Find the Product 3 square root of 10(y^2 square root of 4+ square root of 8y)

$3 \sqrt{10} (y^{2} \sqrt{4} + \sqrt{8 y})$

Simplify each term.

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

$3 \sqrt{10} (y^{2} \sqrt{2^{2}} + \sqrt{8 y})$

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

$3 \sqrt{10} (y^{2} \cdot 2 + \sqrt{8 y})$

Move $2$ to the left of $y^{2}$ .

$3 \sqrt{10} (2 \cdot y^{2} + \sqrt{8 y})$

Rewrite $8 y$ as $2^{2} \cdot (2 y)$ .

Factor $4$ out of $8$ .

$3 \sqrt{10} (2 y^{2} + \sqrt{4 (2) y})$

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

$3 \sqrt{10} (2 y^{2} + \sqrt{2^{2} \cdot 2 y})$

Add parentheses.

$3 \sqrt{10} (2 y^{2} + \sqrt{2^{2} \cdot (2 y)})$

Pull terms out from under the radical.

$3 \sqrt{10} (2 y^{2} + 2 \sqrt{2 y})$

Simplify by multiplying through.

Apply the distributive property.

$3 \sqrt{10} (2 y^{2}) + 3 \sqrt{10} (2 \sqrt{2 y})$

Multiply $2$ by $3$ .

$6 \sqrt{10} y^{2} + 3 \sqrt{10} (2 \sqrt{2 y})$

Multiply $3 \sqrt{10} (2 \sqrt{2 y})$ .

Multiply $2$ by $3$ .

$6 \sqrt{10} y^{2} + 6 \sqrt{10} \sqrt{2 y}$

Combine using the product rule for radicals.

$6 \sqrt{10} y^{2} + 6 \sqrt{2 y \cdot 10}$

Multiply $10$ by $2$ .

$6 \sqrt{10} y^{2} + 6 \sqrt{20 y}$

Simplify each term.

Rewrite $20 y$ as $2^{2} \cdot (5 y)$ .

Factor $4$ out of $20$ .

$6 \sqrt{10} y^{2} + 6 \sqrt{4 (5) y}$

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

$6 \sqrt{10} y^{2} + 6 \sqrt{2^{2} \cdot 5 y}$

Add parentheses.

$6 \sqrt{10} y^{2} + 6 \sqrt{2^{2} \cdot (5 y)}$

Pull terms out from under the radical.

$6 \sqrt{10} y^{2} + 6 (2 \sqrt{5 y})$

Multiply $2$ by $6$ .

$6 \sqrt{10} y^{2} + 12 \sqrt{5 y}$