Simplify ( square root of 40x^4)/( cube root of -x^3)

$\frac{\sqrt{40 x^{4}}}{\sqrt[3]{- x^{3}}}$

Simplify the numerator.

Rewrite $40 x^{4}$ as ${(2 x^{2})}^{2} \cdot 10$ .

Factor $4$ out of $40$ .

$\frac{\sqrt{4 (10) x^{4}}}{\sqrt[3]{- x^{3}}}$

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

$\frac{\sqrt{2^{2} \cdot 10 x^{4}}}{\sqrt[3]{- x^{3}}}$

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

$\frac{\sqrt{2^{2} \cdot 10 {(x^{2})}^{2}}}{\sqrt[3]{- x^{3}}}$

Move $10$ .

$\frac{\sqrt{2^{2} {(x^{2})}^{2} \cdot 10}}{\sqrt[3]{- x^{3}}}$

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

$\frac{\sqrt{{(2 x^{2})}^{2} \cdot 10}}{\sqrt[3]{- x^{3}}}$

Pull terms out from under the radical.

$\frac{2 x^{2} \sqrt{10}}{\sqrt[3]{- x^{3}}}$

Simplify the denominator.

Rewrite $- x^{3}$ as ${(- x)}^{3}$ .

$\frac{2 x^{2} \sqrt{10}}{\sqrt[3]{{(- x)}^{3}}}$

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

$\frac{2 x^{2} \sqrt{10}}{- x}$

Reduce the expression by cancelling the common factors.

Cancel the common factor of $x^{2}$ and $x$ .

Factor $x$ out of $2 x^{2} \sqrt{10}$ .

$\frac{x (2 x \sqrt{10})}{- x}$

Move the negative one from the denominator of $\frac{2 x \sqrt{10}}{- 1}$ .

$- 1 \cdot (2 x \sqrt{10})$

Simplify the expression.

Rewrite $- 1 \cdot (2 x \sqrt{10})$ as $- (2 x \sqrt{10})$ .

$- (2 x \sqrt{10})$

Multiply $2$ by $- 1$ .

$- 2 x \sqrt{10}$