Simplify square root of 60x^4y^7

$\sqrt{60 x^{4} y^{7}}$

Step 1

Rewrite $60 x^{4} y^{7}$ as ${(2 x^{2} y^{3})}^{2} \cdot (15 y)$ .

Factor $4$ out of $60$ .

$\sqrt{4 (15) x^{4} y^{7}}$

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

$\sqrt{2^{2} \cdot 15 x^{4} y^{7}}$

Rewrite $x^{4}$ as ${(x^{2})}^{2}$ .

$\sqrt{2^{2} \cdot 15 {(x^{2})}^{2} y^{7}}$

Factor out $y^{6}$ .

$\sqrt{2^{2} \cdot 15 {(x^{2})}^{2} (y^{6} y)}$

Rewrite $y^{6}$ as ${(y^{3})}^{2}$ .

$\sqrt{2^{2} \cdot 15 {(x^{2})}^{2} ({(y^{3})}^{2} y)}$

Move $15$ .

$\sqrt{2^{2} {(x^{2})}^{2} {(y^{3})}^{2} \cdot 15 y}$

Rewrite $2^{2} {(x^{2})}^{2} {(y^{3})}^{2}$ as ${(2 x^{2} y^{3})}^{2}$ .

$\sqrt{{(2 x^{2} y^{3})}^{2} \cdot 15 y}$

Add parentheses.

$\sqrt{{(2 x^{2} y^{3})}^{2} \cdot (15 y)}$

Step 2

Pull terms out from under the radical.

$2 x^{2} y^{3} \sqrt{15 y}$