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

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

Step 1

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

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

Step 2

Expand the summation.

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

Step 3

Simplify the exponents for each term of the expansion.

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

Step 4

Simplify the polynomial result.

Simplify each term.

Multiply ${(2)}^{4}$ by $1$ .

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

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

$16 \cdot {(4 i)}^{0} + 4 \cdot {(2)}^{3} \cdot {(4 i)}^{1} + 6 \cdot {(2)}^{2} \cdot {(4 i)}^{2} + 4 \cdot {(2)}^{1} \cdot {(4 i)}^{3} + 1 \cdot {(2)}^{0} \cdot {(4 i)}^{4}$

Apply the product rule to $4 i$ .

$16 \cdot (4^{0} i^{0}) + 4 \cdot {(2)}^{3} \cdot {(4 i)}^{1} + 6 \cdot {(2)}^{2} \cdot {(4 i)}^{2} + 4 \cdot {(2)}^{1} \cdot {(4 i)}^{3} + 1 \cdot {(2)}^{0} \cdot {(4 i)}^{4}$

Anything raised to $0$ is $1$ .

$16 \cdot (1 i^{0}) + 4 \cdot {(2)}^{3} \cdot {(4 i)}^{1} + 6 \cdot {(2)}^{2} \cdot {(4 i)}^{2} + 4 \cdot {(2)}^{1} \cdot {(4 i)}^{3} + 1 \cdot {(2)}^{0} \cdot {(4 i)}^{4}$

Multiply $i^{0}$ by $1$ .

$16 \cdot i^{0} + 4 \cdot {(2)}^{3} \cdot {(4 i)}^{1} + 6 \cdot {(2)}^{2} \cdot {(4 i)}^{2} + 4 \cdot {(2)}^{1} \cdot {(4 i)}^{3} + 1 \cdot {(2)}^{0} \cdot {(4 i)}^{4}$

Anything raised to $0$ is $1$ .

$16 \cdot 1 + 4 \cdot {(2)}^{3} \cdot {(4 i)}^{1} + 6 \cdot {(2)}^{2} \cdot {(4 i)}^{2} + 4 \cdot {(2)}^{1} \cdot {(4 i)}^{3} + 1 \cdot {(2)}^{0} \cdot {(4 i)}^{4}$

Multiply $16$ by $1$ .

$16 + 4 \cdot {(2)}^{3} \cdot {(4 i)}^{1} + 6 \cdot {(2)}^{2} \cdot {(4 i)}^{2} + 4 \cdot {(2)}^{1} \cdot {(4 i)}^{3} + 1 \cdot {(2)}^{0} \cdot {(4 i)}^{4}$

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

$16 + 4 \cdot 8 \cdot {(4 i)}^{1} + 6 \cdot {(2)}^{2} \cdot {(4 i)}^{2} + 4 \cdot {(2)}^{1} \cdot {(4 i)}^{3} + 1 \cdot {(2)}^{0} \cdot {(4 i)}^{4}$

Multiply $4$ by $8$ .

$16 + 32 \cdot {(4 i)}^{1} + 6 \cdot {(2)}^{2} \cdot {(4 i)}^{2} + 4 \cdot {(2)}^{1} \cdot {(4 i)}^{3} + 1 \cdot {(2)}^{0} \cdot {(4 i)}^{4}$

Simplify.

$16 + 32 \cdot (4 i) + 6 \cdot {(2)}^{2} \cdot {(4 i)}^{2} + 4 \cdot {(2)}^{1} \cdot {(4 i)}^{3} + 1 \cdot {(2)}^{0} \cdot {(4 i)}^{4}$

Multiply $4$ by $32$ .

$16 + 128 i + 6 \cdot {(2)}^{2} \cdot {(4 i)}^{2} + 4 \cdot {(2)}^{1} \cdot {(4 i)}^{3} + 1 \cdot {(2)}^{0} \cdot {(4 i)}^{4}$

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

$16 + 128 i + 6 \cdot 4 \cdot {(4 i)}^{2} + 4 \cdot {(2)}^{1} \cdot {(4 i)}^{3} + 1 \cdot {(2)}^{0} \cdot {(4 i)}^{4}$

Multiply $6$ by $4$ .

$16 + 128 i + 24 \cdot {(4 i)}^{2} + 4 \cdot {(2)}^{1} \cdot {(4 i)}^{3} + 1 \cdot {(2)}^{0} \cdot {(4 i)}^{4}$

Apply the product rule to $4 i$ .

$16 + 128 i + 24 \cdot (4^{2} i^{2}) + 4 \cdot {(2)}^{1} \cdot {(4 i)}^{3} + 1 \cdot {(2)}^{0} \cdot {(4 i)}^{4}$

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

$16 + 128 i + 24 \cdot (16 i^{2}) + 4 \cdot {(2)}^{1} \cdot {(4 i)}^{3} + 1 \cdot {(2)}^{0} \cdot {(4 i)}^{4}$

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

$16 + 128 i + 24 \cdot (16 \cdot - 1) + 4 \cdot {(2)}^{1} \cdot {(4 i)}^{3} + 1 \cdot {(2)}^{0} \cdot {(4 i)}^{4}$

Multiply $24 (16 \cdot - 1)$ .

Multiply $16$ by $- 1$ .

$16 + 128 i + 24 \cdot - 16 + 4 \cdot {(2)}^{1} \cdot {(4 i)}^{3} + 1 \cdot {(2)}^{0} \cdot {(4 i)}^{4}$

Multiply $24$ by $- 16$ .

$16 + 128 i - 384 + 4 \cdot {(2)}^{1} \cdot {(4 i)}^{3} + 1 \cdot {(2)}^{0} \cdot {(4 i)}^{4}$

Evaluate the exponent.

$16 + 128 i - 384 + 4 \cdot 2 \cdot {(4 i)}^{3} + 1 \cdot {(2)}^{0} \cdot {(4 i)}^{4}$

Multiply $4$ by $2$ .

$16 + 128 i - 384 + 8 \cdot {(4 i)}^{3} + 1 \cdot {(2)}^{0} \cdot {(4 i)}^{4}$

Apply the product rule to $4 i$ .

$16 + 128 i - 384 + 8 \cdot (4^{3} i^{3}) + 1 \cdot {(2)}^{0} \cdot {(4 i)}^{4}$

Raise $4$ to the power of $3$ .

$16 + 128 i - 384 + 8 \cdot (64 i^{3}) + 1 \cdot {(2)}^{0} \cdot {(4 i)}^{4}$

Factor out $i^{2}$ .

$16 + 128 i - 384 + 8 \cdot (64 (i^{2} \cdot i)) + 1 \cdot {(2)}^{0} \cdot {(4 i)}^{4}$

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

$16 + 128 i - 384 + 8 \cdot (64 (- 1 \cdot i)) + 1 \cdot {(2)}^{0} \cdot {(4 i)}^{4}$

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

$16 + 128 i - 384 + 8 \cdot (64 (- i)) + 1 \cdot {(2)}^{0} \cdot {(4 i)}^{4}$

Multiply $- 1$ by $64$ .

$16 + 128 i - 384 + 8 \cdot (- 64 i) + 1 \cdot {(2)}^{0} \cdot {(4 i)}^{4}$

Multiply $- 64$ by $8$ .

$16 + 128 i - 384 - 512 i + 1 \cdot {(2)}^{0} \cdot {(4 i)}^{4}$

Multiply ${(2)}^{0}$ by $1$ .

$16 + 128 i - 384 - 512 i + {(2)}^{0} \cdot {(4 i)}^{4}$

Anything raised to $0$ is $1$ .

$16 + 128 i - 384 - 512 i + 1 \cdot {(4 i)}^{4}$

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

$16 + 128 i - 384 - 512 i + {(4 i)}^{4}$

Apply the product rule to $4 i$ .

$16 + 128 i - 384 - 512 i + 4^{4} i^{4}$

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

$16 + 128 i - 384 - 512 i + 256 i^{4}$

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

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

$16 + 128 i - 384 - 512 i + 256 {(i^{2})}^{2}$

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

$16 + 128 i - 384 - 512 i + 256 {(- 1)}^{2}$

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

$16 + 128 i - 384 - 512 i + 256 \cdot 1$

Multiply $256$ by $1$ .

$16 + 128 i - 384 - 512 i + 256$

Simplify by adding terms.

Subtract $384$ from $16$ .

$- 368 + 128 i - 512 i + 256$

Add $- 368$ and $256$ .

$- 112 + 128 i - 512 i$

Subtract $512 i$ from $128 i$ .

$- 112 - 384 i$

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

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