Algebra 1: Common Core (15th Edition)

a. $18\sqrt 3$ b. $3a^{2}* \sqrt 2$ c. $210x^{3}$ d. Yes, we can simplify to get $42t\sqrt 2t$
a) We can simplify as follows: = $3\sqrt 6*\sqrt 18$ = $3*\sqrt 6*\sqrt 6*\sqrt 3$ = $3*6*\sqrt 3$ = $18\sqrt 3$ a) We can simplify as follows: = $\sqrt 2a*\sqrt 9a^{3}$ = $\sqrt 2a*\sqrt 9*\sqrt a^{3}$ = $\sqrt 2*\sqrt a*3*\sqrt a^{2}*\sqrt a$ = $\sqrt 2*\sqrt a*3*a*\sqrt a$ = $\sqrt 2*a*3*a$ (since $\sqrt a*\sqrt a$ = a) = $3a^{2}\sqrt 2$. a) We can simplify as follows: = $7\sqrt 5x*3\sqrt 20x^{5}$ = $7\sqrt 5*\sqrt x*3\sqrt 5*\sqrt 4*\sqrt x^{4}*\sqrt x$ = $7\sqrt 5*\sqrt x*3\sqrt 5* 2* x^{2}*\sqrt x$ = $7*3* 2* x^{2}*x*5$ (since $\sqrt 5 *\sqrt 5 = 5$ and $\sqrt x *\sqrt x =x$) = $210x^{3}$ a) We can simplify as follows: = $2\sqrt 7t*3\sqrt 14t^{2}$ = $2\sqrt 7*\sqrt t*3\sqrt 7*\sqrt 2*\sqrt t^{2}$ = $2* 7*\sqrt t*3*\sqrt 2*\sqrt t^{2}$ = $42t\sqrt 2t$.