Finite Math Examples

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

Step 1

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

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

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.

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

Step 2.1.2

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

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Step 2.1.2.1

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

$- 3 x \cdot x^{2} + 3 x + x^{4} - 5 x^{2} + 4 = 0$

Step 2.1.2.2

Factor $- 3 x$ out of $3 x$ .

$- 3 x \cdot x^{2} - 3 x \cdot - 1 + x^{4} - 5 x^{2} + 4 = 0$

Step 2.1.2.3

Factor $- 3 x$ out of $- 3 x (x^{2}) - 3 x (- 1)$ .

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

Step 2.1.3

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

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

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 $b = 1$ .

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

Step 2.1.4.2

Remove unnecessary parentheses.

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

Step 2.1.5

Rewrite $x^{4}$ as ${(x^{2})}^{2}$ .

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

Step 2.1.6

Let $u = x^{2}$ . Substitute $u$ for all occurrences of $x^{2}$ .

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

Step 2.1.7

Factor $u^{2} - 5 u + 4$ using the AC method.

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Step 2.1.7.1

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $4$ and whose sum is $- 5$ .

$- 4, - 1$

Step 2.1.7.2

Write the factored form using these integers.

$- 3 x (x + 1) (x - 1) + (u - 4) (u - 1) = 0$

Step 2.1.8

Replace all occurrences of $u$ with $x^{2}$ .

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

Step 2.1.9

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

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

Step 2.1.10

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 $b = 2$ .

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

Step 2.1.11

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

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

Step 2.1.12

Factor.

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Step 2.1.12.1

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 $b = 1$ .

$- 3 x (x + 1) (x - 1) + (x + 2) (x - 2) ((x + 1) (x - 1)) = 0$

Step 2.1.12.2

Remove unnecessary parentheses.

$- 3 x (x + 1) (x - 1) + (x + 2) (x - 2) (x + 1) (x - 1) = 0$

Step 2.1.13

Factor $(x + 1) (x - 1)$ out of $- 3 x (x + 1) (x - 1) + (x + 2) (x - 2) (x + 1) (x - 1)$ .

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Step 2.1.13.1

Factor $(x + 1) (x - 1)$ out of $- 3 x (x + 1) (x - 1)$ .

$(x + 1) (x - 1) (- 3 x) + (x + 2) (x - 2) (x + 1) (x - 1) = 0$

Step 2.1.13.2

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

$(x + 1) (x - 1) (- 3 x) + (x + 1) (x - 1) ((x + 2) (x - 2)) = 0$

Step 2.1.13.3

Factor $(x + 1) (x - 1)$ out of $(x + 1) (x - 1) (- 3 x) + (x + 1) (x - 1) ((x + 2) (x - 2))$ .

$(x + 1) (x - 1) (- 3 x + (x + 2) (x - 2)) = 0$

Step 2.1.14

Expand $(x + 2) (x - 2)$ using the FOIL Method.

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Step 2.1.14.1

Apply the distributive property.

$(x + 1) (x - 1) (- 3 x + x (x - 2) + 2 (x - 2)) = 0$

Step 2.1.14.2

Apply the distributive property.

$(x + 1) (x - 1) (- 3 x + x \cdot x + x \cdot - 2 + 2 (x - 2)) = 0$

Step 2.1.14.3

Apply the distributive property.

$(x + 1) (x - 1) (- 3 x + x \cdot x + x \cdot - 2 + 2 x + 2 \cdot - 2) = 0$

Step 2.1.15

Combine the opposite terms in $x \cdot x + x \cdot - 2 + 2 x + 2 \cdot - 2$ .

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Step 2.1.15.1

Reorder the factors in the terms $x \cdot - 2$ and $2 x$ .

$(x + 1) (x - 1) (- 3 x + x \cdot x - 2 x + 2 x + 2 \cdot - 2) = 0$

Step 2.1.15.2

Add $- 2 x$ and $2 x$ .

$(x + 1) (x - 1) (- 3 x + x \cdot x + 0 + 2 \cdot - 2) = 0$

Step 2.1.15.3

Add $x \cdot x$ and $0$ .

$(x + 1) (x - 1) (- 3 x + x \cdot x + 2 \cdot - 2) = 0$

Step 2.1.16

Simplify each term.

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Step 2.1.16.1

Multiply $x$ by $x$ .

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

Step 2.1.16.2

Multiply $2$ by $- 2$ .

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

Step 2.1.17

Reorder terms.

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

Step 2.1.18

Factor.

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Step 2.1.18.1

Factor $x^{2} - 3 x - 4$ using the AC method.

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Step 2.1.18.1.1

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 4$ and whose sum is $- 3$ .

$- 4, 1$

Step 2.1.18.1.2

Write the factored form using these integers.

$(x + 1) (x - 1) ((x - 4) (x + 1)) = 0$

Step 2.1.18.2

Remove unnecessary parentheses.

$(x + 1) (x - 1) (x - 4) (x + 1) = 0$

Step 2.1.19

Combine exponents.

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Step 2.1.19.1

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

$(x + 1) (x + 1) (x - 1) (x - 4) = 0$

Step 2.1.19.2

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

$(x + 1) (x + 1) (x - 1) (x - 4) = 0$

Step 2.1.19.3

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

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

Step 2.1.19.4

Add $1$ and $1$ .

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

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 $0$ .

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

$x - 1 = 0$

$x - 4 = 0$

Step 2.3

Set ${(x + 1)}^{2}$ equal to $0$ and solve for $x$ .

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Step 2.3.1

Set ${(x + 1)}^{2}$ equal to $0$ .

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

Step 2.3.2

Solve ${(x + 1)}^{2} = 0$ for $x$ .

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Step 2.3.2.1

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

$x + 1 = 0$

Step 2.3.2.2

Subtract $1$ from both sides of the equation.

$x = - 1$

Step 2.4

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

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Step 2.4.1

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

$x - 1 = 0$

Step 2.4.2

Add $1$ to both sides of the equation.

$x = 1$

Step 2.5

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

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Step 2.5.1

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

$x - 4 = 0$

Step 2.5.2

Add $4$ to both sides of the equation.

$x = 4$

Step 2.6

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

$x = - 1, 1, 4$

Step 3