Find the Perfect Square Trinomial (2x-1/x)^2

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

Step 1

Rewrite ${(2 x - \frac{1}{x})}^{2}$ as $(2 x - \frac{1}{x}) (2 x - \frac{1}{x})$ .

$(2 x - \frac{1}{x}) (2 x - \frac{1}{x})$

Step 2

Expand $(2 x - \frac{1}{x}) (2 x - \frac{1}{x})$ using the FOIL Method.

Apply the distributive property.

$2 x (2 x - \frac{1}{x}) - \frac{1}{x} (2 x - \frac{1}{x})$

Apply the distributive property.

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

Apply the distributive property.

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

Step 3

Simplify and combine like terms.

Simplify each term.

Rewrite using the commutative property of multiplication.

$2 \cdot 2 x \cdot x + 2 x (- \frac{1}{x}) - \frac{1}{x} (2 x) - \frac{1}{x} (- \frac{1}{x})$

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

Move $x$ .

$2 \cdot 2 (x \cdot x) + 2 x (- \frac{1}{x}) - \frac{1}{x} (2 x) - \frac{1}{x} (- \frac{1}{x})$

Multiply $x$ by $x$ .

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

Multiply $2$ by $2$ .

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

Cancel the common factor of $x$ .

Move the leading negative in $- \frac{1}{x}$ into the numerator.

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

Factor $x$ out of $2 x$ .

$4 x^{2} + x \cdot 2 \frac{- 1}{x} - \frac{1}{x} (2 x) - \frac{1}{x} (- \frac{1}{x})$

Cancel the common factor.

$4 x^{2} + x \cdot 2 \frac{- 1}{x} - \frac{1}{x} (2 x) - \frac{1}{x} (- \frac{1}{x})$

Rewrite the expression.

$4 x^{2} + 2 \cdot - 1 - \frac{1}{x} (2 x) - \frac{1}{x} (- \frac{1}{x})$

Multiply $2$ by $- 1$ .

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

Cancel the common factor of $x$ .

Move the leading negative in $- \frac{1}{x}$ into the numerator.

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

Factor $x$ out of $2 x$ .

$4 x^{2} - 2 + \frac{- 1}{x} (x \cdot 2) - \frac{1}{x} (- \frac{1}{x})$

Cancel the common factor.

$4 x^{2} - 2 + \frac{- 1}{x} (x \cdot 2) - \frac{1}{x} (- \frac{1}{x})$

Rewrite the expression.

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

Multiply $- 1$ by $2$ .

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

Multiply $- \frac{1}{x} (- \frac{1}{x})$ .

Multiply $- 1$ by $- 1$ .

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

Multiply $\frac{1}{x}$ by $1$ .

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

Multiply $\frac{1}{x}$ by $\frac{1}{x}$ .

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

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

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

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

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

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

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

Add $1$ and $1$ .

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

Subtract $2$ from $- 2$ .

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

Find the Perfect Square Trinomial (2x-1/x)^2

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