Question

1. simplify the radical completely: $sqrt{27x^3y^2}$

a. $8xysqrt{2x}$
b. $3x^2y^2sqrt{7}$
c. $3xysqrt{3x}$
d. $2y^2xsqrt{3}$

Explanation:

Step1: Factor the radicand

Factor \(27x^{3}y^{2}\) into perfect squares and remaining factors: \(27 = 9\times3\), \(x^{3}=x^{2}\times x\), \(y^{2}=y^{2}\). So \(\sqrt{27x^{3}y^{2}}=\sqrt{9\times3\times x^{2}\times x\times y^{2}}\).

Step2: Apply square - root property

Using the property \(\sqrt{ab}=\sqrt{a}\cdot\sqrt{b}\) (\(a\geq0,b\geq0\)) and \(\sqrt{a^{2}} = a\) (\(a\geq0\)):
\(\sqrt{9\times3\times x^{2}\times x\times y^{2}}=\sqrt{9}\times\sqrt{x^{2}}\times\sqrt{y^{2}}\times\sqrt{3x}\)
Since \(\sqrt{9} = 3\), \(\sqrt{x^{2}}=x\) (assuming \(x\geq0\)) and \(\sqrt{y^{2}} = y\) (assuming \(y\geq0\)), we get \(3\times x\times y\times\sqrt{3x}=3xy\sqrt{3x}\).

Answer:

C. \(3xy\sqrt{3x}\)