How do you simplify #2 sqrt 27 - 4 sqrt 3#? Algebra Radicals and Geometry Connections Addition and Subtraction of Radicals 1 Answer Don't Memorise Apr 7, 2016 #= 2sqrt3# Explanation: #2sqrt27 - 4sqrt3# We first simplify #27# by prime factorisation. (express a number as a product of its prime factors) #27 = 3 * 3 * 3 # So, #sqrt(27) = sqrt (3 * 3 * 3 ) = sqrt ( 3^2 * 3 ) = color(green)( 3sqrt3 # #2sqrt27 - 4sqrt3 = 2 * color(green)( 3sqrt3 )- 4sqrt3# #= 6 sqrt3 - 4sqrt3# #= 2sqrt3# Answer link Related questions How do you add and subtract radicals? How is a radical considered a "like term"? How do you simplify #4\sqrt{3}+2\sqrt{12}#? How do you add #3""^3sqrt(2)+5""^3sqrt(16)#? How do you subtract #\sqrt{8x^3}-4x\sqrt{98x}#? How do you combine the radical #\sqrt{6}-\sqrt{27}+2\sqrt{54}+3\sqrt{48}#? How do you simplify #""^3sqrt{\frac{16x^5}{135y^4}}#? What is #sqrt(50)-sqrt(18)#? How do you add #3sqrt2+4sqrt2#? What is the square root of 50 + the square root of 8? See all questions in Addition and Subtraction of Radicals Impact of this question 2371 views around the world You can reuse this answer Creative Commons License