Question

simplify. assume ( f ) is greater than or equal to zero. (sqrt{75f^{9}})

Explanation:

Step1: Factor the radicand

We can factor \(75f^9\) into prime factors and powers of \(f\). First, factor \(75\): \(75 = 25\times3=5^2\times3\). Then, for \(f^9\), we can write it as \(f^{8 + 1}=f^{8}\times f=(f^{4})^2\times f\). So, \(75f^9 = 5^2\times3\times(f^{4})^2\times f\).

Step2: Apply the square - root property

The square - root property states that \(\sqrt{ab}=\sqrt{a}\times\sqrt{b}\) (for \(a\geq0,b\geq0\)) and \(\sqrt{x^{2}} = |x|\). Since we are given that \(f\geq0\), \(|f^{4}| = f^{4}\).
\[

$$\begin{align*} \sqrt{75f^9}&=\sqrt{5^2\times3\times(f^{4})^2\times f}\\ &=\sqrt{5^2}\times\sqrt{(f^{4})^2}\times\sqrt{3f}\\ &= 5\times f^{4}\times\sqrt{3f}\\ &=5f^{4}\sqrt{3f} \end{align*}$$

\]

Answer:

\(5f^{4}\sqrt{3f}\)