# How factor with square root

I have the following equation that I'm trying to factor, but I'm stuck at the end.

$$\frac{zx^{-4}\sqrt{x}(yz^4)^3}{z^7xy}$$

$$\frac{\frac{1}{x^4}\sqrt{x}(yz^4)^3}{z^6xy}$$

$$\frac{(yz^4)^3\frac{1}{x^4}x^{1/2}}{xyz^6}$$

$$\frac{y^3z^{12}x^{-7/2}}{xyz^6}$$

$$\frac{y^2z^{6}x^{-7/2}}{x}$$

I think I can simplify and get rid of the x in the denominator but I'm not sure how with that negative fraction x exponent in the denominator.

• Use this to simplify further$$\frac{x^a}{x^b} = x^{a-b}$$ – zerosofthezeta Apr 16 '14 at 6:23

We have: $$\frac{x^{-7/2}y^2z^6}{x}$$ Remember that $\dfrac{x^a}{x^b}=x^{a-b}$ $$y^2z^6\cdot\frac{x^{-7/2}}{x}$$ $$=y^2z^6x^{-7/2-1}$$ $$=x^{-9/2}y^2z^6$$ $$=\frac{y^2z^6}{x^{9/2}}$$ $$=\frac{y^2z^6}{\sqrt{x^9}}$$ $$=\frac{y^2z^6}{x^4\sqrt{x}}$$ Hope I helped

We use the fact that division by $x$ is the same as multiplication by $x^{-1}$ to combine the remaining appearances of $x$.

\begin{eqnarray*} \frac{y^2x^{6}z^{\frac{-7}{2}}}{x} &=& y^2x^{6}z^{\frac{-7}{2}}x^{-1}\\ &=& y^2x^{6}x^{-1}z^{\frac{-7}{2}}\\ &=& y^2x^{6-1}z^{\frac{-7}{2}}\\ &=& y^2x^{5}z^{\frac{-7}{2}}\\ \end{eqnarray*}