Question

simplify the following expression.
\sqrt{x^{7}y^{7}} \cdot 4\sqrt{2x^{11}y^{2}}

\sqrt{x^{7}y^{7}} \cdot 4\sqrt{2x^{11}y^{2}} = \square
(simplify your answer. type an exact answer, using radicals as needed.)

Explanation:

Step1: Simplify the first square root

The first term is \(\sqrt{x^{7}y^{7}}\). Using the property \(\sqrt{ab}=\sqrt{a}\cdot\sqrt{b}\) and \(\sqrt{a^{m}} = a^{\frac{m}{2}}\), we have:
\(\sqrt{x^{7}y^{7}}=\sqrt{x^{6}\cdot x\cdot y^{6}\cdot y}=\sqrt{(x^{3})^{2}\cdot x\cdot(y^{3})^{2}\cdot y}=x^{3}y^{3}\sqrt{xy}\)

Step2: Simplify the second square root

The second term is \(4\sqrt{2x^{11}y^{2}}\). Using the same properties:
\(\sqrt{2x^{11}y^{2}}=\sqrt{2\cdot x^{10}\cdot x\cdot y^{2}}=\sqrt{2\cdot(x^{5})^{2}\cdot x\cdot y^{2}} = x^{5}y\sqrt{2x}\)
So the second term becomes \(4x^{5}y\sqrt{2x}\)

Step3: Multiply the two simplified terms

Now multiply \(x^{3}y^{3}\sqrt{xy}\) and \(4x^{5}y\sqrt{2x}\):
First, multiply the coefficients and the variables:
Coefficients: \(1\times4 = 4\)
Variables: \(x^{3}\cdot x^{5}=x^{3 + 5}=x^{8}\), \(y^{3}\cdot y=y^{3+1}=y^{4}\)
Radicals: \(\sqrt{xy}\cdot\sqrt{2x}=\sqrt{xy\cdot2x}=\sqrt{2x^{2}y}=x\sqrt{2y}\) (since \(x^{2}\) is a perfect square)

Now combine all parts:
\(4\times x^{8}\times y^{4}\times x\sqrt{2y}=4x^{9}y^{4}\sqrt{2y}\) Wait, no, let's re - check the radical multiplication:

Wait, \(\sqrt{xy}\cdot\sqrt{2x}=\sqrt{2x^{2}y}=x\sqrt{2y}\) (because \(x^{2}\) is non - negative in the context of real - valued square roots, so we can take \(x\) out). Then:

\(x^{3}y^{3}\sqrt{xy}\times4x^{5}y\sqrt{2x}=4x^{3 + 5}y^{3+1}\sqrt{xy\times2x}=4x^{8}y^{4}\sqrt{2x^{2}y}\)

Since \(\sqrt{2x^{2}y}=x\sqrt{2y}\) (assuming \(x\geq0\)), then:

\(4x^{8}y^{4}\times x\sqrt{2y}=4x^{9}y^{4}\sqrt{2y}\)? Wait, no, let's do the exponent for \(x\) in the radical correctly.

Wait, \(xy\times2x = 2x^{2}y\), so \(\sqrt{2x^{2}y}=x\sqrt{2y}\) (when \(x\geq0\)). Then:

\(x^{3}y^{3}\times4x^{5}y=4x^{3 + 5}y^{3+1}=4x^{8}y^{4}\)

And \(\sqrt{xy}\times\sqrt{2x}=\sqrt{2x^{2}y}=x\sqrt{2y}\)

So multiplying these together: \(4x^{8}y^{4}\times x\sqrt{2y}=4x^{9}y^{4}\sqrt{2y}\)? Wait, no, that's a mistake. Wait, \(x^{8}\times x=x^{9}\), \(y^{4}\) remains, and the radical is \(\sqrt{2y}\). But let's check the initial exponents again.

Wait, let's start over with a better approach. Let's use the property \(\sqrt{a}\cdot\sqrt{b}=\sqrt{ab}\) from the beginning.

The original expression is \(\sqrt{x^{7}y^{7}}\cdot4\sqrt{2x^{11}y^{2}}\)

Using \(\sqrt{a}\cdot\sqrt{b}=\sqrt{ab}\), we can rewrite it as \(4\sqrt{x^{7}y^{7}\cdot2x^{11}y^{2}}\)

Now multiply the terms inside the square root:

\(x^{7}\cdot x^{11}=x^{7 + 11}=x^{18}\)

\(y^{7}\cdot y^{2}=y^{7+2}=y^{9}\)

So inside the square root, we have \(2x^{18}y^{9}\)

Now simplify \(\sqrt{2x^{18}y^{9}}\):

\(\sqrt{2x^{18}y^{9}}=\sqrt{2}\cdot\sqrt{x^{18}}\cdot\sqrt{y^{9}}\)

\(\sqrt{x^{18}}=x^{9}\) (since \((x^{9})^{2}=x^{18}\))

\(\sqrt{y^{9}}=\sqrt{y^{8}\cdot y}=y^{4}\sqrt{y}\) (since \((y^{4})^{2}=y^{8}\))

So \(\sqrt{2x^{18}y^{9}}=\sqrt{2}\cdot x^{9}\cdot y^{4}\sqrt{y}=x^{9}y^{4}\sqrt{2y}\)

Then multiply by the coefficient 4:

\(4\times x^{9}y^{4}\sqrt{2y}=4x^{9}y^{4}\sqrt{2y}\)

Answer:

\(4x^{9}y^{4}\sqrt{2y}\)