Expand Using the Binomial Theorem (1+2i)^4

${(1 + 2 i)}^{4}$

Step 1

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

$\sum_{k = 0}^{4} \frac{4!}{(4 - k)! k!} \cdot {(1)}^{4 - k} \cdot {(2 i)}^{k}$

Step 2

Expand the summation.

$\frac{4!}{(4 - 0)! 0!} \cdot {(1)}^{4 - 0} \cdot {(2 i)}^{0} + \frac{4!}{(4 - 1)! 1!} \cdot {(1)}^{4 - 1} \cdot {(2 i)}^{1} + \frac{4!}{(4 - 2)! 2!} \cdot {(1)}^{4 - 2} \cdot {(2 i)}^{2} + \frac{4!}{(4 - 3)! 3!} \cdot {(1)}^{4 - 3} \cdot {(2 i)}^{3} + \frac{4!}{(4 - 4)! 4!} \cdot {(1)}^{4 - 4} \cdot {(2 i)}^{4}$

Step 3

Simplify the exponents for each term of the expansion.

$1 \cdot {(1)}^{4} \cdot {(2 i)}^{0} + 4 \cdot {(1)}^{3} \cdot {(2 i)}^{1} + 6 \cdot {(1)}^{2} \cdot {(2 i)}^{2} + 4 \cdot {(1)}^{1} \cdot {(2 i)}^{3} + 1 \cdot {(1)}^{0} \cdot {(2 i)}^{4}$

Step 4

Simplify the polynomial result.

Simplify each term.

Multiply $1$ by ${(1)}^{4}$ by adding the exponents.

Multiply $1$ by ${(1)}^{4}$ .

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

$1^{1} \cdot {(1)}^{4} \cdot {(2 i)}^{0} + 4 \cdot {(1)}^{3} \cdot {(2 i)}^{1} + 6 \cdot {(1)}^{2} \cdot {(2 i)}^{2} + 4 \cdot {(1)}^{1} \cdot {(2 i)}^{3} + 1 \cdot {(1)}^{0} \cdot {(2 i)}^{4}$

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

$1^{1 + 4} \cdot {(2 i)}^{0} + 4 \cdot {(1)}^{3} \cdot {(2 i)}^{1} + 6 \cdot {(1)}^{2} \cdot {(2 i)}^{2} + 4 \cdot {(1)}^{1} \cdot {(2 i)}^{3} + 1 \cdot {(1)}^{0} \cdot {(2 i)}^{4}$

Add $1$ and $4$ .

$1^{5} \cdot {(2 i)}^{0} + 4 \cdot {(1)}^{3} \cdot {(2 i)}^{1} + 6 \cdot {(1)}^{2} \cdot {(2 i)}^{2} + 4 \cdot {(1)}^{1} \cdot {(2 i)}^{3} + 1 \cdot {(1)}^{0} \cdot {(2 i)}^{4}$

Simplify $1^{5} \cdot {(2 i)}^{0}$ .

$1^{5} + 4 \cdot {(1)}^{3} \cdot {(2 i)}^{1} + 6 \cdot {(1)}^{2} \cdot {(2 i)}^{2} + 4 \cdot {(1)}^{1} \cdot {(2 i)}^{3} + 1 \cdot {(1)}^{0} \cdot {(2 i)}^{4}$

One to any power is one.

$1 + 4 \cdot {(1)}^{3} \cdot {(2 i)}^{1} + 6 \cdot {(1)}^{2} \cdot {(2 i)}^{2} + 4 \cdot {(1)}^{1} \cdot {(2 i)}^{3} + 1 \cdot {(1)}^{0} \cdot {(2 i)}^{4}$

One to any power is one.

$1 + 4 \cdot 1 \cdot {(2 i)}^{1} + 6 \cdot {(1)}^{2} \cdot {(2 i)}^{2} + 4 \cdot {(1)}^{1} \cdot {(2 i)}^{3} + 1 \cdot {(1)}^{0} \cdot {(2 i)}^{4}$

Multiply $4$ by $1$ .

$1 + 4 \cdot {(2 i)}^{1} + 6 \cdot {(1)}^{2} \cdot {(2 i)}^{2} + 4 \cdot {(1)}^{1} \cdot {(2 i)}^{3} + 1 \cdot {(1)}^{0} \cdot {(2 i)}^{4}$

Simplify.

$1 + 4 \cdot (2 i) + 6 \cdot {(1)}^{2} \cdot {(2 i)}^{2} + 4 \cdot {(1)}^{1} \cdot {(2 i)}^{3} + 1 \cdot {(1)}^{0} \cdot {(2 i)}^{4}$

Multiply $2$ by $4$ .

$1 + 8 i + 6 \cdot {(1)}^{2} \cdot {(2 i)}^{2} + 4 \cdot {(1)}^{1} \cdot {(2 i)}^{3} + 1 \cdot {(1)}^{0} \cdot {(2 i)}^{4}$

One to any power is one.

$1 + 8 i + 6 \cdot 1 \cdot {(2 i)}^{2} + 4 \cdot {(1)}^{1} \cdot {(2 i)}^{3} + 1 \cdot {(1)}^{0} \cdot {(2 i)}^{4}$

Multiply $6$ by $1$ .

$1 + 8 i + 6 \cdot {(2 i)}^{2} + 4 \cdot {(1)}^{1} \cdot {(2 i)}^{3} + 1 \cdot {(1)}^{0} \cdot {(2 i)}^{4}$

Apply the product rule to $2 i$ .

$1 + 8 i + 6 \cdot (2^{2} i^{2}) + 4 \cdot {(1)}^{1} \cdot {(2 i)}^{3} + 1 \cdot {(1)}^{0} \cdot {(2 i)}^{4}$

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

$1 + 8 i + 6 \cdot (4 i^{2}) + 4 \cdot {(1)}^{1} \cdot {(2 i)}^{3} + 1 \cdot {(1)}^{0} \cdot {(2 i)}^{4}$

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

$1 + 8 i + 6 \cdot (4 \cdot - 1) + 4 \cdot {(1)}^{1} \cdot {(2 i)}^{3} + 1 \cdot {(1)}^{0} \cdot {(2 i)}^{4}$

Multiply $6 (4 \cdot - 1)$ .

Multiply $4$ by $- 1$ .

$1 + 8 i + 6 \cdot - 4 + 4 \cdot {(1)}^{1} \cdot {(2 i)}^{3} + 1 \cdot {(1)}^{0} \cdot {(2 i)}^{4}$

Multiply $6$ by $- 4$ .

$1 + 8 i - 24 + 4 \cdot {(1)}^{1} \cdot {(2 i)}^{3} + 1 \cdot {(1)}^{0} \cdot {(2 i)}^{4}$

Evaluate the exponent.

$1 + 8 i - 24 + 4 \cdot 1 \cdot {(2 i)}^{3} + 1 \cdot {(1)}^{0} \cdot {(2 i)}^{4}$

Multiply $4$ by $1$ .

$1 + 8 i - 24 + 4 \cdot {(2 i)}^{3} + 1 \cdot {(1)}^{0} \cdot {(2 i)}^{4}$

Apply the product rule to $2 i$ .

$1 + 8 i - 24 + 4 \cdot (2^{3} i^{3}) + 1 \cdot {(1)}^{0} \cdot {(2 i)}^{4}$

Raise $2$ to the power of $3$ .

$1 + 8 i - 24 + 4 \cdot (8 i^{3}) + 1 \cdot {(1)}^{0} \cdot {(2 i)}^{4}$

Factor out $i^{2}$ .

$1 + 8 i - 24 + 4 \cdot (8 (i^{2} \cdot i)) + 1 \cdot {(1)}^{0} \cdot {(2 i)}^{4}$

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

$1 + 8 i - 24 + 4 \cdot (8 (- 1 \cdot i)) + 1 \cdot {(1)}^{0} \cdot {(2 i)}^{4}$

Rewrite $- 1 i$ as $- i$ .

$1 + 8 i - 24 + 4 \cdot (8 (- i)) + 1 \cdot {(1)}^{0} \cdot {(2 i)}^{4}$

Multiply $- 1$ by $8$ .

$1 + 8 i - 24 + 4 \cdot (- 8 i) + 1 \cdot {(1)}^{0} \cdot {(2 i)}^{4}$

Multiply $- 8$ by $4$ .

$1 + 8 i - 24 - 32 i + 1 \cdot {(1)}^{0} \cdot {(2 i)}^{4}$

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

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

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

$1 + 8 i - 24 - 32 i + 1^{1} \cdot {(1)}^{0} \cdot {(2 i)}^{4}$

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

$1 + 8 i - 24 - 32 i + 1^{1 + 0} \cdot {(2 i)}^{4}$

Add $1$ and $0$ .

$1 + 8 i - 24 - 32 i + 1^{1} \cdot {(2 i)}^{4}$

Simplify $1^{1} \cdot {(2 i)}^{4}$ .

$1 + 8 i - 24 - 32 i + {(2 i)}^{4}$

Apply the product rule to $2 i$ .

$1 + 8 i - 24 - 32 i + 2^{4} i^{4}$

Raise $2$ to the power of $4$ .

$1 + 8 i - 24 - 32 i + 16 i^{4}$

Rewrite $i^{4}$ as $1$ .

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

$1 + 8 i - 24 - 32 i + 16 {(i^{2})}^{2}$

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

$1 + 8 i - 24 - 32 i + 16 {(- 1)}^{2}$

Raise $- 1$ to the power of $2$ .

$1 + 8 i - 24 - 32 i + 16 \cdot 1$

Multiply $16$ by $1$ .

$1 + 8 i - 24 - 32 i + 16$

Simplify by adding terms.

Subtract $24$ from $1$ .

$- 23 + 8 i - 32 i + 16$

Add $- 23$ and $16$ .

$- 7 + 8 i - 32 i$

Subtract $32 i$ from $8 i$ .

$- 7 - 24 i$

Expand Using the Binomial Theorem (1+2i)^4

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