Question

1. simplify: \\(sqrt{40x^{10}y^6}\\)

Explanation:

Step1: Factor the radicand

First, factor \(40x^{10}y^{6}\) into perfect - square factors and non - perfect - square factors. We know that \(40 = 4\times10\), \(x^{10}=(x^{5})^{2}\), and \(y^{6}=(y^{3})^{2}\). So, \(40x^{10}y^{6}=4\times10\times(x^{5})^{2}\times(y^{3})^{2}\).

Step2: Apply the square - root property

The square - root property states that \(\sqrt{ab}=\sqrt{a}\times\sqrt{b}\) (where \(a\geq0\) and \(b\geq0\)) and \(\sqrt{a^{2}} = |a|\). For non - negative real numbers (assuming \(x\) and \(y\) are real numbers and we are dealing with the principal square root), when \(a\) is a non - negative real number, \(\sqrt{a^{2}}=a\).
\(\sqrt{40x^{10}y^{6}}=\sqrt{4\times10\times(x^{5})^{2}\times(y^{3})^{2}}\)
Using the property \(\sqrt{ab}=\sqrt{a}\times\sqrt{b}\), we can split the square root:
\(\sqrt{4}\times\sqrt{10}\times\sqrt{(x^{5})^{2}}\times\sqrt{(y^{3})^{2}}\)
We know that \(\sqrt{4} = 2\), \(\sqrt{(x^{5})^{2}}=x^{5}\) (since \(x^{5}\) raised to the power of 2 and then square - rooted, and assuming \(x\) is such that \(x^{5}\) is non - negative, or we are taking the principal square root), and \(\sqrt{(y^{3})^{2}}=y^{3}\) (similarly, assuming \(y\) is such that \(y^{3}\) is non - negative or taking the principal square root).
So, \(\sqrt{40x^{10}y^{6}}=2x^{5}y^{3}\sqrt{10}\)

Answer:

\(2x^{5}y^{3}\sqrt{10}\)