Use the product rule to simplify the expression: the square root of 405x^2.

Step-by-step solution

1. 1

   Find the largest perfect-square factor of 405. Dividing repeatedly, $405=5\cdot 81$, and $81=9^{2}$ is the largest square that divides it. So $\sqrt{405x^{2}}=\sqrt{81\cdot 5\cdot x^{2}} .$

2. 2

   Split the radical with the product rule. Since $\sqrt{ab}=\sqrt a\sqrt b$ for non-negative factors, $\sqrt{81\cdot 5\cdot x^{2}}=\sqrt{81}\cdot\sqrt{5}\cdot\sqrt{x^{2}} .$

3. 3

   Evaluate the numerical square root. $\sqrt{81}=9 .$

4. 4

   Handle the variable part carefully. For an even index, $\sqrt{x^{2}}=|x|$, not $x$: if $x$ were negative, dropping the bars would give a negative value for a quantity that must be non-negative. So the variable contributes $|x|$.

5. 5

   Assemble and test a negative value. $\sqrt{405x^{2}}=9|x|\sqrt5 .$ At $x=-1$ the original is $\sqrt{405}\approx 20.12$, and the answer gives $9\cdot 1\cdot\sqrt5\approx 20.12$, whereas $9x\sqrt5$ would wrongly give $-20.12$.