$ and solve for $x$ .

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Step 5.1

Set $x^{2} - 4 x + 5$ equal to $0$ .

$x^{2} - 4 x + 5 = 0$

Step 5.2

Solve $x^{2} - 4 x + 5 = 0$ for $x$ .

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Step 5.2.1

Use the quadratic formula to find the solutions.

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

Step 5.2.2

Substitute the values $a = 1$ , $b = - 4$ , and $c = 5$ into the quadratic formula and solve for $x$ .

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

Step 5.2.3

Simplify.

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Step 5.2.3.1

Simplify the numerator.

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Step 5.2.3.1.1

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

$x = \frac{4 \pm \sqrt{16 - 4 \cdot 1 \cdot 5}}{2 \cdot 1}$

Step 5.2.3.1.2

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

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Step 5.2.3.1.2.1

Multiply $- 4$ by $1$ .

$x = \frac{4 \pm \sqrt{16 - 4 \cdot 5}}{2 \cdot 1}$

Step 5.2.3.1.2.2

Multiply $- 4$ by $5$ .

$x = \frac{4 \pm \sqrt{16 - 20}}{2 \cdot 1}$

Step 5.2.3.1.3

Subtract $20$ from $16$ .

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

Step 5.2.3.1.4

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

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

Step 5.2.3.1.5

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

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

Step 5.2.3.1.6

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

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

Step 5.2.3.1.7

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

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

Step 5.2.3.1.8

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

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

Step 5.2.3.1.9

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

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

Step 5.2.3.2

Multiply $2$ by $1$ .

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

Step 5.2.3.3

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

$x = 2 \pm i$

Step 5.2.4

The final answer is the combination of both solutions.

$x = 2 + i, 2 - i$

Step 6

The final solution is all the values that make $(x + 1) (x - 3) (x^{2} - 4 x + 5) = 0$ true.

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

Step 7