Question

what is the simplest form of \\(\sqrt{81x^8y^5}\\)?
\\(3x^2\left(\sqrt{y^5}\
ight)\\)
\\(3x^2y\left(\sqrt{y}\
ight)\\)
\\(9x^4y^2\left(\sqrt{y}\
ight)\\)
\\(9x^2y\left(\sqrt{y}\
ight)\\)

Explanation:

Step1: Simplify the square root of 81

We know that $\sqrt{81} = 9$ because $9\times9 = 81$.

Step2: Simplify the square root of $x^{8}$

Using the property $\sqrt{a^{m}}=a^{\frac{m}{2}}$, for $a = x$ and $m = 8$, we have $\sqrt{x^{8}}=x^{\frac{8}{2}}=x^{4}$.

Step3: Simplify the square root of $y^{5}$

We can write $y^{5}$ as $y^{4}\times y$. Then $\sqrt{y^{5}}=\sqrt{y^{4}\times y}=\sqrt{y^{4}}\times\sqrt{y}=y^{2}\times\sqrt{y}$.

Step4: Combine all the simplified parts

Now, $\sqrt{81x^{8}y^{5}}=\sqrt{81}\times\sqrt{x^{8}}\times\sqrt{y^{5}}$. Substituting the values we found above, we get $9\times x^{4}\times y^{2}\sqrt{y}=9x^{4}y^{2}(\sqrt{y})$.

Answer:

$9x^{4}y^{2}(\sqrt{y})$ (the option corresponding to this expression)