Precalculus Examples

$f (x) = x^{3} - 6 x^{2} + 13 x - 20$

Step 1

Set $x^{3} - 6 x^{2} + 13 x - 20$ equal to $0$ .

$x^{3} - 6 x^{2} + 13 x - 20 = 0$

Step 2

Solve for $x$ .

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Step 2.1

Factor $x^{3} - 6 x^{2} + 13 x - 20$ using the rational roots test.

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Step 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.

$p = \pm 1, \pm 20, \pm 2, \pm 10, \pm 4, \pm 5$

$q = \pm 1$

Step 2.1.2

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

$\pm 1, \pm 20, \pm 2, \pm 10, \pm 4, \pm 5$

Step 2.1.3

Substitute $4$ and simplify the expression. In this case, the expression is equal to $0$ so $4$ is a root of the polynomial.

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Step 2.1.3.1

Substitute $4$ into the polynomial.

$4^{3} - 6 \cdot 4^{2} + 13 \cdot 4 - 20$

Step 2.1.3.2

Raise $4$ to the power of $3$ .

$64 - 6 \cdot 4^{2} + 13 \cdot 4 - 20$

Step 2.1.3.3

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

$64 - 6 \cdot 16 + 13 \cdot 4 - 20$

Step 2.1.3.4

Multiply $- 6$ by $16$ .

$64 - 96 + 13 \cdot 4 - 20$

Step 2.1.3.5

Subtract $96$ from $64$ .

$- 32 + 13 \cdot 4 - 20$

Step 2.1.3.6

Multiply $13$ by $4$ .

$- 32 + 52 - 20$

Step 2.1.3.7

Add $- 32$ and $52$ .

$20 - 20$

Step 2.1.3.8

Subtract $20$ from $20$ .

$0$

Step 2.1.4

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.

$\frac{x^{3} - 6 x^{2} + 13 x - 20}{x - 4}$

Step 2.1.5

Divide $x^{3} - 6 x^{2} + 13 x - 20$ by $x - 4$ .

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Step 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 $0$ .

$x$

$4$

$x^{3}$

$6 x^{2}$

$13 x$

$20$

Step 2.1.5.2

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

					$x^{2}$
	$x$	-	$4$		$x^{3}$	-	$6 x^{2}$	+	$13 x$	-	$20$

Step 2.1.5.3

Multiply the new quotient term by the divisor.

				$x^{2}$
$x$	-	$4$		$x^{3}$	-	$6 x^{2}$	+	$13 x$	-	$20$
			+	$x^{3}$	-	$4 x^{2}$

Step 2.1.5.4

The expression needs to be subtracted from the dividend, so change all the signs in $x^{3} - 4 x^{2}$

				$x^{2}$
$x$	-	$4$		$x^{3}$	-	$6 x^{2}$	+	$13 x$	-	$20$
			-	$x^{3}$	+	$4 x^{2}$

Step 2.1.5.5

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

				$x^{2}$
$x$	-	$4$		$x^{3}$	-	$6 x^{2}$	+	$13 x$	-	$20$
			-	$x^{3}$	+	$4 x^{2}$
					-	$2 x^{2}$

Step 2.1.5.6

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

				$x^{2}$
$x$	-	$4$		$x^{3}$	-	$6 x^{2}$	+	$13 x$	-	$20$
			-	$x^{3}$	+	$4 x^{2}$
					-	$2 x^{2}$	+	$13 x$

Step 2.1.5.7

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

				$x^{2}$	-	$2 x$
$x$	-	$4$		$x^{3}$	-	$6 x^{2}$	+	$13 x$	-	$20$
			-	$x^{3}$	+	$4 x^{2}$
					-	$2 x^{2}$	+	$13 x$

Step 2.1.5.8

Multiply the new quotient term by the divisor.

				$x^{2}$	-	$2 x$
$x$	-	$4$		$x^{3}$	-	$6 x^{2}$	+	$13 x$	-	$20$
			-	$x^{3}$	+	$4 x^{2}$
					-	$2 x^{2}$	+	$13 x$
					-	$2 x^{2}$	+	$8 x$

Step 2.1.5.9

The expression needs to be subtracted from the dividend, so change all the signs in $- 2 x^{2} + 8 x$

				$x^{2}$	-	$2 x$
$x$	-	$4$		$x^{3}$	-	$6 x^{2}$	+	$13 x$	-	$20$
			-	$x^{3}$	+	$4 x^{2}$
					-	$2 x^{2}$	+	$13 x$
					+	$2 x^{2}$	-	$8 x$

Step 2.1.5.10

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

				$x^{2}$	-	$2 x$
$x$	-	$4$		$x^{3}$	-	$6 x^{2}$	+	$13 x$	-	$20$
			-	$x^{3}$	+	$4 x^{2}$
					-	$2 x^{2}$	+	$13 x$
					+	$2 x^{2}$	-	$8 x$
							+	$5 x$

Step 2.1.5.11

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

				$x^{2}$	-	$2 x$
$x$	-	$4$		$x^{3}$	-	$6 x^{2}$	+	$13 x$	-	$20$
			-	$x^{3}$	+	$4 x^{2}$
					-	$2 x^{2}$	+	$13 x$
					+	$2 x^{2}$	-	$8 x$
							+	$5 x$	-	$20$

Step 2.1.5.12

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

				$x^{2}$	-	$2 x$	+	$5$
$x$	-	$4$		$x^{3}$	-	$6 x^{2}$	+	$13 x$	-	$20$
			-	$x^{3}$	+	$4 x^{2}$
					-	$2 x^{2}$	+	$13 x$
					+	$2 x^{2}$	-	$8 x$
							+	$5 x$	-	$20$

Step 2.1.5.13

Multiply the new quotient term by the divisor.

				$x^{2}$	-	$2 x$	+	$5$
$x$	-	$4$		$x^{3}$	-	$6 x^{2}$	+	$13 x$	-	$20$
			-	$x^{3}$	+	$4 x^{2}$
					-	$2 x^{2}$	+	$13 x$
					+	$2 x^{2}$	-	$8 x$
							+	$5 x$	-	$20$
							+	$5 x$	-	$20$

Step 2.1.5.14

The expression needs to be subtracted from the dividend, so change all the signs in $5 x - 20$

				$x^{2}$	-	$2 x$	+	$5$
$x$	-	$4$		$x^{3}$	-	$6 x^{2}$	+	$13 x$	-	$20$
			-	$x^{3}$	+	$4 x^{2}$
					-	$2 x^{2}$	+	$13 x$
					+	$2 x^{2}$	-	$8 x$
							+	$5 x$	-	$20$
							-	$5 x$	+	$20$

Step 2.1.5.15

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

				$x^{2}$	-	$2 x$	+	$5$
$x$	-	$4$		$x^{3}$	-	$6 x^{2}$	+	$13 x$	-	$20$
			-	$x^{3}$	+	$4 x^{2}$
					-	$2 x^{2}$	+	$13 x$
					+	$2 x^{2}$	-	$8 x$
							+	$5 x$	-	$20$
							-	$5 x$	+	$20$
										$0$

Step 2.1.5.16

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

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

Step 2.1.6

Write $x^{3} - 6 x^{2} + 13 x - 20$ as a set of factors.

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

Step 2.2

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

$x - 4 = 0$

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

Step 2.3

Set $x - 4$ equal to $0$ and solve for $x$ .

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Step 2.3.1

Set $x - 4$ equal to $0$ .

$x - 4 = 0$

Step 2.3.2

Add $4$ to both sides of the equation.

$x = 4$

Step 2.4

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

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Step 2.4.1

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

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

Step 2.4.2

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

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Step 2.4.2.1

Use the quadratic formula to find the solutions.

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

Step 2.4.2.2

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 2.4.2.3

Simplify.

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Step 2.4.2.3.1

Simplify the numerator.

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Step 2.4.2.3.1.1

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

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

Step 2.4.2.3.1.2

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

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Step 2.4.2.3.1.2.1

Multiply $- 4$ by $1$ .

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

Step 2.4.2.3.1.2.2

Multiply $- 4$ by $5$ .

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

Step 2.4.2.3.1.3

Subtract $20$ from $4$ .

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

Step 2.4.2.3.1.4

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

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

Step 2.4.2.3.1.5

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

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

Step 2.4.2.3.1.6

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

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

Step 2.4.2.3.1.7

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

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

Step 2.4.2.3.1.8

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

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

Step 2.4.2.3.1.9

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

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

Step 2.4.2.3.2

Multiply $2$ by $1$ .

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

Step 2.4.2.3.3

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

$x = 1 \pm 2 i$

Step 2.4.2.4

The final answer is the combination of both solutions.

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

Step 2.5

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

$x = 4, 1 + 2 i, 1 - 2 i$

Step 3