Expand Using the Binomial Theorem (1-2x)^2

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

Step 1

Use the binomial expansion theorem to find each term. The binomial theorem states ${(a + b)}^{n} = \sum_{k = 0}^{n} n C k \cdot (a^{n - k} b^{k})$ .

$\sum_{k = 0}^{2} \frac{2!}{(2 - k)! k!} \cdot {(1)}^{2 - k} \cdot {(- 2 x)}^{k}$

Step 2

Expand the summation.

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

Step 3

Simplify the exponents for each term of the expansion.

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

Step 4

Simplify each term.

Multiply $1$ by ${(1)}^{2}$ by adding the exponents.

Multiply $1$ by ${(1)}^{2}$ .

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

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

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

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

Add $1$ and $2$ .

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

Simplify $1^{3} \cdot {(- 2 x)}^{0}$ .

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

One to any power is one.

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

Evaluate the exponent.

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

Multiply $2$ by $1$ .

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

Simplify.

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

Multiply $- 2$ by $2$ .

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

Multiply $1$ by ${(1)}^{0}$ by adding the exponents.

Multiply $1$ by ${(1)}^{0}$ .

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

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

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

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

Add $1$ and $0$ .

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

Simplify $1^{1} \cdot {(- 2 x)}^{2}$ .

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

Apply the product rule to $- 2 x$ .

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

Raise $- 2$ to the power of $2$ .

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

Expand Using the Binomial Theorem (1-2x)^2

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