Simplify - square root of 12k^7q^8

$- \sqrt{12 k^{7} q^{8}}$

Step 1

Rewrite $12 k^{7} q^{8}$ as ${(2 k^{3} q^{4})}^{2} \cdot (3 k)$ .

Factor $4$ out of $12$ .

$- \sqrt{4 (3) k^{7} q^{8}}$

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

$- \sqrt{2^{2} \cdot 3 k^{7} q^{8}}$

Factor out $k^{6}$ .

$- \sqrt{2^{2} \cdot 3 (k^{6} k) q^{8}}$

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

$- \sqrt{2^{2} \cdot 3 ({(k^{3})}^{2} k) q^{8}}$

Rewrite $q^{8}$ as ${(q^{4})}^{2}$ .

$- \sqrt{2^{2} \cdot 3 ({(k^{3})}^{2} k) {(q^{4})}^{2}}$

Move $k$ .

$- \sqrt{2^{2} \cdot 3 ({(k^{3})}^{2}) {(q^{4})}^{2} k}$

Move $3$ .

$- \sqrt{(2^{2} ({(k^{3})}^{2})) {(q^{4})}^{2} \cdot 3 k}$

Rewrite $(2^{2} ({(k^{3})}^{2})) {(q^{4})}^{2}$ as ${(2 k^{3} q^{4})}^{2}$ .

$- \sqrt{{(2 k^{3} q^{4})}^{2} \cdot 3 k}$

Add parentheses.

$- \sqrt{{(2 k^{3} q^{4})}^{2} \cdot (3 k)}$

Step 2

Pull terms out from under the radical.

$- (2 k^{3} q^{4} \sqrt{3 k})$

Step 3

Multiply $2$ by $- 1$ .

$- 2 k^{3} q^{4} \sqrt{3 k}$