Simplify square root of -(x+2)^8

$\sqrt{- {(x + 2)}^{8}}$

Rewrite $- {(x + 2)}^{8}$ as ${({({(x + 2)}^{2})}^{2})}^{2} \cdot - 1$ .

Rewrite ${(x + 2)}^{8}$ as ${({(x + 2)}^{4})}^{2}$ .

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

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

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

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

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

Pull terms out from under the radical.

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

Rewrite $\sqrt{- 1}$ as $i$ .

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

Multiply the exponents in ${({(x + 2)}^{2})}^{2}$ .

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

${(x + 2)}^{2 \cdot 2} i$

Multiply $2$ by $2$ .

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

Use the Binomial Theorem.

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

Simplify terms.

Simplify each term.

Multiply $2$ by $4$ .

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

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

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

Multiply $4$ by $6$ .

$(x^{4} + 8 x^{3} + 24 x^{2} + 4 x \cdot 2^{3} + 2^{4}) i$

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

$(x^{4} + 8 x^{3} + 24 x^{2} + 4 x \cdot 8 + 2^{4}) i$

Multiply $8$ by $4$ .

$(x^{4} + 8 x^{3} + 24 x^{2} + 32 x + 2^{4}) i$

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

$(x^{4} + 8 x^{3} + 24 x^{2} + 32 x + 16) i$

Apply the distributive property.

$x^{4} i + 8 x^{3} i + 24 x^{2} i + 32 x i + 16 i$