Finite Math Examples

$x^{3} - 7 x^{2} + 6 x + 20 = 0$

Step 1

Factor $x^{3} - 7 x^{2} + 6 x + 20$ using the rational roots test.

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Step 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 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 1.3

Substitute $5$ and simplify the expression. In this case, the expression is equal to $0$ so $5$ is a root of the polynomial.

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Step 1.3.1

Substitute $5$ into the polynomial.

$5^{3} - 7 \cdot 5^{2} + 6 \cdot 5 + 20$

Step 1.3.2

Raise $5$ to the power of $3$ .

$125 - 7 \cdot 5^{2} + 6 \cdot 5 + 20$

Step 1.3.3

Raise $5$ to the power of $2$ .

$125 - 7 \cdot 25 + 6 \cdot 5 + 20$

Step 1.3.4

Multiply $- 7$ by $25$ .

$125 - 175 + 6 \cdot 5 + 20$

Step 1.3.5

Subtract $175$ from $125$ .

$- 50 + 6 \cdot 5 + 20$

Step 1.3.6

Multiply $6$ by $5$ .

$- 50 + 30 + 20$

Step 1.3.7

Add $- 50$ and $30$ .

$- 20 + 20$

Step 1.3.8

Add $- 20$ and $20$ .

$0$

Step 1.4

Since $5$ is a known root, divide the polynomial by $x - 5$ to find the quotient polynomial. This polynomial can then be used to find the remaining roots.

$\frac{x^{3} - 7 x^{2} + 6 x + 20}{x - 5}$

Step 1.5

Divide $x^{3} - 7 x^{2} + 6 x + 20$ by $x - 5$ .

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Step 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 $0$ .

$x$

$5$

$x^{3}$

$7 x^{2}$

$6 x$

$20$

Step 1.5.2

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

					$x^{2}$
	$x$	-	$5$		$x^{3}$	-	$7 x^{2}$	+	$6 x$	+	$20$

Step 1.5.3

Multiply the new quotient term by the divisor.

				$x^{2}$
$x$	-	$5$		$x^{3}$	-	$7 x^{2}$	+	$6 x$	+	$20$
			+	$x^{3}$	-	$5 x^{2}$

Step 1.5.4

The expression needs to be subtracted from the dividend, so change all the signs in $x^{3} - 5 x^{2}$

				$x^{2}$
$x$	-	$5$		$x^{3}$	-	$7 x^{2}$	+	$6 x$	+	$20$
			-	$x^{3}$	+	$5 x^{2}$

Step 1.5.5

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

				$x^{2}$
$x$	-	$5$		$x^{3}$	-	$7 x^{2}$	+	$6 x$	+	$20$
			-	$x^{3}$	+	$5 x^{2}$
					-	$2 x^{2}$

Step 1.5.6

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

				$x^{2}$
$x$	-	$5$		$x^{3}$	-	$7 x^{2}$	+	$6 x$	+	$20$
			-	$x^{3}$	+	$5 x^{2}$
					-	$2 x^{2}$	+	$6 x$

Step 1.5.7

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

				$x^{2}$	-	$2 x$
$x$	-	$5$		$x^{3}$	-	$7 x^{2}$	+	$6 x$	+	$20$
			-	$x^{3}$	+	$5 x^{2}$
					-	$2 x^{2}$	+	$6 x$

Step 1.5.8

Multiply the new quotient term by the divisor.

				$x^{2}$	-	$2 x$
$x$	-	$5$		$x^{3}$	-	$7 x^{2}$	+	$6 x$	+	$20$
			-	$x^{3}$	+	$5 x^{2}$
					-	$2 x^{2}$	+	$6 x$
					-	$2 x^{2}$	+	$10 x$

Step 1.5.9

The expression needs to be subtracted from the dividend, so change all the signs in $- 2 x^{2} + 10 x$

				$x^{2}$	-	$2 x$
$x$	-	$5$		$x^{3}$	-	$7 x^{2}$	+	$6 x$	+	$20$
			-	$x^{3}$	+	$5 x^{2}$
					-	$2 x^{2}$	+	$6 x$
					+	$2 x^{2}$	-	$10 x$

Step 1.5.10

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

				$x^{2}$	-	$2 x$
$x$	-	$5$		$x^{3}$	-	$7 x^{2}$	+	$6 x$	+	$20$
			-	$x^{3}$	+	$5 x^{2}$
					-	$2 x^{2}$	+	$6 x$
					+	$2 x^{2}$	-	$10 x$
							-	$4 x$

Step 1.5.11

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

				$x^{2}$	-	$2 x$
$x$	-	$5$		$x^{3}$	-	$7 x^{2}$	+	$6 x$	+	$20$
			-	$x^{3}$	+	$5 x^{2}$
					-	$2 x^{2}$	+	$6 x$
					+	$2 x^{2}$	-	$10 x$
							-	$4 x$	+	$20$

Step 1.5.12

Divide the highest order term in the dividend $- 4 x$ by the highest order term in divisor $x$ .

				$x^{2}$	-	$2 x$	-	$4$
$x$	-	$5$		$x^{3}$	-	$7 x^{2}$	+	$6 x$	+	$20$
			-	$x^{3}$	+	$5 x^{2}$
					-	$2 x^{2}$	+	$6 x$
					+	$2 x^{2}$	-	$10 x$
							-	$4 x$	+	$20$

Step 1.5.13

Multiply the new quotient term by the divisor.

				$x^{2}$	-	$2 x$	-	$4$
$x$	-	$5$		$x^{3}$	-	$7 x^{2}$	+	$6 x$	+	$20$
			-	$x^{3}$	+	$5 x^{2}$
					-	$2 x^{2}$	+	$6 x$
					+	$2 x^{2}$	-	$10 x$
							-	$4 x$	+	$20$
							-	$4 x$	+	$20$

Step 1.5.14

The expression needs to be subtracted from the dividend, so change all the signs in $- 4 x + 20$

				$x^{2}$	-	$2 x$	-	$4$
$x$	-	$5$		$x^{3}$	-	$7 x^{2}$	+	$6 x$	+	$20$
			-	$x^{3}$	+	$5 x^{2}$
					-	$2 x^{2}$	+	$6 x$
					+	$2 x^{2}$	-	$10 x$
							-	$4 x$	+	$20$
							+	$4 x$	-	$20$

Step 1.5.15

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

				$x^{2}$	-	$2 x$	-	$4$
$x$	-	$5$		$x^{3}$	-	$7 x^{2}$	+	$6 x$	+	$20$
			-	$x^{3}$	+	$5 x^{2}$
					-	$2 x^{2}$	+	$6 x$
					+	$2 x^{2}$	-	$10 x$
							-	$4 x$	+	$20$
							+	$4 x$	-	$20$
										$0$

Step 1.5.16

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

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

Step 1.6

Write $x^{3} - 7 x^{2} + 6 x + 20$ as a set of factors.

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

Step 2

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

$x - 5 = 0$

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

Step 3

Set $x - 5$ equal to $0$ and solve for $x$ .

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Step 3.1

Set $x - 5$ equal to $0$ .

$x - 5 = 0$

Step 3.2

Add $5$ to both sides of the equation.

$x = 5$

Step 4

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

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Step 4.1

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

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

Step 4.2

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

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Step 4.2.1

Use the quadratic formula to find the solutions.

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

Step 4.2.2

Substitute the values $a = 1$ , $b = - 2$ , and $c = - 4$ into the quadratic formula and solve for $x$ .

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

Step 4.2.3

Simplify.

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Step 4.2.3.1

Simplify the numerator.

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Step 4.2.3.1.1

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

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

Step 4.2.3.1.2

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

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Step 4.2.3.1.2.1

Multiply $- 4$ by $1$ .

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

Step 4.2.3.1.2.2

Multiply $- 4$ by $- 4$ .

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

Step 4.2.3.1.3

Add $4$ and $16$ .

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

Step 4.2.3.1.4

Rewrite $20$ as $2^{2} \cdot 5$ .

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Step 4.2.3.1.4.1

Factor $4$ out of $20$ .

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

Step 4.2.3.1.4.2

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

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

Step 4.2.3.1.5

Pull terms out from under the radical.

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

Step 4.2.3.2

Multiply $2$ by $1$ .

$x = \frac{2 \pm 2 \sqrt{5}}{2}$

Step 4.2.3.3

Simplify $\frac{2 \pm 2 \sqrt{5}}{2}$ .

$x = 1 \pm \sqrt{5}$

Step 4.2.4

The final answer is the combination of both solutions.

$x = 1 + \sqrt{5}, 1 - \sqrt{5}$

Step 5

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

$x = 5, 1 + \sqrt{5}, 1 - \sqrt{5}$

Step 6

The result can be shown in multiple forms.

Exact Form:

$x = 5, 1 + \sqrt{5}, 1 - \sqrt{5}$

Decimal Form:

$x = 5, 3.23606797 \dots, - 1.23606797 \dots$