Calculus Examples

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

Step 1

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

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

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.

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

Step 2.1.2

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

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Step 2.1.2.1

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

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

Step 2.1.2.2

Factor $2$ out of $- 2$ .

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

Step 2.1.2.3

Factor $2$ out of $2 (x^{4}) + 2 (- 1)$ .

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

Step 2.1.3

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

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

Step 2.1.4

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

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

Step 2.1.5

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

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

Step 2.1.6

Factor.

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Step 2.1.6.1

Simplify.

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Step 2.1.6.1.1

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

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

Step 2.1.6.1.2

Factor.

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Step 2.1.6.1.2.1

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 $b = 1$ .

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

Step 2.1.6.1.2.2

Remove unnecessary parentheses.

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

Step 2.1.6.2

Remove unnecessary parentheses.

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

Step 2.1.7

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

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Step 2.1.7.1

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

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

Step 2.1.7.2

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

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

Step 2.1.7.3

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

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

Step 2.1.8

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

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

Step 2.1.9

Factor.

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Step 2.1.9.1

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 $b = 1$ .

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

Step 2.1.9.2

Remove unnecessary parentheses.

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

Step 2.1.10

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

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Step 2.1.10.1

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

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

Step 2.1.10.2

Factor $(x + 1) (x - 1)$ out of $5 x (x + 1) (x - 1)$ .

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

Step 2.1.10.3

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

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

Step 2.1.11

Apply the distributive property.

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

Step 2.1.12

Multiply $2$ by $1$ .

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

Step 2.1.13

Reorder terms.

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

Step 2.1.14

Factor.

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Step 2.1.14.1

Factor by grouping.

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Step 2.1.14.1.1

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = 2 \cdot 2 = 4$ and whose sum is $b = 5$ .

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Step 2.1.14.1.1.1

Factor $5$ out of $5 x$ .

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

Step 2.1.14.1.1.2

Rewrite $5$ as $1$ plus $4$

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

Step 2.1.14.1.1.3

Apply the distributive property.

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

Step 2.1.14.1.1.4

Multiply $x$ by $1$ .

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

Step 2.1.14.1.2

Factor out the greatest common factor from each group.

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Step 2.1.14.1.2.1

Group the first two terms and the last two terms.

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

Step 2.1.14.1.2.2

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

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

Step 2.1.14.1.3

Factor the polynomial by factoring out the greatest common factor, $2 x + 1$ .

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

Step 2.1.14.2

Remove unnecessary parentheses.

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

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 $0$ .

$x + 1 = 0$

$x - 1 = 0$

$2 x + 1 = 0$

$x + 2 = 0$

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 $x - 1$ 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 $2 x + 1$ equal to $0$ and solve for $x$ .

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Step 2.5.1

Set $2 x + 1$ equal to $0$ .

$2 x + 1 = 0$

Step 2.5.2

Solve $2 x + 1 = 0$ for $x$ .

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Step 2.5.2.1

Subtract $1$ from both sides of the equation.

$2 x = - 1$

Step 2.5.2.2

Divide each term in $2 x = - 1$ by $2$ and simplify.

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Step 2.5.2.2.1

Divide each term in $2 x = - 1$ by $2$ .

$\frac{2 x}{2} = \frac{- 1}{2}$

Step 2.5.2.2.2

Simplify the left side.

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Step 2.5.2.2.2.1

Cancel the common factor of $2$ .

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Step 2.5.2.2.2.1.1

Cancel the common factor.

$\frac{2 x}{2} = \frac{- 1}{2}$

Step 2.5.2.2.2.1.2

Divide $x$ by $1$ .

$x = \frac{- 1}{2}$

Step 2.5.2.2.3

Simplify the right side.

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Step 2.5.2.2.3.1

Move the negative in front of the fraction.

$x = - \frac{1}{2}$

Step 2.6

Set $x + 2$ equal to $0$ and solve for $x$ .

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Step 2.6.1

Set $x + 2$ equal to $0$ .

$x + 2 = 0$

Step 2.6.2

Subtract $2$ from both sides of the equation.

$x = - 2$

Step 2.7

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

$x = - 1, 1, - \frac{1}{2}, - 2$

Step 3