How do you simplify #\frac { 6n ^ { 4} } { 4n ^ { 2} }#? Prealgebra Exponents, Radicals and Scientific Notation Exponents 1 Answer Shwetank Mauria Jun 9, 2017 #(6n^4)/(4n^2)=(3n^2)/2# Explanation: #(6n^4)/(4n^2)# = #(2xx3xxnxxnxxnxxn)/(2xx2xxnxxn)# = #(cancel2xx3xxcancelnxxcancelnxxnxxn)/(2xxcancel2xxcancelnxxcanceln)# = #(3xxnxxn)/2# = #(3n^2)/2# Answer link Related questions How do you simplify #c^3v^9c^-1c^0#? How do you simplify #(- 1/5)^-2 + (-2)^-2#? How do you simplify #(4^6)^2 #? How do you simplify #3x^(2/3) y^(3/4) (2x^(5/3) y^(1/2))^3 #? How do you simplify #4^3ยท4^5#? How do you simplify #(5^-2)^-3#? How do you simplify and write #(-5.3)^0# with positive exponents? How do you factor #12j^2k - 36j^6k^6 + 12j^2#? How do you simplify the expression #2^5/(2^3 times 2^8)#? When can I add exponents? See all questions in Exponents Impact of this question 1419 views around the world You can reuse this answer Creative Commons License