#6c^3+11c^2-10c = c(6c^2+11x-10)#

#=c(2c+5)(3c-2)#

I first noticed that all the terms were divisible by #c#, so separated that out as a factor.

I then looked for a factorisation of the form:

#6c^2+11x-10 = (2c+a)(3c+b)#

#= 6c^2+(3a+2b)c+ab#

So I looked for #a# and #b# such that #3a+2b = 11# and #ab=-10#

I could tell that #a# must be odd, because otherwise both #2c# and #a# would be even and all the resulting coefficients would be even.

So #a=+-1# or #a=+-5#.

In order to get #3a+b=11# then the best choice to try seemed to be #a=5#, which forces #b=(-10)/a=(-10)/5=-2#.

So try #(2c+5)(3c-2)# - works.