The equation can be solved algebraically - it is a cubic equation.
The equation can be divided by 5 and #x# which will make the numbers smaller. However we do not know what the value of #x# is so we cannot divide by #x# in case it is equal to 0.
#40x^3-10x^2 -5x =0" "larr div# 5
=#8x^3-2x^2 -x =0" "# #x# is a common factor - divide it out.
=#x(8x^2-2x -1) = 0" "larr# factor the quadratic equation
=#x(4x+1)(2x-1)=0" "larr# 3 factors
#x =0" "larr x= 0 # is a solution
#"if "4x+1 =0, " then " 4x=1 rarr x = -1/4#
#"if " 2x-1 = 0 " then " x = 1/2#
#x = 0, x= -1/4, x = 1/2#
