How do you simplify #sqrt32*sqrt144#? Algebra Radicals and Geometry Connections Multiplication and Division of Radicals 1 Answer Ratnaker Mehta Jul 1, 2016 #48sqrt2.# Explanation: We use prime factorisation of #32=2^5# and #144=2^4*3^2.# Hence the expression, #=sqrt32*sqrt144,# #=(2^5)^(1/2)*(2^4*3^2)^(1/2)#=#(2^4*2)^(1/2)*(2^4*3^2)^(1/2)# #={(2^4)^(1/2)2^(1/2)}{(2^4)^(1/2)*(3^2)^(1/2)}#=#2^(4*1/2)*2^(1/2)*2^(4*1/2)*3^(2*1/2)=2^2*2^(1/2)*2^2*3^1=4*sqrt2*4*3=48sqrt2.# Answer link Related questions How do you simplify #\frac{2}{\sqrt{3}}#? How do you multiply and divide radicals? How do you rationalize the denominator? What is Multiplication and Division of Radicals? How do you simplify #7/(""^3sqrt(5)#? How do you multiply #(sqrt(a) +sqrt(b))(sqrt(a)-sqrt(b))#? How do you rationalize the denominator for #\frac{2x}{\sqrt{5}x}#? Do you always have to rationalize the denominator? How do you simplify #sqrt(5)sqrt(15)#? How do you simplify #(7sqrt(13) + 2sqrt(6))(2sqrt(3)+3sqrt(6))#? See all questions in Multiplication and Division of Radicals Impact of this question 916 views around the world You can reuse this answer Creative Commons License