Precalculus Examples

$f (- x) = \frac{4 x^{2} - 1}{x^{2}}$

Step 1

Since $x$ is on the right side of the equation, switch the sides so it is on the left side of the equation.

$\frac{4 x^{2} - 1}{x^{2}} = f (- x)$

Step 2

Simplify the numerator.

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Step 2.1

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

$\frac{{(2 x)}^{2} - 1}{x^{2}} = f (- x)$

Step 2.2

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

$\frac{{(2 x)}^{2} - 1^{2}}{x^{2}} = f (- x)$

Step 2.3

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

$\frac{(2 x + 1) (2 x - 1)}{x^{2}} = f (- x)$

Step 3

Rewrite using the commutative property of multiplication.

$\frac{(2 x + 1) (2 x - 1)}{x^{2}} = - f x$

Step 4

Move all terms containing $x$ to the left side of the equation.

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Step 4.1

Add $f x$ to both sides of the equation.

$\frac{(2 x + 1) (2 x - 1)}{x^{2}} + f x = 0$

Step 4.2

To write $f x$ as a fraction with a common denominator, multiply by $\frac{x^{2}}{x^{2}}$ .

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

Step 4.3

Combine the numerators over the common denominator.

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

Step 4.4

Simplify the numerator.

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Step 4.4.1

Expand $(2 x + 1) (2 x - 1)$ using the FOIL Method.

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Step 4.4.1.1

Apply the distributive property.

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

Step 4.4.1.2

Apply the distributive property.

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

Step 4.4.1.3

Apply the distributive property.

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

Step 4.4.2

Combine the opposite terms in $2 x (2 x) + 2 x \cdot - 1 + 1 (2 x) + 1 \cdot - 1$ .

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Step 4.4.2.1

Reorder the factors in the terms $2 x \cdot - 1$ and $1 (2 x)$ .

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

Step 4.4.2.2

Add $- 1 \cdot 2 x$ and $1 \cdot 2 x$ .

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

Step 4.4.2.3

Add $2 x (2 x)$ and $0$ .

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

Step 4.4.3

Simplify each term.

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Step 4.4.3.1

Rewrite using the commutative property of multiplication.

$\frac{2 \cdot 2 x \cdot x + 1 \cdot - 1 + f x \cdot x^{2}}{x^{2}} = 0$

Step 4.4.3.2

Multiply $x$ by $x$ by adding the exponents.

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Step 4.4.3.2.1

Move $x$ .

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

Step 4.4.3.2.2

Multiply $x$ by $x$ .

$\frac{2 \cdot 2 x^{2} + 1 \cdot - 1 + f x \cdot x^{2}}{x^{2}} = 0$

Step 4.4.3.3

Multiply $2$ by $2$ .

$\frac{4 x^{2} + 1 \cdot - 1 + f x \cdot x^{2}}{x^{2}} = 0$

Step 4.4.3.4

Multiply $- 1$ by $1$ .

$\frac{4 x^{2} - 1 + f x \cdot x^{2}}{x^{2}} = 0$

Step 4.4.4

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

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Step 4.4.4.1

Move $x^{2}$ .

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

Step 4.4.4.2

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

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Step 4.4.4.2.1

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

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

Step 4.4.4.2.2

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

$\frac{4 x^{2} - 1 + f x^{2 + 1}}{x^{2}} = 0$

Step 4.4.4.3

Add $2$ and $1$ .

$\frac{4 x^{2} - 1 + f x^{3}}{x^{2}} = 0$