Write in Standard Form square root of -8( square root of -3- square root of 5)

$\sqrt{- 8} (\sqrt{- 3} - \sqrt{5})$

Step 1

To write a polynomial in standard form, simplify and then arrange the terms in descending order.

$a x^{2} + b x + c$

Step 2

Rewrite $- 8$ as $- 1 (8)$ .

$\sqrt{- 1 (8)} (\sqrt{- 3} - \sqrt{5})$

Step 3

Rewrite $\sqrt{- 1 (8)}$ as $\sqrt{- 1} \cdot \sqrt{8}$ .

$\sqrt{- 1} \cdot \sqrt{8} (\sqrt{- 3} - \sqrt{5})$

Step 4

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

$i \cdot \sqrt{8} (\sqrt{- 3} - \sqrt{5})$

Step 5

Rewrite $- 3$ as $- 1 (3)$ .

$i \cdot \sqrt{8} (\sqrt{- 1 (3)} - \sqrt{5})$

Step 6

Rewrite $\sqrt{- 1 (3)}$ as $\sqrt{- 1} \cdot \sqrt{3}$ .

$i \cdot \sqrt{8} (\sqrt{- 1} \cdot \sqrt{3} - \sqrt{5})$

Step 7

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

$i \cdot \sqrt{8} (i \cdot \sqrt{3} - \sqrt{5})$

Step 8

Rewrite $8$ as $2^{2} \cdot 2$ .

Factor $4$ out of $8$ .

$i \cdot \sqrt{4 (2)} (i \cdot \sqrt{3} - \sqrt{5})$

Rewrite $4$ as $2^{2}$ .

$i \cdot \sqrt{2^{2} \cdot 2} (i \cdot \sqrt{3} - \sqrt{5})$

Step 9

Pull terms out from under the radical.

$i \cdot (2 \sqrt{2}) (i \cdot \sqrt{3} - \sqrt{5})$

Step 10

Move $2$ to the left of $i$ .

$2 \cdot i \sqrt{2} (i \cdot \sqrt{3} - \sqrt{5})$

Step 11

Apply the distributive property.

$2 i \sqrt{2} (i \sqrt{3}) + 2 i \sqrt{2} (- \sqrt{5})$

Step 12

Multiply $2 i \sqrt{2} (i \sqrt{3})$ .

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

$2 (i^{1} i) \sqrt{2} \sqrt{3} + 2 i \sqrt{2} (- \sqrt{5})$

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

$2 (i^{1} i^{1}) \sqrt{2} \sqrt{3} + 2 i \sqrt{2} (- \sqrt{5})$

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

$2 i^{1 + 1} \sqrt{2} \sqrt{3} + 2 i \sqrt{2} (- \sqrt{5})$

Add $1$ and $1$ .

$2 i^{2} \sqrt{2} \sqrt{3} + 2 i \sqrt{2} (- \sqrt{5})$

Combine using the product rule for radicals.

$2 i^{2} \sqrt{3 \cdot 2} + 2 i \sqrt{2} (- \sqrt{5})$

Multiply $3$ by $2$ .

$2 i^{2} \sqrt{6} + 2 i \sqrt{2} (- \sqrt{5})$

Step 13

Multiply $2 i \sqrt{2} (- \sqrt{5})$ .

Multiply $- 1$ by $2$ .

$2 i^{2} \sqrt{6} - 2 i \sqrt{2} \sqrt{5}$

Combine using the product rule for radicals.

$2 i^{2} \sqrt{6} - 2 i \sqrt{5 \cdot 2}$

Multiply $5$ by $2$ .

$2 i^{2} \sqrt{6} - 2 i \sqrt{10}$

Step 14

Simplify each term.

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

$2 \cdot - 1 \sqrt{6} - 2 i \sqrt{10}$

Multiply $2$ by $- 1$ .

$- 2 \sqrt{6} - 2 i \sqrt{10}$

Write in Standard Form square root of -8( square root of -3- square root of 5)

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