question\nexpress in simplest radical form.\n$sqrt{180}$

Explanation:

Step1: Prime - factorize 180

$180=2\times2\times3\times3\times5 = 2^{2}\times3^{2}\times5$

Step2: Apply square - root property

$\sqrt{180}=\sqrt{2^{2}\times3^{2}\times5}$ According to the property $\sqrt{ab}=\sqrt{a}\cdot\sqrt{b}$ ($a = 2^{2}\times3^{2}$, $b = 5$), we have $\sqrt{2^{2}\times3^{2}\times5}=\sqrt{2^{2}}\times\sqrt{3^{2}}\times\sqrt{5}$

Step3: Simplify square - roots of perfect squares

$\sqrt{2^{2}} = 2$, $\sqrt{3^{2}}=3$, so $\sqrt{2^{2}}\times\sqrt{3^{2}}\times\sqrt{5}=2\times3\times\sqrt{5}$

Step4: Calculate the product

$2\times3\times\sqrt{5}=6\sqrt{5}$

Answer:

$6\sqrt{5}$