Finite Math Examples

$f (x) = x^{4} - 5 x^{3} - x^{2} - 25 x - 30$

Step 1

Set $x^{4} - 5 x^{3} - x^{2} - 25 x - 30$ equal to $0$ .

$x^{4} - 5 x^{3} - x^{2} - 25 x - 30 = 0$

Step 2

Solve for $x$ .

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Step 2.1

Factor the left side of the equation.

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Step 2.1.1

Regroup terms.

$x^{4} - x^{2} - 5 x^{3} - 25 x - 30 = 0$

Step 2.1.2

Factor $x^{2}$ out of $x^{4} - x^{2}$ .

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Step 2.1.2.1

Factor $x^{2}$ out of $x^{4}$ .

$x^{2} x^{2} - x^{2} - 5 x^{3} - 25 x - 30 = 0$

Step 2.1.2.2

Factor $x^{2}$ out of $- x^{2}$ .

$x^{2} x^{2} + x^{2} \cdot - 1 - 5 x^{3} - 25 x - 30 = 0$

Step 2.1.2.3

Factor $x^{2}$ out of $x^{2} x^{2} + x^{2} \cdot - 1$ .

$x^{2} (x^{2} - 1) - 5 x^{3} - 25 x - 30 = 0$

Step 2.1.3

Rewrite $1$ as $1^{2}$ .

$x^{2} (x^{2} - 1^{2}) - 5 x^{3} - 25 x - 30 = 0$

Step 2.1.4

Factor.

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Step 2.1.4.1

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = x$ and $b = 1$ .

$x^{2} ((x + 1) (x - 1)) - 5 x^{3} - 25 x - 30 = 0$

Step 2.1.4.2

Remove unnecessary parentheses.

$x^{2} (x + 1) (x - 1) - 5 x^{3} - 25 x - 30 = 0$

Step 2.1.5

Factor $5$ out of $- 5 x^{3} - 25 x - 30$ .

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Step 2.1.5.1

Factor $5$ out of $- 5 x^{3}$ .

$x^{2} (x + 1) (x - 1) + 5 (- x^{3}) - 25 x - 30 = 0$

Step 2.1.5.2

Factor $5$ out of $- 25 x$ .

$x^{2} (x + 1) (x - 1) + 5 (- x^{3}) + 5 (- 5 x) - 30 = 0$

Step 2.1.5.3

Factor $5$ out of $- 30$ .

$x^{2} (x + 1) (x - 1) + 5 (- x^{3}) + 5 (- 5 x) + 5 (- 6) = 0$

Step 2.1.5.4

Factor $5$ out of $5 (- x^{3}) + 5 (- 5 x)$ .

$x^{2} (x + 1) (x - 1) + 5 (- x^{3} - 5 x) + 5 (- 6) = 0$

Step 2.1.5.5

Factor $5$ out of $5 (- x^{3} - 5 x) + 5 (- 6)$ .

$x^{2} (x + 1) (x - 1) + 5 (- x^{3} - 5 x - 6) = 0$

Step 2.1.6

Factor.

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Step 2.1.6.1

Factor $- x^{3} - 5 x - 6$ using the rational roots test.

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Step 2.1.6.1.1

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.

$p = \pm 1, \pm 6, \pm 2, \pm 3$

$q = \pm 1$

Step 2.1.6.1.2

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

$\pm 1, \pm 6, \pm 2, \pm 3$

Step 2.1.6.1.3

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 2.1.6.1.3.1

Substitute $- 1$ into the polynomial.

$- {(- 1)}^{3} - 5 \cdot - 1 - 6$

Step 2.1.6.1.3.2

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

$- - 1 - 5 \cdot - 1 - 6$

Step 2.1.6.1.3.3

Multiply $- 1$ by $- 1$ .

$1 - 5 \cdot - 1 - 6$

Step 2.1.6.1.3.4

Multiply $- 5$ by $- 1$ .

$1 + 5 - 6$

Step 2.1.6.1.3.5

Add $1$ and $5$ .

$6 - 6$

Step 2.1.6.1.3.6

Subtract $6$ from $6$ .

$0$

Step 2.1.6.1.4

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.

$\frac{- x^{3} - 5 x - 6}{x + 1}$

Step 2.1.6.1.5

Divide $- x^{3} - 5 x - 6$ by $x + 1$ .

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Step 2.1.6.1.5.1

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

$x$

$1$

$x^{3}$

$0 x^{2}$

$5 x$

$6$

Step 2.1.6.1.5.2

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

				-	$x^{2}$
	$x$	+	$1$	-	$x^{3}$	+	$0 x^{2}$	-	$5 x$	-	$6$

Step 2.1.6.1.5.3

Multiply the new quotient term by the divisor.

			-	$x^{2}$
$x$	+	$1$	-	$x^{3}$	+	$0 x^{2}$	-	$5 x$	-	$6$
			-	$x^{3}$	-	$x^{2}$

Step 2.1.6.1.5.4

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

			-	$x^{2}$
$x$	+	$1$	-	$x^{3}$	+	$0 x^{2}$	-	$5 x$	-	$6$
			+	$x^{3}$	+	$x^{2}$

Step 2.1.6.1.5.5

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

			-	$x^{2}$
$x$	+	$1$	-	$x^{3}$	+	$0 x^{2}$	-	$5 x$	-	$6$
			+	$x^{3}$	+	$x^{2}$
					+	$x^{2}$

Step 2.1.6.1.5.6

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

			-	$x^{2}$
$x$	+	$1$	-	$x^{3}$	+	$0 x^{2}$	-	$5 x$	-	$6$
			+	$x^{3}$	+	$x^{2}$
					+	$x^{2}$	-	$5 x$

Step 2.1.6.1.5.7

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

			-	$x^{2}$	+	$x$
$x$	+	$1$	-	$x^{3}$	+	$0 x^{2}$	-	$5 x$	-	$6$
			+	$x^{3}$	+	$x^{2}$
					+	$x^{2}$	-	$5 x$

Step 2.1.6.1.5.8

Multiply the new quotient term by the divisor.

			-	$x^{2}$	+	$x$
$x$	+	$1$	-	$x^{3}$	+	$0 x^{2}$	-	$5 x$	-	$6$
			+	$x^{3}$	+	$x^{2}$
					+	$x^{2}$	-	$5 x$
					+	$x^{2}$	+	$x$

Step 2.1.6.1.5.9

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

			-	$x^{2}$	+	$x$
$x$	+	$1$	-	$x^{3}$	+	$0 x^{2}$	-	$5 x$	-	$6$
			+	$x^{3}$	+	$x^{2}$
					+	$x^{2}$	-	$5 x$
					-	$x^{2}$	-	$x$

Step 2.1.6.1.5.10

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

			-	$x^{2}$	+	$x$
$x$	+	$1$	-	$x^{3}$	+	$0 x^{2}$	-	$5 x$	-	$6$
			+	$x^{3}$	+	$x^{2}$
					+	$x^{2}$	-	$5 x$
					-	$x^{2}$	-	$x$
							-	$6 x$

Step 2.1.6.1.5.11

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

			-	$x^{2}$	+	$x$
$x$	+	$1$	-	$x^{3}$	+	$0 x^{2}$	-	$5 x$	-	$6$
			+	$x^{3}$	+	$x^{2}$
					+	$x^{2}$	-	$5 x$
					-	$x^{2}$	-	$x$
							-	$6 x$	-	$6$

Step 2.1.6.1.5.12

Divide the highest order term in the dividend $- 6 x$ by the highest order term in divisor $x$ .

			-	$x^{2}$	+	$x$	-	$6$
$x$	+	$1$	-	$x^{3}$	+	$0 x^{2}$	-	$5 x$	-	$6$
			+	$x^{3}$	+	$x^{2}$
					+	$x^{2}$	-	$5 x$
					-	$x^{2}$	-	$x$
							-	$6 x$	-	$6$

Step 2.1.6.1.5.13

Multiply the new quotient term by the divisor.

			-	$x^{2}$	+	$x$	-	$6$
$x$	+	$1$	-	$x^{3}$	+	$0 x^{2}$	-	$5 x$	-	$6$
			+	$x^{3}$	+	$x^{2}$
					+	$x^{2}$	-	$5 x$
					-	$x^{2}$	-	$x$
							-	$6 x$	-	$6$
							-	$6 x$	-	$6$

Step 2.1.6.1.5.14

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

			-	$x^{2}$	+	$x$	-	$6$
$x$	+	$1$	-	$x^{3}$	+	$0 x^{2}$	-	$5 x$	-	$6$
			+	$x^{3}$	+	$x^{2}$
					+	$x^{2}$	-	$5 x$
					-	$x^{2}$	-	$x$
							-	$6 x$	-	$6$
							+	$6 x$	+	$6$

Step 2.1.6.1.5.15

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

			-	$x^{2}$	+	$x$	-	$6$
$x$	+	$1$	-	$x^{3}$	+	$0 x^{2}$	-	$5 x$	-	$6$
			+	$x^{3}$	+	$x^{2}$
					+	$x^{2}$	-	$5 x$
					-	$x^{2}$	-	$x$
							-	$6 x$	-	$6$
							+	$6 x$	+	$6$
										$0$

Step 2.1.6.1.5.16

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

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

Step 2.1.6.1.6

Write $- x^{3} - 5 x - 6$ as a set of factors.

$x^{2} (x + 1) (x - 1) + 5 ((x + 1) (- x^{2} + x - 6)) = 0$

Step 2.1.6.2

Remove unnecessary parentheses.

$x^{2} (x + 1) (x - 1) + 5 (x + 1) (- x^{2} + x - 6) = 0$

Step 2.1.7

Factor $x + 1$ out of $x^{2} (x + 1) (x - 1) + 5 (x + 1) (- x^{2} + x - 6)$ .

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Step 2.1.7.1

Factor $x + 1$ out of $x^{2} (x + 1) (x - 1)$ .

$(x + 1) (x^{2} (x - 1)) + 5 (x + 1) (- x^{2} + x - 6) = 0$

Step 2.1.7.2

Factor $x + 1$ out of $5 (x + 1) (- x^{2} + x - 6)$ .

$(x + 1) (x^{2} (x - 1)) + (x + 1) (5 (- x^{2} + x - 6)) = 0$

Step 2.1.7.3

Factor $x + 1$ out of $(x + 1) (x^{2} (x - 1)) + (x + 1) (5 (- x^{2} + x - 6))$ .

$(x + 1) (x^{2} (x - 1) + 5 (- x^{2} + x - 6)) = 0$

Step 2.1.8

Apply the distributive property.

$(x + 1) (x^{2} x + x^{2} \cdot - 1 + 5 (- x^{2} + x - 6)) = 0$

Step 2.1.9

Multiply $x^{2}$ by $x$ by adding the exponents.

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Step 2.1.9.1

Multiply $x^{2}$ by $x$ .

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Step 2.1.9.1.1

Raise $x$ to the power of $1$ .

$(x + 1) (x^{2} x + x^{2} \cdot - 1 + 5 (- x^{2} + x - 6)) = 0$

Step 2.1.9.1.2

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$(x + 1) (x^{2 + 1} + x^{2} \cdot - 1 + 5 (- x^{2} + x - 6)) = 0$

Step 2.1.9.2

Add $2$ and $1$ .

$(x + 1) (x^{3} + x^{2} \cdot - 1 + 5 (- x^{2} + x - 6)) = 0$

Step 2.1.10

Move $- 1$ to the left of $x^{2}$ .

$(x + 1) (x^{3} - 1 x^{2} + 5 (- x^{2} + x - 6)) = 0$

Step 2.1.11

Rewrite $- 1 x^{2}$ as $- x^{2}$ .

$(x + 1) (x^{3} - x^{2} + 5 (- x^{2} + x - 6)) = 0$

Step 2.1.12

Apply the distributive property.

$(x + 1) (x^{3} - x^{2} + 5 (- x^{2}) + 5 x + 5 \cdot - 6) = 0$

Step 2.1.13

Simplify.

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Step 2.1.13.1

Multiply $- 1$ by $5$ .

$(x + 1) (x^{3} - x^{2} - 5 x^{2} + 5 x + 5 \cdot - 6) = 0$

Step 2.1.13.2

Multiply $5$ by $- 6$ .

$(x + 1) (x^{3} - x^{2} - 5 x^{2} + 5 x - 30) = 0$

Step 2.1.14

Subtract $5 x^{2}$ from $- x^{2}$ .

$(x + 1) (x^{3} - 6 x^{2} + 5 x - 30) = 0$

Step 2.1.15

Factor.

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Step 2.1.15.1

Rewrite $x^{3} - 6 x^{2} + 5 x - 30$ in a factored form.

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Step 2.1.15.1.1

Factor out the greatest common factor from each group.

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Step 2.1.15.1.1.1

Group the first two terms and the last two terms.

$(x + 1) ((x^{3} - 6 x^{2}) + 5 x - 30) = 0$

Step 2.1.15.1.1.2

Factor out the greatest common factor (GCF) from each group.

$(x + 1) (x^{2} (x - 6) + 5 (x - 6)) = 0$

Step 2.1.15.1.2

Factor the polynomial by factoring out the greatest common factor, $x - 6$ .

$(x + 1) ((x - 6) (x^{2} + 5)) = 0$

Step 2.1.15.2

Remove unnecessary parentheses.

$(x + 1) (x - 6) (x^{2} + 5) = 0$

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 $0$ .

$x + 1 = 0$

$x - 6 = 0$

$x^{2} + 5 = 0$

Step 2.3

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

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Step 2.3.1

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

$x + 1 = 0$

Step 2.3.2

Subtract $1$ from both sides of the equation.

$x = - 1$

Step 2.4

Set $x - 6$ equal to $0$ and solve for $x$ .

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Step 2.4.1

Set $x - 6$ equal to $0$ .

$x - 6 = 0$

Step 2.4.2

Add $6$ to both sides of the equation.

$x = 6$

Step 2.5

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

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Step 2.5.1

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

$x^{2} + 5 = 0$

Step 2.5.2

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

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Step 2.5.2.1

Subtract $5$ from both sides of the equation.

$x^{2} = - 5$

Step 2.5.2.2

Take the specified root of both sides of the equation to eliminate the exponent on the left side.

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

Step 2.5.2.3

Simplify $\pm \sqrt{- 5}$ .

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Step 2.5.2.3.1

Rewrite $- 5$ as $- 1 (5)$ .

$x = \pm \sqrt{- 1 (5)}$

Step 2.5.2.3.2

Rewrite $\sqrt{- 1 (5)}$ as $\sqrt{- 1} \cdot \sqrt{5}$ .

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

Step 2.5.2.3.3

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

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

Step 2.5.2.4

The complete solution is the result of both the positive and negative portions of the solution.

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Step 2.5.2.4.1

First, use the positive value of the $\pm$ to find the first solution.

$x = i \sqrt{5}$

Step 2.5.2.4.2

Next, use the negative value of the $\pm$ to find the second solution.

$x = - i \sqrt{5}$

Step 2.5.2.4.3

The complete solution is the result of both the positive and negative portions of the solution.

$x = i \sqrt{5}, - i \sqrt{5}$

Step 2.6

The final solution is all the values that make $(x + 1) (x - 6) (x^{2} + 5) = 0$ true. The multiplicity of a root is the number of times the root appears.

$x = - 1$ (Multiplicity of $1$ )

$x = 6$ (Multiplicity of $1$ )

$x = i \sqrt{5}$ (Multiplicity of $1$ )

$x = - i \sqrt{5}$ (Multiplicity of $1$ )

$x = - 1$ (Multiplicity of $1$ )

$x = 6$ (Multiplicity of $1$ )

$x = i \sqrt{5}$ (Multiplicity of $1$ )

$x = - i \sqrt{5}$ (Multiplicity of $1$ )

Step 3