Write in Standard Form (3+ square root of -5)(7- square root of -10)

$(3 + \sqrt{- 5}) (7 - \sqrt{- 10})$

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

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

Rewrite $- 5$ as $- 1 (5)$ .

$(3 + \sqrt{- 1 (5)}) (7 - \sqrt{- 10})$

Rewrite $\sqrt{- 1 (5)}$ as $\sqrt{- 1} \cdot \sqrt{5}$ .

$(3 + \sqrt{- 1} \cdot \sqrt{5}) (7 - \sqrt{- 10})$

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

$(3 + i \cdot \sqrt{5}) (7 - \sqrt{- 10})$

Rewrite $- 10$ as $- 1 (10)$ .

$(3 + i \cdot \sqrt{5}) (7 - \sqrt{- 1 (10)})$

Rewrite $\sqrt{- 1 (10)}$ as $\sqrt{- 1} \cdot \sqrt{10}$ .

$(3 + i \cdot \sqrt{5}) (7 - (\sqrt{- 1} \cdot \sqrt{10}))$

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

$(3 + i \cdot \sqrt{5}) (7 - (i \cdot \sqrt{10}))$

Remove parentheses.

$(3 + i \sqrt{5}) (7 - i \sqrt{10})$

Expand $(3 + i \sqrt{5}) (7 - i \sqrt{10})$ using the FOIL Method.

Apply the distributive property.

$3 (7 - i \sqrt{10}) + i \sqrt{5} (7 - i \sqrt{10})$

Apply the distributive property.

$3 \cdot 7 + 3 (- i \sqrt{10}) + i \sqrt{5} (7 - i \sqrt{10})$

Apply the distributive property.

$3 \cdot 7 + 3 (- i \sqrt{10}) + i \sqrt{5} \cdot 7 + i \sqrt{5} (- i \sqrt{10})$

Simplify terms.

Simplify each term.

Multiply $3$ by $7$ .

$21 + 3 (- i \sqrt{10}) + i \sqrt{5} \cdot 7 + i \sqrt{5} (- i \sqrt{10})$

Multiply $- 1$ by $3$ .

$21 - 3 (i \sqrt{10}) + i \sqrt{5} \cdot 7 + i \sqrt{5} (- i \sqrt{10})$

Move $7$ to the left of $i \sqrt{5}$ .

$21 - 3 i \sqrt{10} + 7 \cdot (i \sqrt{5}) + i \sqrt{5} (- i \sqrt{10})$

Multiply $i \sqrt{5} (- i \sqrt{10})$ .

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

$21 - 3 i \sqrt{10} + 7 i \sqrt{5} + \sqrt{5} (- (i^{1} i) \sqrt{10})$

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

$21 - 3 i \sqrt{10} + 7 i \sqrt{5} + \sqrt{5} (- (i^{1} i^{1}) \sqrt{10})$

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

$21 - 3 i \sqrt{10} + 7 i \sqrt{5} + \sqrt{5} (- i^{1 + 1} \sqrt{10})$

Add $1$ and $1$ .

$21 - 3 i \sqrt{10} + 7 i \sqrt{5} + \sqrt{5} (- i^{2} \sqrt{10})$

Combine using the product rule for radicals.

$21 - 3 i \sqrt{10} + 7 i \sqrt{5} - i^{2} \sqrt{5 \cdot 10}$

Multiply $5$ by $10$ .

$21 - 3 i \sqrt{10} + 7 i \sqrt{5} - i^{2} \sqrt{50}$

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

$21 - 3 i \sqrt{10} + 7 i \sqrt{5} - - 1 \sqrt{50}$

Multiply $- 1$ by $- 1$ .

$21 - 3 i \sqrt{10} + 7 i \sqrt{5} + 1 \sqrt{50}$

Multiply $\sqrt{50}$ by $1$ .

$21 - 3 i \sqrt{10} + 7 i \sqrt{5} + \sqrt{50}$

Rewrite $50$ as $5^{2} \cdot 2$ .

Factor $25$ out of $50$ .

$21 - 3 i \sqrt{10} + 7 i \sqrt{5} + \sqrt{25 (2)}$

Rewrite $25$ as $5^{2}$ .

$21 - 3 i \sqrt{10} + 7 i \sqrt{5} + \sqrt{5^{2} \cdot 2}$

Pull terms out from under the radical.

$21 - 3 i \sqrt{10} + 7 i \sqrt{5} + 5 \sqrt{2}$

Reorder $21 - 3 i \sqrt{10} + 7 i \sqrt{5}$ and $5 \sqrt{2}$ .

$5 \sqrt{2} + 21 - 3 i \sqrt{10} + 7 i \sqrt{5}$

Write in Standard Form (3+ square root of -5)(7- square root of -10)

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