Question

-5\sqrt4{405x}

Explanation:

Step1: Factor 405

We know that \(405 = 81\times5\), and \(81 = 3^4\). So we can rewrite \(405\) as \(3^4\times5\).
So the expression \(-5\sqrt[4]{405x}\) becomes \(-5\sqrt[4]{3^4\times5\times x}\)

Step2: Simplify the fourth - root

Using the property of radicals \(\sqrt[n]{ab}=\sqrt[n]{a}\cdot\sqrt[n]{b}\) (\(a\geq0,b\geq0\) for even \(n\)) and \(\sqrt[n]{a^n}=a\) (when \(n\) is even and \(a\geq0\)), we have:
\(-5\sqrt[4]{3^4\times5x}=-5\times\sqrt[4]{3^4}\times\sqrt[4]{5x}\)
Since \(\sqrt[4]{3^4} = 3\) (because \(3\geq0\) and we are taking the fourth - root), then:
\(-5\times3\times\sqrt[4]{5x}=-15\sqrt[4]{5x}\)

Answer:

\(-15\sqrt[4]{5x}\)