Write in Standard Form square root of -6 square root of -3

$\sqrt{- 6} \sqrt{- 3}$

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

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

Rewrite $- 6$ as $- 1 (6)$ .

$\sqrt{- 1 (6)} \sqrt{- 3}$

Rewrite $\sqrt{- 1 (6)}$ as $\sqrt{- 1} \cdot \sqrt{6}$ .

$\sqrt{- 1} \cdot \sqrt{6} \sqrt{- 3}$

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

$i \cdot \sqrt{6} \sqrt{- 3}$

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

$i \cdot \sqrt{6} \sqrt{- 1 (3)}$

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

$i \cdot \sqrt{6} (\sqrt{- 1} \cdot \sqrt{3})$

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

$i \cdot \sqrt{6} (i \cdot \sqrt{3})$

Multiply $i \sqrt{6} (i \sqrt{3})$ .

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

$i^{1} i \sqrt{6} \sqrt{3}$

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

$i^{1} i^{1} \sqrt{6} \sqrt{3}$

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

$i^{1 + 1} \sqrt{6} \sqrt{3}$

Add $1$ and $1$ .

$i^{2} \sqrt{6} \sqrt{3}$

Combine using the product rule for radicals.

$i^{2} \sqrt{3 \cdot 6}$

Multiply $3$ by $6$ .

$i^{2} \sqrt{18}$

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

$- 1 \sqrt{18}$

Rewrite $18$ as $3^{2} \cdot 2$ .

Factor $9$ out of $18$ .

$- 1 \sqrt{9 (2)}$

Rewrite $9$ as $3^{2}$ .

$- 1 \sqrt{3^{2} \cdot 2}$

Pull terms out from under the radical.

$- 1 (3 \sqrt{2})$

Multiply $3$ by $- 1$ .

$- 3 \sqrt{2}$

The result can be shown in multiple forms.

Exact Form:

$- 3 \sqrt{2}$

Decimal Form:

$- 4.24264068 \dots$