Write in Standard Form ( square root of 3+4i)( square root of 3-4i)

$(\sqrt{3} + 4 i) (\sqrt{3} - 4 i)$

Step 1

Expand $(\sqrt{3} + 4 i) (\sqrt{3} - 4 i)$ using the FOIL Method.

Apply the distributive property.

$\sqrt{3} (\sqrt{3} - 4 i) + 4 i (\sqrt{3} - 4 i)$

Apply the distributive property.

$\sqrt{3} \sqrt{3} + \sqrt{3} (- 4 i) + 4 i (\sqrt{3} - 4 i)$

Apply the distributive property.

$\sqrt{3} \sqrt{3} + \sqrt{3} (- 4 i) + 4 i \sqrt{3} + 4 i (- 4 i)$

Step 2

Simplify and combine like terms.

Simplify each term.

Combine using the product rule for radicals.

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

Multiply $3$ by $3$ .

$\sqrt{9} + \sqrt{3} (- 4 i) + 4 i \sqrt{3} + 4 i (- 4 i)$

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

$\sqrt{3^{2}} + \sqrt{3} (- 4 i) + 4 i \sqrt{3} + 4 i (- 4 i)$

Pull terms out from under the radical, assuming positive real numbers.

$3 + \sqrt{3} (- 4 i) + 4 i \sqrt{3} + 4 i (- 4 i)$

Multiply $4 i (- 4 i)$ .

Multiply $- 4$ by $4$ .

$3 + \sqrt{3} (- 4 i) + 4 i \sqrt{3} - 16 i i$

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

$3 + \sqrt{3} (- 4 i) + 4 i \sqrt{3} - 16 (i^{1} i)$

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

$3 + \sqrt{3} (- 4 i) + 4 i \sqrt{3} - 16 (i^{1} i^{1})$

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

$3 + \sqrt{3} (- 4 i) + 4 i \sqrt{3} - 16 i^{1 + 1}$

Add $1$ and $1$ .

$3 + \sqrt{3} (- 4 i) + 4 i \sqrt{3} - 16 i^{2}$

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

$3 + \sqrt{3} (- 4 i) + 4 i \sqrt{3} - 16 \cdot - 1$

Multiply $- 16$ by $- 1$ .

$3 + \sqrt{3} (- 4 i) + 4 i \sqrt{3} + 16$

Add $3$ and $16$ .

$\sqrt{3} (- 4 i) + 4 i \sqrt{3} + 19$

Reorder the factors of $\sqrt{3} (- 4 i)$ .

$- 4 i \sqrt{3} + 4 i \sqrt{3} + 19$

Add $- 4 i \sqrt{3}$ and $4 i \sqrt{3}$ .

$0 + 19$

Add $0$ and $19$ .

$19$